the analysis of linear circuits and design ch02

The Analysis and Design of Linear Circuits

2 2.1

Seventh Edition

Basi Ba sic c Ci Circ rcui uitt An Anal alys ysis is Exerci Exe rcise se Sol Soluti utions ons

Exercise 2–1. A 6-V lantern battery powers a light bulb that draws 3 mA of current. What is the resistance of the lamp? How much power does the lantern use? V Using Ohm’s law, we have v have v = = iR iR or or R R = = vi , so we can compute the resistance as R as R = = 36mA = 2kΩ. The power is p = = v vii = (6 V) V)(3 (3 mA mA)) = 18 mW. Exercise 2–2. What is the maximum current that can flow through a 18 -W, 6.8-kΩ resistor? resistor? What is the maximum voltage that can be across it? The resistor resistor can dissi dissipate pate up to 0.125 W of pow power. er. We have pMAX = i2MAX R, which we can solved for iMAX and then substitute in values for the power and resistance iMAX = Similarly, we can use p use p MAX =

2 vMAX R



pMAX = R



0.125 = 4.287 28755 mA 6800

to solve for for the maximum maximum voltage voltage as follows: follows:

vMAX =



R pMAX =



(6800)(0..125) = 29. (6800)(0 29.155V

Exercise 2–3. A digital clock is a voltage that switches between two values at a constant rate that is used to time digital circuits. A particular clock switches between 0 V and 5 V every 10 µ µs. s. Sketch the clock’s i clock’s i--v characteristics for the times when the clock is at 0 V and at 5 V. When the clock has a value of 0 V, its voltage is constant and zero for a wide range of currents. In this case, the i the i--v characteristic is a vertical line at 0 V. Likewise, when the clock has a value of 5 V, the voltage is constant at 5 V for a wide range of currents. In this case, the i-v characteristic is a vertical line at 5 V. Exercise 2–4. Refer to Figure 2–12.

Figure 2–12

(a). Write KCL equations equations at nodes A, B, C, and D. KCL sta KCL states tes that the sum of the currents currents enter entering ing a node is zero at ev every ery instan instant. t. As we sum the currents at a node, if the current enters that node, it is positive and if the current leaves the node, it is negative. At node A, both currents i currents i 1 and and i i 2 are leaving the node, so the equation is i1 i2 = 0. At node B, current i2 enters the node and currents i3 and and i i 4 leave the node, so we have i2 i3 i4 = 0. At node C, current i current i 4 enters the node and currents i currents i 5 and and i i 6 leave the node, so we have i have i 4 i5 i6 = 0. At node D, currents i currents i 1 , i 3 , i 5 , and i and i 6 enter the node, so we have i 1 + + i i3 + + i i5 + + i i6 = 0.

− − − − − −

(b). Giv Given en i i 1 =

−1 mA, i = 0.5 mA, i = 0.2 mA, find i , i , and i . Applying the KCL equation for node A, we can find i = −i = 1 mA. Applying the KCL equation for node B, we have i have i = i − i = 1 − 0.5 = 0.5 mA. Finally, applying the KCL equation for node C, we have i have i = i − i = 0.5 − 0.2 = 0.3 mA. 3

6

2

4

2

4

5

4

2

5

1

3

6

Solution Manual Chapter 2

Page 2-1


Seventh Edition

Figure 2–14

Exercise 2–5. Find the voltages v voltages v x and and v v y in Figure 2–14. To find v find v x , write the KVL equation around Loop 1 as vx + 2 + 6 = 0 and and solv solvee for for v x = +8 V. To find vy , write the KVL equation around Loop 2 as v y + 1 6 = 0 and solve for v for v y = +5 V.

−

−

Exercise 2–6. Find the voltages v voltages v x , vy , and vz in Figure 2–15.

Figure 2–15

In Figure 2–15, some of the unknown voltages do not appear across elements, but we can still write KVL equations. equati ons. For Loop 1 starting with the low lowest est element, element, the KVL equation is 10 4 0 + 5 + vx = 0, which can be solved to yield v x = 25 V. For Loop 2, the KVL equation is vx + 20 + v + vy = 0, which can be solved for vy = 25 20 = 5 V. Finally, for Loop 3, the KVL equation is vy 5 + vz = 0, which yields v yields vz = 5+ 5 = 10 V. V.

− − −

−

−

Exercise 2–7. Identify 2–7. Identify the elements connected in series or parallel when a short circuit is connected between nodes A and B in each of the circuits in Figure 2–18.

(a)

(b)

(c)

Figure 2–18


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Seventh Edition

Figure 2–14

Exercise 2–5. Find the voltages v voltages v x and and v v y in Figure 2–14. To find v find v x , write the KVL equation around Loop 1 as vx + 2 + 6 = 0 and and solv solvee for for v x = +8 V. To find vy , write the KVL equation around Loop 2 as v y + 1 6 = 0 and solve for v for v y = +5 V.

−

−

Exercise 2–6. Find the voltages v voltages v x , vy , and vz in Figure 2–15.

Figure 2–15

In Figure 2–15, some of the unknown voltages do not appear across elements, but we can still write KVL equations. equati ons. For Loop 1 starting with the low lowest est element, element, the KVL equation is 10 4 0 + 5 + vx = 0, which can be solved to yield v x = 25 V. For Loop 2, the KVL equation is vx + 20 + v + vy = 0, which can be solved for vy = 25 20 = 5 V. Finally, for Loop 3, the KVL equation is vy 5 + vz = 0, which yields v yields vz = 5+ 5 = 10 V. V.

− − −

−

−

Exercise 2–7. Identify 2–7. Identify the elements connected in series or parallel when a short circuit is connected between nodes A and B in each of the circuits in Figure 2–18.

(a)

(b)

(c)

Figure 2–18


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Seventh Edition

In the solution, the short circuit has been applied in each of the circuits and Element 2 has been shorted out of the circu circuit. it. For the circuit in Figure 2–18(a), 2–18(a), all of the elements share share the same two two nodes, A and C, so Eleme Elements nts 1, 2, and 3 are in parallel. For the circuit in Figur Figuree 2–18(b), Elements Elements 1 and 3 share nodes A and C, so they are in paral parallel. lel. In addition, addition, Elements Elements 4 and 5 are the only elements elements connects to node D, so they are in series. For the circuit in Figure 2–18(c), Elements 1 and 3 are in parallel because they share nodes A and C. In addition, Elements 4 and 6 are in parallel, because they share nodes A and D.

Exercise 2–8. Identify the elements in Figure 2–19 that are connected in (a) parallel, (b) series, or (c) neither. Refer to the figure in the textbook.

(a).. Elemen (a) Elements ts 1, 8, and 11 share share the uppe upperr left node and gro ground und,, so they are in par parall allel. el. In addition addition,, Elements 3, 4, and 5 share the center node and ground, so they are in parallel.

(b). Elements Elements 9 and 10 are in serie series, s, because they share a single node and no other elements elements with current current connect to that node. Likewise, Elements 6 and 7 share a single node with no other elements, so they are also in series.

(c). The remaining remaining element, element, Element 2, is neith neither er in serie seriess nor paral parallel lel with any other elements. elements.

Exercise 2–9. A 1-kΩ resistor is added between nodes A and B in Figure 2–20. Find i x , v x , i O , and v and v O if iS = 1 mA and R = 2 kΩ.

Figure 2–20

The resulti resulting ng cir circui cuitt is shown shown in Figure Figure 2–2 2–20. 0. Not Notee tha thatt the 1-kΩ resistor resistor has been inserte inserted d and the current through it labeled as i1 and the voltage across it labeled as v1 . Using KCL KCL at the current source, source, we have ix = iS . Writin ritingg KCL at the top node, we have have ix i1 iO = 0. Writin ritingg KVL aroun around d the left loop yields vx + + v v1 = 0. Writing KVL around the right loop yields v1 + + v vO = 0. Alternately, we can see that the three elements all share the top and bottom nodes, so they are all in parallel and have the same voltage, v voltage, v x = v 1 = v O . Using these equations and Ohm’s law, v = Ri Ri,, we can solve for the unknown values

−


− − − −

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Seventh Edition

as follows: ix

= iS = 1 mA mA

v1

= vO

R1 i1

= RO iO

1000ii1 = 1000 i1 ix + + i i1 + + i iO

2000iO

= 2iO = 0

i1 + + i iO

=

2iO + + i iO

=

3iO

=

iO

=

vx

=

−i = −1 mA −1 mA −1 mA −333 µA v = (2kΩ (2kΩ)( )(−333 µA) x

O

=

−667mV

Exercise 2–10. The wire connecting R 1 to node B in Figure 2-21 is broken. What would you measure for iA , v 1 , i 2 , and v and v 2 ? Is KVL violated? Where does the source voltage appear across?

Figure 2–21

Figuree 2–21 shows the resulting circuit. Figur circuit. If the circu circuit it is brok broken en betw b etween een R R 1 and node B, then no current can flow in the circuit and all currents are zero, i A = i 1 = i 2 = 0. Using Ohm’s law, v = Ri Ri,, for the voltages across the resistors, the current is zero, so the voltages must also be zero and we have v1 = v2 = 0. No Note te that a new voltage, v voltage, v x , has been labeled across the gap where the circuit is broken. We can now write KVL as vA + v1 + vx + v2 = 0. With v With v 1 = v2 = 0, we get v get v x = vA = = V V O . KVL is not violated because the voltage from the source now appears across the gap in the open (broken) circuit.

−


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Seventh Edition

Exercise 2–11. Repeat the problem of Example 2-10 if the 30-V voltage source is replaced with a 2-mA current source with the arrow pointed up towards node A.

Figure Ex2–11

Figure Ex2–11 shows the resulting circuit. The description of the circuit requires four element equations and four connection equations. The element equations are v1 =

100i1

v2 =

200i2

v3 =

300i3

iA

−2 mA

=

The four connection equations are KCL : Node A

−i − i − i = 0 i − i = 0 −v + v = 0 −v + v + v = 0 A

KCL : Node B KVL : Loop 1 KVL : Loop 2

3

1

3

1

2

A

3

1

2

The KCL equation at node B implies i 1 = i 2 . We can then start with the KVL equation around loop 2 and solve as follows:

−v + v + v 3

2

=

0

v1 + v2

=

v3

1

100i1 + 200i2 =

300i3

100i1 + 200i1 =

300i3

300i1 =

300i3

i1


=

i3

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Seventh Edition

Now using the KCL equation at node A, we have

−i − i − i A

1

= 0

3

i1 + i3

=

−i

= 2 mA

A

i1 + i1 =

2 mA

2i1 =

2 mA

i1

= i3 = i2 = 1 mA

Now apply Ohm’s law to solve for the voltages v1 =

100i1 = 100mV

v2 =

200i2 = 200mV

v3 =

300i3 = 300mV

Exercise 2–12. In Figure 2–24, i 1 = 200 mA and i 3 = 100 mA. Find the voltage v x . The KCL equation at the center node is i1 i 2 i 3 = 0. Solving for i2 , we have i2 = i1 i 3 = 200 + 100 = 300 mA. Apply Ohm’s law to solve for v1 = 100i1 = (100)(200 mA) = 20 V, and v2 = 50i2 = (50)(300 mA) = 15 V. Write the KVL equation around the left loop as vx + v 1 + v 2 = 0. Solve for vx = v 1 + v2 = 20 + 15 = 35 V.

− −

−

−

−

Exercise 2–13. In Figure 2–25(a), the 2-A source is replaced by a 100-V source with the + terminal at the top, and the 3-A source is removed. Find the current and its direction through the voltage source.

Figure Ex2–13 Figure Ex2–13 shows the resulting circuit. Writing KCL at node C, we have i3 5 = 0, which yields i3 = 5 A. Write the KCL equation at node B to get i1 i2 i3 = 0, which can be solved for i1 = i 2 +i3 = i 2 +5. Write the KVL equation around loop 1 to get 100 + v1 + v2 = 0, which yields the following

−


−

− −

v1 + v2 =

100

100i1 + 50i2 =

100

100(i2 + 5) + 50i2 =

100

100i2 + 500 + 50i2 =

100

150i2

=

i2

=

−400 −2.667A

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Seventh Edition

We can then solve for i1 = i2 + 5 = 2.333 A and iA = i1 = 2.333 A. Since iA is negative, the current follows in the opposite direction through the voltage source, which is up, and has a magnitude of 2.333 A.

−

−

Exercise 2–14. Find the equivalent resistance for the circuit in Figure 2–29. Redraw the original circuit to an equivalent circuit without the diagonal resistor. Starting from the right side, combine resistors in series or parallel as appropriate to reduce the circuit to a single resistor. The following sequence of circuits shows the progress in reducing the circuit.

Figure Ex2–14 Starting at the far right, combine the 500-Ω and 1-kΩ resistors in series to get a 1.5-kΩ resistor. Next, combine the three 1.5-kΩ resistors in parallel to get a 500-Ω resistor. Combine the two 500-Ω resistors in series to get a 1-kΩ resistor. Combine the two 1-kΩ resistors in parallel to get the final equivalent resistance of 500-Ω. Exercise 2–15. Find the equivalent resistance between terminals A–C, B–D, A–D, and B–C in the circuit in Figure 2–30.

Figure 2–30 If current flows only between terminals A and C, then no current flows through terminals B and D and resistors R2 and R3 are not active in the circuit. The equivalent resistance RA C = R1 . If current flows −


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Seventh Edition

only between terminals B and D, then no current flows through terminals A and C and none of the resistors are active in the circuit. The equivalent resistance RB D = 0. If current flows between terminals A and D, resistors R2 and R3 are in parallel and that combination is in series with R1 . The equivalent resistance RA D = R 1 + R2 R3 = R1 + RR + RR . If current flows between terminals B and C, then no current flows through R 1 and it is not part of the circuit. The equivalent resistance is the parallel combination of R 2 and R3 or R B C = R 2 R3 = RR + RR . −

−

−

 

2

2

2

2

3

3

3

3

Exercise 2–16. Find the equivalent resistance between terminals A–B, A–C, A–D, B–C, B–D, and C–D in the circuit of Figure 2-31. For example: RA B = (80 80) + 60 = 100 Ω.



−

Figure 2–31

If current flows between terminals A and C, then no current flows through the 60-Ω and the 25-Ω resistors and they are not part of the circuit. The two 80-Ω resistors are in parallel and that combination is in series with the 30-Ω resistor, so we have RA C = (80 80) + 30 = 70 Ω. If current flows between terminals A and D, then no current flows through the 60-Ω and the 30-Ω resistors and they are not part of the circuit. Again, the two 80-Ω resistors are in parallel and that combination is in series with the 25-Ω resistor, so we have RA D = (80 80) + 25 = 65 Ω. If current flows between terminals B and C, then no current flows through the 25-Ω resistor and it is not part of the circuit. In addition, in the remaining circuit, the two 80-Ω resistors are shorted out. The resulting circuit is a series combination of the 60-Ω and 30-Ω resistors, which yields RB C = 60 + 30 = 90 Ω. If current flows between terminals B and D, then no current flows through the 30-Ω resistor and it is not part of the circuit. In addition, in the remaining circuit, the two 80-Ω resistors are again shorted out. The resulting circuit is a series combination of the 60-Ω and 25-Ω resistors, which yields RB D = 60 + 25 = 85 Ω. Finally, with current flowing between terminals C and D, the 60-Ω resistor is not part of the circuit and the two 80-Ω resistors are shorted out. The equivalent resistance is the series combination of the 30-Ω and 25-Ω resistors, which yields RC D = 25 + 30 = 55 Ω. −

−





−

−

−

Exercise 2–17. A practical current source consists of a 2-mA ideal current source in parallel with a 500-Ω resistance. Find the equivalent practical voltage source. The equivalent practical voltage source will have the same 500-Ω resistance. To transform the current source into a voltage source, we compute vS = i S R = (2mA)(500 Ω) = 1 V. Exercise 2–18. Find the equivalent circuit for each of the following (a). Three ideal 1.5-V batteries connected in series. For voltage sources connected in series, the voltages add. Assuming all three sources are oriented in the same direction, the equivalent voltage is 1.5 + 1.5 + 1.5 = 4.5 V. (b). A 5-mA current source in series with a 100-kΩ resistor. A current source in series with a resistor acts as a current source without the resistor, so the equivalent circuits is a single 5-mA current source.


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Seventh Edition

(c). A 40-A ideal current source in parallel with an ideal 10-A current source. For ideal current sources in parallel, the currents add, so the equivalent circuit is a 50-A current source. (d). A 100-V source in parallel with two 10-kΩ resistors. A voltage source in parallel with any resistance acts like a voltage source, so the equivalent circuits is a single 100-V voltage source. (e). An ideal 15-V source in series with an ideal 10-mA source. This is not a valid combination of sources and the two cannot be combined in a theoretical perspective. (f). A 15-V ideal source and a 5-V ideal source connected in parallel. This is not a valid combination of voltage sources, since a parallel combination of elements must have the same voltage. Exercise 2–19. Find the voltages v x , v y , and v z in the circuit of Figure 2–39. Show that the sum of all the voltages across each of the individual resistors equals the source voltage.

Figure 2–39

For each resistor, use voltage division to find its corresponding voltage. vx

=

vy

=

vO

=

vz

=

   

100 100 + 560 + 330 + 220 560 100 + 560 + 330 + 220 330 100 + 560 + 330 + 220 220 100 + 560 + 330 + 220

   

24 = 1.9835 V

24 = 11.1074 V

24 = 6.5455 V

24 = 4.3636 V

Sum the voltages to get 1.9835 + 11.107 + 6.5455 + 4.3636 = 24 V, which matches the source voltage.


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Seventh Edition

Exercise 2–20. In Figure 2-40, Rx = 10 kΩ. The output voltage vO = 20 V. Find the voltage source that would produce that output. (Hint: It is not 10 V.)

Figure 2–40

Combine the two 10-kΩ resistors in parallel to get a single 5-kΩ resistor in series with the 2-kΩ resistor. The 5-kΩ resistor still has 20 V across it. Use the voltage division equation to solve for the voltage of the source as follows: 20 =

vs

=

 

5000 5000 + 2000 5000 + 2000 5000

 

vs 20 = 28 V

Exercise 2–21. In Figure 2–41, suppose that a resistor R4 is connected across the output. What value should R 4 be if we want 12 vS to appear between node A and ground? Using the concept of voltage division, for one-half of vS to appear between node A and ground, the resistance between node A and ground will have to match R1 so that the source voltage divides equally between the two parts of the circuit. The equivalent resistance between node A and ground is the series combination of R3 and R4 in parallel with R2 or REQ = R 2 (R3 + R4 ). Setting REQ = R 1 we can solve for R 4 as follows:



R1 R1 (R2 + R3 + R4 ) R1 R4

−R R R (R − R ) 4

2

1

4

2

R4

= R2

 (R + R ) 3

4

R2 (R3 + R4 ) R2 R3 + R2 R4 = R2 + R3 + R4 R2 + R3 + R4

=

= R2 R3 + R2 R4 = R2 R3 = R2 R3 =

R2 R3

−R R −R R −R R −R R −R R −R R R −R 1

1

2

1

3

1

2

1

3

1

2

1

3

2

=

R1 R3 + R1 R2 R3 R2 R2 R1

−

−

Exercise 2–22. Ten volts (vs ) are connected acorss the 10-kΩ potentiometer (RTOTAL) shown in Figure 2–42(c). A load resistor of 10 kΩ is connected across its output. At what resistance should the wiper (RTOTAL R1 ) be set so that 2 V appears at the output, vO ?

−

To solve this problem, first define R2 = RTOTAL R1 , which is the resistance we want to find. For a 10-kΩ potentiometer, R1 + R2 = 10 kΩ, so R1 = 10kΩ R2 . The equivalent resistance of the output is REQ = R 2 10 kΩ. Now use the voltage division equation and the specified source and output voltages to




−

−

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Seventh Edition

solve for R 2 as follows:



2 =



4

2 10 (104

REQ R1 + REQ

− R + 2



      10 =

104 R2 104 + R2

4

4

− R )(10 + R ) + 10 R −R + 10 + 10 R R + (4 × 10 )R − 10 2

2

2 2

8

2 2

4

4

2

−



=

104 R2 104 + R2

=

(104 R2 )(5)

= (5

2 8

2

104

104 R2 104 + R2 104 R2 R2 + 104 + R2

4

× 10 )R



  

10

10

2

= 0

Solving for the positive root of the quadratic equation, we get R 2 = 2.36 kΩ. The other root is negative, so it is not a valid solution for a resistance. Exercise 2–23. (a). Find i x and iz in the circuit of Figure 2–46(a). Use current division to find all of the currents. Note that iz flows through an equivalent resistance of 10 Ω. ix

iy

iz

  

=

=

=

1 20

1 20

1 20

+

1 20 1 1 + 10 20

+

1 20 1 1 + 10 20

+

1 10 1 1 + 10 20

  

5 = 1.25 A

5 = 1.25 A

5 = 2.5 A

(b). Show that the sum of i x , i y , and i z equals the source current. Sum the currents found in part (a), i x + iy + iz = 1.25 + 1.25 + 2.5 = 5 A. Exercise 2–24. The circuit in Figure 2-47 shows a delicate device that is modeled by a 90-Ω equivalent resistance. The device requires a current of 1 mA to operate properly. A 1.5-mA fuse is inserted in series with the device to protect it from overheating. The resistance of the fuse is 10 Ω. Without the shunt resistance Rx , the source would deliver 5 mA to the device, causing the fuse to blow. Inserting a shunt resistor Rx diverts a portion of the available source current around the fuse and device. Select a value of Rx so only 1 mA is delivered to the device. The equivalent resistance of the device and its fuse is 100 Ω. Write the current division equation such that the current through the device is 1 mA and then solve for the shunt resistance R x . 1 =



1 + 100

1 100 1 Rx

+

1 100



10 =



Rx Rx + 100 + Rx



10 =



Rx 2Rx + 100



10

2Rx + 100 = 10Rx 8Rx = Rx

100

= 12.5 Ω


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Seventh Edition

Tracing back to the first circuit with the voltage source, we see that ix enters a circuit where the current divides equally between two 30-Ω resistors. The current through each resistor is half of the original or 222 mA. Therefore, 222 mA flows through each 15-Ω resistor in the original circuit. We can then compute vx = (15)(0.222) = 3.33 V. Exercise 2–29. Find v x and v y using circuit reduction on the circuit in Figure 2–55. Combine the two voltage sources in series together to get a single 12-V source. To solve for vx , first note that the voltage source is in parallel with the series combination of resistors on the far right. From the perspective of the left side of the circuit, we can safely ignore the resistors to the right of the voltage source. Perform a source transformation on the 12-V source and the 1-kΩ to its left to get a 12-mA current source in parallel with a 1-kΩ resistor. Now perform current division to find the current through the 1.5-kΩ resistor as follows 1 1.5 + 2.2 ( 12) = 2.0614 mA ix = 1 1 1 + + 1.5 + 2.2 3.3 1 Note the sign convention for vx introduces the negative sign for the current. Apply Ohm’s law to find the voltage v x = (1500)( 0.0020614) = 3.092 V. To solve for v y , perform voltage division using the equivalent 12-V source as follows:

 

−

  −

−

−

vy =



3.3 1 + 3.3



12 = 9.2093V

Exercise 2–30. Find the voltage across the current source in Figure 2–57. Combine the resistors on the left in series to get an equivalent resistance of 2 .2 + 1.5 + 1 = 4.7kΩ. Combine the resistors on the right in series to get an equivalent resistance of 1 + 3.3 = 4.3 kΩ. Combine the two equivalent resistance in parallel to get a final equivalent resistance of 4 .7 4.3 = 2.2456 kΩ. The current flowing through the equivalent resistance yields a voltage of (2.2456 kΩ)(0.1 mA) = 224.56 mV. Given the sign convention in Figure 2–57, the source voltage is negative and v S = 224.56 mV.



−


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Seventh Edition

Problem Solutions

Problem 2–1. The current through a 56-kΩ resistor is 2.2 mA. Find the voltage across the resistor. Using Ohm’s law we have v = Ri = (56 103 )(2.2 10 3 ) = 123.2 V.

×

×

−

Problem 2–2. The voltage across a particular resistor is 6.23 V and the current is 2.75 mA. What is the actual resistance of the resistor? Using the inside back cover, what is the likely standard value of the resistor? Using Ohm’s law we can solve for resistance as R = v/i = (6.23)/(2.75 10 3 ) = 2.2655 kΩ. Using the table of standard values, the resistor is likely marked as a 2.2-kΩ resistor.

×

−

Problem 2–3. A 100-kΩ resistor dissipates 100 mW. Find the current through the resistor. The power dissipated by a resistor is p = i 2 R. Solving for current, we have i = p/R = (10 1mA.

 

−

1 )/(105 )

=

Problem 2–4. The conductance of a particular resistor is 0.5 mS. Find the current through the resistor when connected across a 5-V source. Using the conductance version of Ohm’s law, we have i = Gv = (0.5 10 3 )(5) = 2.5mA.

×

−

Problem 2–5. In Figure P2–5 the resistor dissipates 25 mW. Find R x . The power dissipated by a resistor can be written as p = v2 /R. Solving for the resistance, we have Rx = v 2 /px = (152 )/(25 10 3 ) = 9kΩ.

×

−

Problem 2–6. In Figure P2–6 find R x and the power delivered to the resistor. Using Ohm’s law to solve for resistance, we have Rx = v/i = 100/(10 2 ) = 10 kΩ. The power delivered to the resistor is p = vi = (100)(10 2 ) = 1 W. −

−

Problem 2–7. A resistor found in the lab has three orange stripes followed by a gold stripe. An ohmmeter measures its resistance as 34.9 kΩ. Is the resistor properly color coded? (See inside back cover for color code.) Since there are three colored stripes and a gold stripe, the first two stripes are the significant digits, the third stripe is the multiplier, and the gold stripe is the tolerance. Using the color code table, the significant digits for the first two stripes are 3 and 3. The multiplier associated with orange is 1 k, so we have 33 1000 = 33 kΩ. The tolerance associated with the gold stripe is 5%, which gives a range of resistances from 31.35 to 34.65 kΩ. The resistor measured outside of this range, but its measured value is within 10% of 33 kΩ, so it should have a silver tolerance stripe in place of the gold one.

×

±

Problem 2–8. The i-v characteristic of a nonlinear resistor is v = 82i + 0.18i3 . (a). Calculate v and p for i =

±0.5, ±1, ±2, ±5, and ±10 A.

Since there are 10 values for the current to examine, this problem is best solved with MATLAB. We will solve the problem for one current and then automate the process. Given i = +0.5 A, use the expression to calculate the voltage as v = 82i + 0.18i3 = (82)(0.5) + (0.18)(0.5)3 = 41.023 V. We can now calculate the power as p = vi = (41.023)(0.5) = 20.511 W. To compute the other solutions, we use MATLAB as shown below. ii = [-10, -5, -2, -1, -0.5, 0.5, 1, 2, 5, 10]; v = 8 2*ii + 0.18*ii.ˆ3; p = v .*ii; Results = [ii' v' p']

The corresponding MATLAB output is shown below. Results = -10.0000e+000 -5.0000e+000 -2.0000e+000 -1.0000e+000 -500.0000e-003

-1.0000e+003 -432.5000e+000 -165.4400e+000 -82.1800e+000 -41.0225e+000


10.0000e+003 2.1625e+003 330.8800e+000 82.1800e+000 20.5113e+000

Page 2-14


500.0000e-003 1.0000e+000 2.0000e+000 5.0000e+000 10.0000e+000

41.0225e+000 82.1800e+000 165.4400e+000 432.5000e+000 1.0000e+003

Seventh Edition

20.5113e+000 82.1800e+000 330.8800e+000 2.1625e+003 10.0000e+003

The results are summarized in the following table. i (A) -10 -5 -2 -1 -0.5 0.5 1 2 5 10

v (V) -1000.00 -432.50 -165.44 -82.18 -41.02 41.02 82.18 165.44 432.50 1000.00

p (W) 10000.00 2162.50 330.88 82.18 20.51 20.51 82.18 330.88 2162.50 10000.00

(b). Find the maximum error in v when the device is treated as a 82-Ω linear resistance on the range i < 0.5 A.

||

The term 0.18i3 makes the expression for v nonlinear and represents the difference between the actual device and an 82-Ω resistor. The maximum error will occur when the absolute value of the cubic term is maximized. That occurs at the extreme values for i, which are 0.5 A, in this case. At +0.5 A, the actual voltage is 41.0225 V and the voltage across a 82-Ω resistor is exactly 41 V. The error is 22.5 mV. At 0.5 A, the actual voltage is 41.0225 V and the voltage across a 82-Ω resistor is exactly 41 V, so the error is 22.5 mV. In both cases, the percentage of error is 0.055%, which is exceptionally small.

±

−

−

−

−

Problem 2–9. A 100-kΩ resistor has a power rating of 0.125 W. Find the maximum voltage that can be applied to the resistor. The power dissipated by a resistor can be expressed as p = v2 /R. Solving for the voltage, we have v = pR = (0.125)(105 ) = 111.803 V.

√



Problem 2–10. A certain type of film resistor is available with resistance values between 10 Ω and 100 MΩ. The maximum ratings for all resistors of this type are 500 V and 0.25 W. Show that the voltage rating is the controlling limit for R > 1 MΩ, and that the power rating is the controlling limit when R < 1 MΩ. The power dissipated by a resistor can be expressed as p = v 2 /R. Solving for the resistance, we have R = v2 /p. With both maximum ratings applied, the resistance is R = (500)2 /(0.25) = 1 MΩ. Therefore at R = 1 MΩ there are no issues with either type of rating. If the resistance increases above 1 MΩ, then using R = v 2 /p, the maximum power must be less than 0.25 W. Therefore, the resistor will never dissipate 0.25 W and the voltage rating will be the only active constraint. If the resistance is less than 1 MΩ, then the maximum voltage must be less than 500 V and power rating will be the only active constraint. Problem 2–11. Figure P2–11 shows the circuit symbol for a class of two-terminal devices called diodes. The i-v relationship for a specific pn junction diode is i = 2 10 16 e40v 1 A.





× − (a). Use this equation to find i and p for v = 0, ±0.1, ±0.2, ±0.4, and ±0.8 V. Use these data to plot the −

i-v characteristic of the element.

For each voltage, use the given equation to compute the current and then use p = vi to compute the associated power. MATLAB is appropriate for these calculations and plotting. v = [-0.8, -0.4, -0.2, -0.1, 0 0.1, 0.2, 0.4, 0.8]; ii = 2e-16 *(exp(40 *v)-1); p = v .*ii; Results = [v' ii' p']


Page 2-15


Seventh Edition

The corresponding MATLAB output is shown below followed by a plot of the data in Figure P2–11. Results = -800.0000e-003 -400.0000e-003 -200.0000e-003 -100.0000e-003 0.0000e+000 100.0000e-003 200.0000e-003 400.0000e-003 800.0000e-003

-200.0000e-018 -200.0000e-018 -199.9329e-018 -196.3369e-018 0.0000e+000 10.7196e-015 595.9916e-015 1.7772e-009 15.7926e-003

16

x 10

160.0000e-018 80.0000e-018 39.9866e-018 19.6337e-018 0.0000e+000 1.0720e-015 119.1983e-015 710.8888e-012 12.6341e-003

−3

14

12

10

) A ( t n e r r u C

8

6

4

2

0

−2 −0.8

−0.6

−0.4

−0.2

0 Voltage (V)

0.2

0.4

0.6

0.8

Figure P2–11

(b). Is the diode linear or nonlinear, bilateral or nonbilateral, and active or passive? The plot in Part (a) shows that the device is nonlinear and nonbilateral. The power for the device is always positive, so it is passive. (c). Use the diode model to predict i and p for v = 5 V. Do you think the model applies to voltages in this range? Explain. For v = 5 V, i = 1.45 power are too large.

× 10

71

A and p = 7.23

−5 V. For v = −5 V, i = −2.00 × 10

× 10

71

W. The model is not valid because the current and

(d). Repeat (c) for v =

16

A and p = 1.00 and power are both essentially zero. −

× 10

−

15

W. The model is valid because the current

Problem 2–12. A thermistor is a temperature-sensing element composed of a semiconductor material which exhibits a large change in resistance proportional to a small change in temperature. A particular thermistor has a resistance of 5 kΩ at 25 C. Its resistance is 340 Ω at 100 C. Assuming a straight-line relationship between these two values, at what temperature will the thermistor’s resistance equal 1 kΩ? Find the rate at which the resistance changes for each degree of temperature. ◦

◦

∆Ω =


5000 340 4660 = = 25 100 75

− −

−

−62.13 Ω/ C ◦

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Seventh Edition

To go from 5 kΩ to 1 kΩ, the resistance changes by 4 kΩ, which means the temperature change is 4000/( 62.13) = 64.38 C. The final temperature is 25 + 64.38 = 89.38 C.

−

−

−

◦

◦

Problem 2–13. In Figure P2–13 i 2 = 5 A and i 3 = 2 A. Find i 1 and i4 . The KCL equations for nodes B and C are

−

−i − i = 0 i + i − i = 0 Using the first equation, we can solve for i = − i = 5 A. Using the second equation, we can solve for i = i + i = −5 + 2 = −3 A. 2

1

2

3

4

1

4

2

2

3

Problem 2–14. In Figure P2–14 v 1 = 3 V and v 3 = 5 V. Find v 2 , v 4 and v 5 . We can use a KVL equation on the left loop and the two given voltages to solve for v2 . The KVL equation is v1 + v2 + v3 = 0. Solving for v 2 = v1 v3 = 3 5 = 2 V. In examining the circuit, there is a ground on each side of v 5 , so the voltage difference across this element is zero, v 5 = 0 V. We can now use KVL around the right loop to solve for v 4 . The KVL equation is v3 + v4 + v5 = 0. Solve for v 4 = v 3 v5 = 5 0 = 5 V.

−

−

− − −

−

−

Problem 2–15. For the circuit in Figure P2–15: (a). Identify the nodes and at least two loops. The circuit has three nodes and three loops. The nodes are labeled A, B, and C. The first loop contains elements 1 and 2, the second loop contains elements 2, 3, and 4, and the third loop contains elements 1, 3 and 4. (b). Identify any elements connected in series or in parallel. Elements 3 and 4 are connected in series. Elements 1 and 2 are connected in parallel. (c). Write KCL and KVL connection equations for the circuit. The KCL equations are Node A

−i − i − i = 0 i − i = 0 1

Node B

2

3

Node C

3

4

i1 + i2 + i4 = 0

The KVL equations are Loop 12

−v + v = 0 1

Loop 2 3 4

2

−v + v + v = 0 −v + v + v = 0

Loop 1 3 4

2

3

4

1

3

4

Problem 2–16. In Figure P2–15, i 2 = 20 mA and i 4 = 10 mA. Find i 1 and i 3 . The KCL equations for the circuit are

−

Node A Node B Node C Using the equation for node C, we can solve i 1 = B, we can solve i3 = i 4 = 10 mA.


−i − i − i = 0 i − i = 0 1

2

3

3

4

i1 + i2 + i4 = 0

−i − i = 20 − 10 = 10 mA. Using the equation for node 2

4

Page 2-17


Seventh Edition

Problem 2–17. For the circuit in Figure P2–17: (a). Identify the nodes and at least three loops in the circuit. The are four nodes and at least five loops. There are only three independent KVL equations. The nodes are labeled A, B, C, and D. Valid loops include the following sequences of elements: (1, 3, 2), (2, 4, 5), (3, 6, 4), (1, 6, 5), and (2, 3, 6, 5). (b). Identify any elements connected in series or in parallel. In this circuit, none of the elements are connected in series and none of them are connected in parallel. (c). Write KCL and KVL connection equations for the circuit. The KCL equations are Node A

−i − i − i = 0 −i + i − i = 0

Node B

2

3

4

1

3

6

Node C

i1 + i2 + i5 = 0

Node D

i4

− i + i = 0 5

6

Three independent KVL equations are Loop 1 3 2

−v − v + v = 0 Loop 2 4 5 −v + v + v = 0 Loop 3 6 4 v + v − v = 0 = 10 V, v = −10 V, and v = 3 V. Find v , v , and v . 1

3

2

2

4

5

3

Problem 2–18. In Figure P2–17 v 2 The KVL equations are

6

4

3

4

Loop 1 3 2

1

5

6

−v − v + v = 0 Loop 2 4 5 −v + v + v = 0 Loop 3 6 4 v + v − v = 0 Using the first equation, we can solve for v = v − v = 10 + 10 = 20 V. Using the second equation, we can solve for v = v − v = 10 − 3 = 7 V. Using the third equation, we can solve v = v − v = 3 + 10 = 13 V. 1

3

2

2

4

5

3

1

5

2

2

6

4

3

4

6

4

3

Problem 2–19. In many circuits the ground is often the metal case that houses the circuit. Occasionally a failure occurs whereby a wire connected to a particular node touches the case causing that node to become connected to ground. Suppose that in Figure P2–17 Node B accidently touches ground. How would that affect the voltages found in Problem 2–18? If Node B is connected to ground, then element 6 is connected to ground on both sides, so its voltage is v6 = 0 V. If we define v6 = 0 V, all of the original KVL equations found in Problem 2–17 are still valid. Even though the equations are valid, Problem 2–18 is no longer valid because there is a conflict with the given voltages. Using the KVL equation v3 + v6 v4 = 0 and substituting in v6 = 0, we get v3 = v4 . In Problem 2–18, the given values are v 3 = 10 V and v 4 = 3 V, which is not possible if Node B is connected to ground.

−

−

Problem 2–20. The circuit in Figure P2–20 is organized around the three signal lines A, B, and C. (a). Identify the nodes and at least three loops in the circuit. The are four nodes and at least five loops. The nodes are labeled A, B, C, and D. Valid loops include the following sequences of elements: (1, 3, 2), (2, 4, 5), (3, 6, 4), (1, 6, 5), and (2, 3, 6, 5).


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Seventh Edition

(b). Write KCL connection equations for the circuit. The KCL equations are Node A

−i − i − i = 0 −i + i − i = 0

Node B

(c). If i1 =

−20 mA, i

2 =

2

3

4

1

3

6

Node C

i1 + i2 + i5 = 0

Node D

i4

−12 mA, and i

3 =

− i + i = 0 5

6

50 mA, find i 4 , i 5 , and i 6 .

Using the KCL equation at node A, we can solve for i4 = i2 i3 = 12 50 = 38 mA. Using the KCL equation are node C, we can solve for i 5 = i1 i2 = 20 + 12 = 32 mA. Using the KCL equation at node D, we can solve for i 6 = i 5 i4 = 32 + 38 = 70 mA.

− −

− −

−

−

−

(d). Show that the circuit in Figure P2–20 is identical to that in Figure P2–17. The circuits have the same nodes, connections, and current directions, so they must be equivalent. Problem 2–21. In Figure P2–21 v 2 = 10 V, v 4 = 5 V, and v 5 = 15 V. Find v 1 , v 3 , and v6 . The KVL equations for the circuit are Loop 1 2 3 Loop 3 4 5 Loop 2 6 4

−v + v + v = 0 −v + v + v = 0 −v + v − v = 0 1

2

3

3

4

5

2

6

4

Using the second loop equation, we can solve for v3 = v4 + v 5 = 5 + 15 = 20 V. Using the first loop equation, we can solve for v1 = v2 + v3 = 10 + 20 = 30 V. Finally, using the third loop equation, we can solve for v 6 = v 2 + v4 = 10 + 5 = 15 V. Problem 2–22. In Figure P2–22 i 1 = 25 mA, i 2 = 10 mA, and i 3 = The KCL equations for the circuit are Node A i1

−15 mA. Find i

4

and i 5 .

− i + i − i = 0 Node B −i + i − i = 0 Node C −i + i + i = 0 Using the first node equation, we can solve for i = i − i + i = 25 − 10 − 15 = 0 mA. Using the second node equation, we can solve for i = −i + i = −25 + 10 = −15 mA. 2

1

4

1

2

5

3

4

5

4

5

3

1

2

3

2

Problem 2–23. (a). Use the passive sign convention to assign voltage variables consistent with the currents in Figure P2–22. Write three KVL connection equations using these voltage variables. Figure P2–23 shows the original Figure P–22 with the voltages labeled following the passive sign convention. The KVL equations for the circuit are Loop 12 Loop 2 4 5 Loop 34


v1 + v2 = 0

−v + v − v = 0 2

4

5

v3 + v4 = 0

Page 2-19


Seventh Edition

Figure P2–23

(b). If v4 = 0 V, what can be said about the voltages across all the other elements? If v 4 = 0 V, then the third loop equation indicates that v 3 = 0 V. Applying these voltages to the other two loop equations, we have v 1 = v2 and v 2 = v5 .

−

−

Problem 2–24. The KCL equations for a three-node circuit are: Node A Node B Node C

−i + i − i = 0 −i − i + i = 0 i + i + i − i = 0 1

1

2

4

2

3

5

3

4

5

Draw the circuit diagram and indicate the reference directions for the element currents. There are many equivalent diagrams to solve this problem. One possible solution is shown in Figure P2– 24. Problem 2–25. Find v x and i x in Figure P2–25. The current ix points in the opposite direction as the 500-µA current source, so ix = Ohm’s law, we have v x = Ri x = (68 103 )( 500 10 6 ) = 34 V.

×

− ×

−

−

−500 µA.

Using

Problem 2–26. Find v x and i x in Figure P2–26. The circuit has a single current, ix , which flows clockwise. Label the 22-kΩ resistor as R1 with voltage drop v 1 following the passive sign convention (positive on left and negative on right). The KVL equation for


Page 2-20


Seventh Edition

Figure P2–24

the circuit combined with Ohm’s law provides the solution.

−18 + v + v 1

(22

3

= 0

x

v1 + vx

= 18

R1 ix + Rx ix

= 18

3

× 10 )i + (68 × 10 )i (90 × 10 )i x

3

x

= 18

x

= 18

ix =

200 µA

vx =

(68

3

× 10 )i

x

= 13.6 V

Problem 2–27. Find vx and ix in Figure P2–27. Compare the results of your answers with those in Problem 2–26. What effect did adding the 33-kΩ resistor have on the overall circuit? Why isn’t i y zero? Label the 22-kΩ resistor as R1 with voltage drop v1 following the passive sign convention (positive on left and negative on right). Label the 33-kΩ resistor with voltage vy following the passive sign convention (positive on top and negative on bottom). Label the source current as iS following the passive sign convention (flowing into the positive terminal of the voltage source). The KVL equations for the circuit are:

−18 + v

−v + v + v y

1

= 0

y

= 0

x

The KCL equation for the circuit is

−i − i − i = 0 S

x

y

Solving the first KVL equation, we have v y = 18 V. Substituting this result into the second KVL equation,


Page 2-21


Seventh Edition

we get v 1 + vx = 18. Current i x flows through both R 1 and R x , so we can solve as follows:

(22

3

v1 + vx

= 18

R1 ix + Rx ix

= 18

3

× 10 )i + (68 × 10 )i (90 × 10 )i x

3

x

= 18

x

= 18

ix =

200 µA

vx =

(68

3

× 10 )i

x

= 13.6 V

These results for v x and i x match those in Problem 2–26. We can also find i y using Ohm’s law, i y = v y /Ry = (18)/(33 103 ) = 545.5 µA. Applying the KCL equation, we get iS = ix iy = 200 545.5 = 745.5 µA. Adding the 33-kΩ resistor increased the amount of current flowing from the source. The current iy is not zero because there is a voltage across the 33-kΩ resistor.

×

− −

− −

−

Problem 2–28. A modeler wants to light his model building using miniature grain-of-wheat light bulbs connected in parallel as shown in Figure P2–28. He uses two 1.5-V “C-cells” to power his lights. He wants to use as many lights as possible, but wants to limit his current drain to 500 µA to preserve the batteries. If each light has a resistance of 36 kΩ, how many lights can he install and still be under his current limit? The two 1.5-V batteries are connected in series to provide a total of 3 V to the circuit. Since the light bulbs are connected in parallel, the entire 3 V appears across each one. Using Ohm’s law, the current through each bulb is i = v /R = 3/(36 103 ) = 83.3 µA. The design requires the batteries to provide no more the 500 µA, so we can connect up to 500/83.3 = 6 bulbs in parallel across the batteries.

×

Problem 2–29. Find v x and i x in Figure P2–29. In the circuit, 0.5 A flows through the 10-Ω resistor in the center. The voltage drop across this resistor is v = Ri = (10)(0.5) = 5 V. The 10-Ω resistor is connected in parallel to the 5-Ω resistor, so they have the same voltage drop. The associated KVL equation verifies this fact. With 5 V across the 5-Ω resistor, the current is ix = v /R = 5/5 = 1 A. KCL at the top node requires that the current entering the node equal the current leaving the node. Since we have 0.5 + 1.0 = 1.5 A leaving the node, 1.5 A must enter the node through the 4 Ω resistor. The voltage drop across the 4 Ω resistor is v = Ri = (4)(1.5) = 6 V. We can now write a KVL equation around the first loop to get vx + 6 + 5 = 0, which implies v x = 11 V.

−

Problem 2–30. In Figure P2–30: (a). Assign a voltage and current variable to every element. Figure P2–30 shows the voltage and current labels following the passive sign convention. (b). Use KVL to find the voltage across each resistor. The KVL equations are

−v −v

S1 + v1 + vS3

= 0

S1 + v2

S2

= 0

vS2 + v3 + vS3

= 0

−v

Solving the first equation, we have v 1 = v S1 vS3 = 10 15 = 5 V. Solving the second equation, we have v 2 = v S1 + vS2 = 10+5 = 15 V. Solving the third equation, we have v 3 = vS2 vS3 = 5 15 = 20 V.

−

−

−

−

− −

−−

(c). Use Ohm’s law to find the current through each resistor.


Page 2-22


Seventh Edition

Figure P2–30

Applying i = v/R to each resistor, we have i1

=

v1 5 = = R1 200

i2

=

v2 15 = = 75 mA R2 200

i3

=

v3 = R3

−

−20 200

=

−25mA

−100mA

(d). Use KCL to find the current through each voltage source. The KCL equations are

−i − i − i = 0 i − i + i = 0 i + i − i = 0 Solving the first equation, we have i = −i − i = 25 − 75 = −50 mA. Solving the second equation, we have i = i − i = −100 − 75 = −175 mA. Solving the third equation, we have i = i + i = −25 − 100 = −125 mA. S1

S2

3

2


1

2

S1

2

3

S2

1

3

S3

1

2

S3

1

3

Page 2-23


Seventh Edition

Problem 2–31. Find the power provided by the source in Figure P2–31. Figure P2–31 shows the voltage and current labels following the passive sign convention. The KCL

Figure P2–31

equations are iS

−i −i i −i 1

2

= 0

2

3

= 0

The KVL equations are vS + v1

−v + v + v 1

2

3

= 0 = 0

The current source requires i S = 5 mA. The first KCL equation implies i 1 = 5mA i2 and the second implies i2 = i3 . Using Ohm’s law and substituting these equations into the second KVL equation, we can solve for the source power as follows:

−

v1

=

v2 + v3

R1 i1

=

R2 i2 + R3 i3

−i ) = 1000(0.005 − i ) = 5 − 1000i = 2

R2 i2 + R3 i2

2

500i2 + 1500i2

R1 (0.005

2

2000i2

3000i2

=

5

i2

=

1.667mA

i3

=

i2 = 1.667mA

i1

=

5

vS

=

pS

=

− i = 5 − 1.667 = 3.333mA −v = −R i = −(1000)(0.003333) = −3.333V v i = (−3.333)(0.005) = −16.667mW 2

1

S

1 1

S

Problem 2–32. Figure P2–32 shows a subcircuit connected to the rest of the circuit at four points.


Page 2-24


Seventh Edition

(a). Use element and connection constraints to find v x and i x . Label the 5-kΩ resistor as R 1 with the current flowing from left to right. Label the 2-kΩ resistor as R 2 with the positive sign at the bottom. Using Ohm’s law, we can compute i1 = v 1 /R1 = 20/5000 = 4 mA. The KCL equation at the center node is 4 mA + i1 i2 ix = 0. Substituting in the known values, we can solve for ix as ix = 4+i1 i2 = 4+4 6 = 2 mA. Using Ohm’s law vx = R x i x = (8000)(0.002) = 16 V.

−

− −

−

(b). Show that the sum of the currents into the rest of the circuit is zero. The sum of the currents entering the rest of the circuit is

−i + i − 4 + i 1

2

x

=

−4 + 6 − 4 + 2 = 0 mA.

(c). Find the voltage vA with respect to the ground in the circuit. From the ground to vA there are three voltages. First, there is an increase across the voltage source of 12 V. Next, there is an increase across Rx of 16 V. Finally, there is a decrease across R2 of v2 = R2 i2 = (2000)(0.006) = 12 V. Therefore, v A = 12 + 16 12 = 16 V.

−

Problem 2–33. In Figure P2–33, i x = 0.5 mA. Find the value of R. Label the left 10-kΩ resistor as Rx , with voltage vx . Label the unknown resistor as R1 , with current i1 flowing down. Label the right 10-kΩ resistor as R2 with the current flowing right to left. Apply the passive sign convention to label the voltages. Use Ohm’s law to solve for v x = R x ix = (10000)(0.0005) = 5 V. Write the KVL equation for the left side as 4 vx + v1 = 0 and solve for v 1 as v 1 = 4 + vx = 4 + 5 = 9 V. Write the KVL equation for the right side as v1 v2 + 15 = 0 and solve for v2 as v 2 = 15 v1 = 15 9 = 6 V. Use Ohm’s law to solve for i2 = v2 /R2 = 6/10000 = 0.6 mA. Write the KCL equation for the center node as i1 + i 2 ix = 0 and solve for i1 as i1 = i2 ix = 0.6 0.5 = 0.1 mA. Use Ohm’s law to find R1 = v 1 /i1 = 9/(0.0001) = 90 kΩ.

− − − −

−

−

−

−

−

−

Problem 2–34. Figure P2–34 shows a resistor with one terminal connected to ground and the other connected to an arrow. The arrow symbol is used to indicate a connection to one terminal of a voltage source whose other terminal is connected to ground. The label next to the arrow indicates the source voltage at the ungrounded terminal. Find the voltage across, current through, and power dissipated in the resistor. The voltage across the resistor is the voltage on the left side minus the voltage on the right side. Therefore, vx = 0 ( 12) = 0 + 12 = 12 V. Use Ohm’s law to find ix = v x /Rx = 12/(39 103 ) = 307.692 µA. The power dissipated by the resistor is p x = v x ix = (12)(307.692 10 6 ) = 3.6923 mW.

−−

×

−

×

Problem 2–35. Find the equivalent resistance R EQ in Figure P2–35. The 10-Ω resistor and the 30-Ω resistor are in parallel. That combination is in series with the 7.5-Ω resistor. We can calculate the equivalent resistance as follows: REQ = 7.5 + (30

 10) = 7.5 +

1 1 1 + 30 10

= 7.5 +

(30)(10) = 7.5 + 7.5 = 15 Ω 30 + 10

Problem 2–36. Find the equivalent resistance R EQ in Figure P2–36. Combine the 33-kΩ and 47-kΩ resistors in series to get an equivalent resistance of 33+47 = 80 kΩ. The 80kΩ resistance is in parallel with the 100-kΩ resistor, which yields an equivalent resistance of 100 80 = 44.4 kΩ. That resistance is in series with the 68-kΩ resistor, which yields REQ = 68 + 44.4 = 112.4 kΩ.



Problem 2–37. Find the equivalent resistance R EQ in Figure P2–37. Working from the right to the left, combine the 10-kΩ resistor in parallel with the 15-kΩ resistor to get an equivalent resistance of 6 kΩ. That resistance is in series with the 33-kΩ resistor, which yields an equivalent resistance of 39 kΩ. Finally, combine the 39-kΩ resistance in parallel with the 56-kΩ resistor to get REQ = 22.99 kΩ. Problem 2–38. Equivalent resistance is defined at a particular pair of terminals. In the following figure the same circuit is looked at from two different terminal pairs. Find the equivalent resistances REQ1 and REQ2 in Figure P2–38. Note that in calculating R EQ2 the 33-kΩ resistor is connected to an open circuit and therefore doesn’t affect the calculation.


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Seventh Edition

For REQ1, ignore the two terminals on the right and collapse the circuit from right to left. The 10-kΩ and the two 22-kΩ resistors are in series; that result is in parallel with the 56-kΩ resistor; and that result is in series with the 33-kΩresistor. We can calculate the equivalent resistance as follows: REQ1 = 33 + [56 (10 + 22 + 22)] = 33 + [56



 54] = 33 + 27.49 = 60.49kΩ

For REQ2 , ignore the two terminals on the left and the 33-kΩ resistor. Collapse the circuit from left to right. The 10, 56, and lower 22-kΩ resistors are in series and that result is in parallel with the right 22-kΩ resistor. We can calculate the equivalent resistance as follows: REQ2 = [(10 + 56 + 22)

 22] = [88  22] = 17.6 kΩ

Problem 2–39. Find R EQ in Figure P2–39 when the switch is open. Repeat when the switch is closed. When the switch is open, the two 100-Ω resistors are in parallel and that result is in series with the two 50-Ω resistors. We can calculate REQ = 50 + (100 100) + 50 = 50 + 50 + 50 = 150 Ω. With the switch closed, the wire shorts out the two 100-Ω resistors, so they do not contribute to the equivalent resistance. The results is that the two 50-Ω resistors are in series, so R EQ = 50 + 50 = 100 Ω.



Problem 2–40. Find REQ between nodes A and B for each of the circuits in Figure P2–40. What conclusion can you draw about resistors of the same value connected in parallel? We can calculate the equivalent resistance for Circuit (a) as follows: REQ =

1 R R = = 1 1 1 1+1+1 3 + + R R R

For Circuit (b), we have: 1 R R = = 1 1 1 1 1 1+1+1+1+1 5 + + + + R R R R R In general, for Circuit (c), we have: REQ =

1 R R = = 1 1 1 1 1+1+1+ +1 n + + + + R R R R We can conclude that for identical resistors connected in the parallel, the equivalent resistance is the value of one resistor divided by the number of resistors. REQ =

···

···

Problem 2–41. Show how the circuit in Figure P2–41 could be connected to achieve a resistance of 100 Ω, 200 Ω, 150 Ω, 50 Ω, 25 Ω, 33.3 Ω, and 133.3 Ω. For 100 Ω, we need a single 100-Ω resistor, which is a connection between terminals A and D. For 200 Ω, we need two 100-Ω resistors in series, which is a connection between terminals A and B. For 150 Ω, we need a 100-Ω resistor in series with a 50-Ω resistor, which is a connection between terminals A and C. For 50 Ω, we need a single 50-Ω resistor, which is a connection between terminals C and D. We can get 25 Ω by connecting the two 100-Ω resistors in parallel, which yields 50 Ω, and then connecting that result in parallel with the 50-Ω resistor, to get 25 Ω. The required combination is to connect the A, B, and C terminals together on one side and have the D terminal on the other. For 33.3 Ω, connect the 100 and 50-Ω resistors in parallel, which requires B and C to be connected on one side and the D terminal on the other. Finally, to get 133.3 Ω, connect a 100-Ω resistor in series with a parallel combination of a 100 and a 50-Ω resistor. This requires a connection to the A terminal on one side and the B and C terminals connected on the other. The following table summarizes the results. Resistance (Ω) 100 200 150 50 25 33.3 133.3


Terminal 1 A A A C A+B+C B+C A

Terminal 2 D B C D D D B+C

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Seventh Edition

Problem 2–42. In Figure P2–42 find the equivalent resistance between terminals A-B, A-C, A-D, B-C, B-D, and C-D. For A-B, ignore the 20-Ω resistor and the 10-Ω resistor connected to terminal D. We then have RAB = [100

 (60 + 40)] + 30 = [100  100] + 30 = 50 + 30 = 80 Ω

For A-C, ignore the 30-Ω resistor and the 10-Ω resistor connected to terminal D. We then have RAC = [60

 (100 + 40)] + 20 = [60  140] + 20 = 42 + 20 = 62 Ω

For A-D, ignore the 30-Ω resistor and the 20-Ω resistor. We then have RAD == [60

 (100 + 40)] + 10 = [60  140] + 10 = 42 + 10 = 52 Ω

For B-C, ignore the A terminal and the 10-Ω resistor. We then have RBC = 30 + [40 (100 + 60)] + 20 = 30 + [40



 160] + 20 = 30 + 32 + 20 = 82 Ω

For B-D, ignore the A terminal and the 20-Ω resistor. We then have RBD = 30 + [40 (100 + 60)] + 10 = 30 + [40



 160] + 10 = 30 + 32 + 10 = 72 Ω

For C-D, ignore the A terminal and the 30-Ω resistor. In the center of the circuit, the wire shorts out the 60, 100, and 40-Ω resistors, so we then have RCD = 20 + 0 + 10 = 30 Ω Problem 2–43. In Figure P2–43 find the equivalent resistance between terminals A-B, A-C, A-D, B-C, B-D, and C-D. For RAB , only the 33-kΩ resistor is active, so R AB = 33 kΩ. Similarly for RAC, only the 33-kΩ resistor is active, so R AC = 33 kΩ. For R AD , the two 100-kΩ resistors are in parallel and that result is in series with the 33-kΩ resistor, so R AD = 33+ (100 100) = 33 + 50 = 83 kΩ. For R BC , there is a path between the two terminals with no resistors, so R BC = 0 Ω. For R BD , ignore the 33-kΩ resistor, and the two 100-kΩ resistors are in parallel to give R BD = 100 100 = 50 kΩ. Similarly for R CD , ignore the 33-kΩ resistor, and the two 100-kΩ resistors are in parallel to give R CD = 100 100 = 50 kΩ.







Problem 2–44. Select a value of R L in Figure P2–44 so that R EQ = 6 kΩ. Repeat for REQ = 5 kΩ. Create an expression for R EQ in terms of R L and then solve for RL . Use the new expression to find the appropriate values for R L for the given values of R EQ. All resistance are in kilohms. REQ REQ (20 + RL ) 20REQ + REQ RL REQ RL (REQ

− 10R − 10)R

= 10

 (10 + R

L

) =

10(10 + RL ) 100 + 10RL = 10 + 10 + RL 20 + RL

= 100 + 10RL = 100 + 10RL

L

= 100

− 20R 100 − 20R 100 − 20R R − 10

EQ

L

=

EQ

RL

=

EQ

EQ

For R EQ = 6 kΩ, we have

− (20)(6) = −20 = 5kΩ −4 6 − 10 100 − (20)(5) 0 = = 5 − 10 −5 = 0kΩ

RL = For R EQ = 5 kΩ, we have RL


100

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Seventh Edition

Problem 2–45. Using no more than four 1-kΩ resistors, show how the following equivalent resistors can be constructed: 2 kΩ, 500 Ω, 1.5 kΩ, 333 Ω, 250 Ω, and 400 Ω. The following table presents the solutions. REQ (Ω) 2000 500 1500 333 250 400

Combination of 1-kΩ Resistors Two resistors in series: R + R Two resistors in parallel: R R One resistor in series with a parallel combination of two resistors: R + (R R) Three resistors in parallel: R R R Four resistors in parallel: R R R R Two resistors in series in parallel with two resistors in parallel: (R + R) R R





    

 

Problem 2–46. Do a source transformation at terminals A and B for each practical source in Figure P2–46. (a). After the transformation, we will have a voltage source in series with a resistor. The resistance will not change, so R = 100 Ω. Apply v S = i S R to find the voltage source vS = (0.005)(100) = 500 mV. Figure P2–46 (a) shows the results. (b). After the transformation, we will have a current source in parallel with a resistor. The resistance will not change, so R = 5 kΩ. Apply iS = vS /R to find the current source iS = 5/5000 = 1 mA. Figure P2–46 (b) shows the results.

(a)

(b) Figure P2–46

Problem 2–47. Find the equivalent practical voltage source at terminals A and B in Figure P2–47. A current source in series with a resistor is equivalent to just the current source, so we can remove the 5-Ω resistor without affecting the performance of the circuit between terminals A and B. That leaves a 5-A current source in parallel with a 10-Ω resistor. The current source and parallel resistor can be converted into a voltage source in series with the same resistor. The value for the voltage source follows Ohm’s Law, so v S = i S R = (5)(10) = 50 V. Figure P2–47 shows the resulting circuit.

Figure P2–47


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Seventh Edition

Problem 2–48. In Figure P2–48, the i-v characteristic of network N is v + 50i = 5 V. Find the equivalent practical current source for the network. When the circuit is open between nodes A and B, there is no current, i = 0 A, and the voltage must be v = 5 V in order to satisfy the i-v characteristic. When a short is placed between nodes A and B, the voltage is zero, v = 0 V, and the current is i = 100 mA in order to satisfy the i-v characteristic. The corresponding practical current source will have a current iS = 100 mA and a parallel resistance R = vS /iS = (5)/(0.1) = 50 Ω. Problem 2–49. Select the value of R x in Figure P2–49 so that R EQ = 75 kΩ. Combining the resistors from right to left, we can find the following expression for REQ , where all resistances are in kΩ. REQ REQ (32 + Rx )REQ 32REQ + REQ Rx (84

−R

EQ )Rx

Rx

= 15 + 47 + [22

 (R + 10)] x



22(Rx + 10) = 62 + 22 + Rx + 10





22Rx + 220 = 62 + 32 + Rx



= (32 + Rx )62 + 22Rx + 220 = 2204 + 84Rx = 2204 + 84Rx = 32REQ =

− 2204 − 2204

32REQ 84 REQ

−

For R EQ = 75 kΩ, we have Rx =

(32)(75) 2204 196 = = 21.78 kΩ 84 75 9

−

−

Problem 2–50. Two 10-kΩ potentiometers (a variable resistor whose value between the two ends is 10 kΩ and between one end and the wiper—the third terminal—can range from 0 Ω to 10 kΩ) are connected as shown in Figure P2–50. What is the range of R EQ ? At the limits of their settings, the two poteniometers are either in series or parallel. These represent the maximum and minimum equivalent resistances that the combination can take. When the poteniometers are arranged in parallel, the equivalent resistance is REQ = 10 10 = 5 kΩ. When the poteniometers are arranged in series, the equivalent resistance is REQ = 10 + 10 = 20 kΩ. The equivalent resistance ranges between 5 and 20 kΩ.



Problem 2–51. Select the value of R in Figure P2–51 so that RAB = RL . Find an expression for R AB in terms of R and R L . Set R AB equal to R L . Solve for R in terms of R L and choose the positive solution for the resistance. RAB (5R + RL )RAB (5R + RL )RL 5RRL + RL2


= R + [4R

 (R + R

L )]

= R +

4R(R + RL ) 5R + RL

= (5R + RL )R + 4R2 + 4RRL = 5R2 + RRL + 4R2 + 4RRL = 9R2 + 5RRL

R2

=

RL2 9

R

=

RL 3

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Seventh Edition

Problem 2–52. What is the range of R EQ in Figure P2–52? The poteniometer can range from 0 Ω to 10 kΩ and it is in parallel with a 10-kΩ resistor. That parallel combination is in series with a 5.6-kΩ resistor. When the poteniometer has a value of 0 Ω, it shorts out the 10-kΩ resistor, so only the 5.6-kΩ resistor is active and R EQ = 5.6 kΩ. When the poteniometer has a value of 10 kΩ, the parallel combination is 10 10 = 5 kΩ. That result is in series with the 5.6-kΩ resistor, so we have R EQ = 5.6 + 5 = 10.6 kΩ. REQ varies between 5.6 and 10.6 kΩ.



Problem 2–53. Find the equivalent resistance between terminals A and B in Figure P2–53. Place a voltage source, vS, between terminals A and B and redraw the circuit as the equivalent circuit shown in Figure P2–53. The figure is labeled with currents through and voltages across each of the resistors.

Figure P2–53

Using KVL, we can show that the voltage drop across each resistor is vS and it appears in the direction labeled in the figure. Since the resistors are all equal, the current through each resistor is vS /R. Applying KCL at the node above the voltage source, we have iS vS /R vS /R vS /R = 0, which implies iS = 3vS /R. The equivalent resistance is the ratio of v S to the current flowing into the circuit, which iS . Therefore, we have vS vS R REQ = = = 3v iS 3 S R

− −

−

−

−

−

−

Problem 2–54. Use voltage division in Figure P2–54 to find v x . Apply the equation for voltage division to get vx =



4 2+8+4



(12) = 3.4286V

Problem 2–55. Use voltage division in Figure P2–55 to obtain an expression for v L in terms of R, R L , and vS. The two right resistors are in parallel and the voltage v L appears across that combination. Combine the parallel resistors and then use voltage division to develop the expression for vL . REQ

=

R

R

L

vL

=



vL

=

RL vS R + 2RL

=



RRL R + RL

REQ vS R + REQ

=

 

RRL R + RL RRL R + R + RL

 

vS

=





RRL vS 2 R + RRL + RRL

Problem 2–56. Use current division in Figure P2–56 to find i x and v x .


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Seventh Edition

Combine the 500-Ω and the 1.5-kΩ resistors in series to get an equivalent resistance of 2 kΩ. Now apply current division as follows: 1 2

   

1 2

ix

=

vx

= (1500)(1.5) = 2250 V = 2.25kV

1 1 + 2 2

(3) =



(3) = 1.5 A

Problem 2–57. Use current division in Figure P2–57 to find an expression for vL in terms of R, RL , and iS . Combine the two right resistors in series to get an equivalent resistance REQ = R + R L . Apply the two-path current division rule to solve for the current through R L . iL =

R R R (iS ) = (iS ) = (iS ) R + REQ R + R + RL 2R + RL

Apply Ohm’s law to solve for the voltage v L vL = RL iL =

RRL iS 2R + RL

Problem 2–58. Find i x , i y , and i z in Figure P2–58. Combine the 20-Ω and 5-Ω resistors in parallel to get an equivalent resistance of 4 Ω. Combine that result with the 6-Ω resistor in series to get a total equivalent resistance of 10 Ω in the right branch. Apply the two-path current division rule to solve for i x and i z . ix

=

10 (200) = 80mA 15 + 10

iz

=

15 (200) = 120 mA 15 + 10

Apply the two-path current division rule again to solve for i y by dividing i z iy =

20 (120) = 96 mA 20 + 5

Problem 2–59. Find v O in the circuit of Figure P2–59. The circuit can be treated as a voltage source in series with three resistors, so voltage division applies. The output voltage, vO , appears across 75% of the 5-kΩ poteniometer or, equivalently, 3.75 kΩ. The 5-kΩ resistor and the remaining 25% of the potentiometer, or 1.25 kΩ, are the other two resistors in the circuit. Compute v O directly as follows: vO =

3.75 3.75 (5) = (5) = 1.875V 5 + 1.25 + 3.75 10

Problem 2–60. (A) The 1-kΩ load in Figure P2–60 needs 5 V across it to operate correctly. Where should the wiper on the potentiometer be set (RX ) to obtain the desired output voltage? Figure P2–60 shows an equivalent circuit with the poteniometer split into its two equivalent components. To solve the problem, find an equivalent resistance for the parallel combination of resistors and then apply


Page 2-31


Seventh Edition

Figure P2–60

voltage division to find an expression for R x . Solve for Rx and select the positive result. REQ 5V

= Rx

 1000

=

5 =

1 = 2 x

x

2 x

x

1000Rx 1000 + Rx

REQ (24V) 5000 Rx + REQ

−

 

1000Rx 1000 + Rx 1000Rx 5000 Rx + 1000 + Rx

5

×

−

 

(24) =

(5000

−

24000Rx Rx )(1000 + Rx ) + 1000Rx

4800Rx 106 + 4000Rx Rx2 + 1000Rx

6

= 4800Rx

6

= 0

−R + 5000R + 5 × 10 R − 200R − 5 × 10

=

Rx

=

Rx

= 2.338kΩ

−

−2138 or 2338Ω

Problem 2–61. Find the range of values of v O in Figure P2–61. If we combine the right resistor with the poteniometer in parallel, we can use voltage division to solve for vO . The poteniometer takes on values from 0 Ω to 1.5 kΩ. When the poteniometer is 0 Ω, the output is shorted out and the voltage is vO = 0 V. When the poteniometer is 1.5 kΩ, the parallel combination is 1500 1500 = 750 Ω. The output voltage is therefore



vO =

750 (50) = 21.4286V 1000 + 750

Problem 2–62. Use current division in the circuit of Figure P2–62 to find RX so that the voltage out is 3 V. If the output voltage is 3 V, then the current flowing through the right branch in the circuit is ix = v/R = 3/10 = 300 mA. Note that Rx is in series with the right 10 Ω resistor. Apply the two-path current


Page 2-32


Seventh Edition

division rule to solve for R x .

0.3 =

10 (1) 10 + Rx + 10

(20 + Rx )(0.3)

=

10

20 + Rx

=

33.33

Rx

=

13.33 Ω

Problem 2–63. (A) Figure P2–63 shows a voltage bridge circuit, that is, two voltage dividers in parallel with a source v S. One resistor R X is variable. The goal is often to “balance” the bridge by making v x = 0 V. Derive an expression for R X in terms of the other resistors when the bridge is balanced. Let the node between resistors R A and R B have a voltage v 1 and let the node between resistors R C and RX have a voltage v2 . The goal is to make v1 equal v2 so that vx is zero. Use voltage division to derive expressions for v1 and v 2 , set those expressions equal, and solve for RX .

v1

=

RB (vS ) RA + RB

v2

=

RX (vS ) RC + RX

RB vS RA + RB

=

RX v S RC + RX

RB (RC + RX ) RB RC + RB RX RX

= RX (RA + RB ) = RA RX + RB RX =

RB RC RA

Problem 2–64. (A) Ideally, a voltmeter has infinite internal resistance and can be placed across any device to read the voltage without affecting the result. A particular digital multimeter (DMM), a common laboratory tool, is connected across the circuit shown in Figure P2–64. The expected voltage was 10.2 V. However, the DMM reads 7.73 V. The large, but finite, internal resistance of the DMM was ”loading” the circuit and causing a wrong measurement to be made. Find the value of the internal resistance RM of this DMM. Apply voltage division to find the equivalent resistance of the parallel combination of the 10-MΩ resistor with the DMM.

7.73 V =


REQ (15V) 4.7 + REQ

7.73(4.7 + REQ )

=

15REQ

7.27REQ

=

36.331

REQ

=

4.99739 MΩ

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Seventh Edition

Now use the expression for a parallel combination of resistors to find the internal resistance of the DMM.

=

10RM 10 + RM

4.99739 =

10RM 10 + RM

REQ

4.99739(10 + RM )

=

10RM

5.00261RM

=

49.9739

RM

=

9.98955 MΩ

Problem 2–65. (D) Select values for R 1 , R 2 , and R 3 in Figure P2–65 so the voltage divider produces the two output voltages shown. There are many valid solutions to this problem. One approach is to constrain the resistor values so that the series combination has an equivalent resistance of R1 + R 2 + R 3 = 5 kΩ. Then the current will be i = v /REQ = 5/5000 = 1 mA. With a current of 1 mA, we must have R3 = 1 kΩ to get a voltage drop of 1 V. The second resistor, R2 , increases the voltage drop by 2.3 V, so we must have R2 = 2.3 kΩ. Finally, the resistors must sum to 5 kΩ, so R 1 = 1.7 kΩ.

Problem 2–66. (D) Select a value of R x in Figure P2–66 so that vL = 2 V. Combine the two right 1-kΩ resistors in parallel to get an equivalent resistance of 500 Ω. Voltage vL appears across the parallel combination, so apply voltage division to solve for R x .

vL = 2 V 2(1500 + Rx )

=

500 (12 V) 1000 + Rx + 500

= 6000

2Rx = Rx =

3000 1500 Ω = 1.5 kΩ

Problem 2–67. (D) Select a value of Rx in Figure P2–67 so that vL = 2 V. Repeat for 4 V and 6 V. Caution: Rx must be positive. First, combine R x in parallel with the 50-Ω resistor. Use voltage division with the equivalent resistance


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Seventh Edition

to find a general expression for R x in terms of v L and then substitute in the desired values for v L . REQ

=

50Rx 50 + Rx

vL

=

REQ (12) 100 + REQ

vL (100 + REQ )

= 12REQ

100vL =

(12

−v

L )REQ

100vL =

(12

−v

L)

100vL (50 + Rx )

= 50Rx (12

5000vL + 100vL Rx = (600

− 150v

L )Rx

=

Rx

=

600Rx

50Rx 50 + Rx

−v

L)

− 50v

L Rx

5000vL 5000vL 600 150vL

−

For vL = 2 V, we get Rx = 10000/300 = 33.3 Ω. For vL = 4 V, we get Rx = 20000/0 = Ω, which is an open circuit. For vL = 6 V, we get Rx = 30000/( 300) = 100 Ω, which is not possible, so there is no solution for vL = 6 V.

−

∞

−

Problem 2–68. Use circuit reduction to find v x and i x in Figure P2–68. Find vx by combining the 2.2-kΩ and 1-kΩ resistors in series to get REQ1 = 2.2 + 1 = 3.2 kΩ and then combine that result in parallel with the 3.3-kΩ resistor to get a total equivalent resistance of REQ2 = 3.3 3.2 = 1.6246 kΩ. The voltage v x appears across this equivalent resistance with a current of 300 mA, so we have vx = R EQ2 i = (1624.6)(0.300) = 487.4 V. To solve for ix , perform a source transformation to convert the current source in parallel with a 3.3-kΩ resistor into a voltage source v S = i S R = (0.300)(3300) = 990 V in series with a 3.3-kΩ resistor. Combine the resulting resistors in series and solve for ix using Ohm’s law, ix = v S /(3300 + 2200 + 1000) = 990/6500 = 152.3 mA.



Problem 2–69. Use circuit reduction to find v x , ix , and p x in Figure P2–69. Repeat using OrCAD. Find px first by finding an equivalent resistance for all of the resistors combined. To do so, collapse the circuit from right to left to develop the following expression: REQ

= 3.3

 {2.2 + [2  (1 + 1)]}

REQ

= 1.6246 kΩ

= 3.3

 {2.2 + [2  2]}

= 3.3

 {2.2 + 1}

= 3.3

 3.2

Solve for the power px = i 2 REQ = (0.5)2 (1624.6) = 406.15 W. To solve for vx , perform a source transformation on the left and combine the three resistors on the right to get the circuit shown in Figure P2–69(a). The source transformation yields vS = i S R = (0.500)(3300) = 1650 V in series with a 3.3 kΩ resistor. Use voltage division to calculate v x as follows vx =

2.2 (1650) = 558.46 V 3.3 + 2.2 + 1

To find i x , leave the resistors on the right intact and perform a source transformation as described above. Combine the 3.3-kΩ and 2.2-kΩ resistors in series to yield the circuit shown in Figure P2–69(b). Perform another source transformation to get a 300-mA current source in parallel with a 5.5-kΩ resistor as shown in Figure P2–69(c). Also in Figure P2–69(c), the two 1-kΩ resistors in series have been combined, since ix


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Seventh Edition

(a)

(b)

(c)

(d) Figure P2–69

flows through both of them. Now combine the 5.5-kΩ and 2-kΩ resistors in parallel to get the circuit in Figure P2–69(d). Apply the two-path current division rule to calculate i x as follows ix =

1.4667 (300 mA) = 126.92mA 1.4667 + 2

The following OrCAD circuit confirms the solution. In Figure P2–69(e), voltage vx appears across R1 and has a value vx = 812.3 253.8 = 558.5 V. Current i x flows through R 5 and has a value i x = 126.9 mA. The power p x = (0.5)(812.3) = 406.15 W.

−

(e) Figure P2–69


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Seventh Edition

Problem 2–70. Use circuit reduction to find v x and i x in Figure P2–70. Figure P2–70(a) displays a circuit that is electrically equivalent to the original circuit. Combine the two sets of R and 2R resistors in parallel to get the circuit in Figure P2–70(b), which retains vx . Apply voltage division to solve for vx as follows 2R 2 vS 3 vx = (vS ) = (vS ) = 2R 2R 6+2+2 5 2R + + 3 3

(a)

(b)

(c) Figure P2–70

To solve for i x , combine only the top resistors in parallel to get an equivalent resistance of R EQ1 = 2R/3 in series with the top 2R resistor and the source. Combine those two resistors in series to get an equivalent resistance of REQ2 = 8R/3 in series with the source. Perform a source transformation to get the circuit in Figure P2–70(c) with iS = 3vS /8R. Apply current division to solve for ix , noting that its direction is oppostive that of the source, so it will be negative. 1 3vS 4 3vS vS 2R ix = = = 3 1 1 8R 3+8+4 8R 10R + + 8R R 2R Problem 2–71. Use circuit reduction to find v x , ix , and p x in Figure P2–71. To find px , collapse the resistors working from right to left. The two 1-kΩ resistors are in parallel and combine to yield a 500-Ω resistor. That equivalent resistance is in series with the 1.5-kΩ resistor, which combine to yield a 2-kΩ resistor. That equivalent resistance is in parallel with the 3-kΩ resistor, which yields a 1.2-kΩ resistor. That resistor is in series with the 2-kΩ resistor, which yields a total equivalent resistance of 3.2 kΩ. The following expression summarizes these calculations, where all resistances are in kΩ.

− 

− 

−

REQ = 2 + 3

{  [1.5 + (1  1)]} = 2 + {3  [1.5 + 0.5]} = 2 + {3  2} = 2 + 1.2 = 3.2 kΩ


Page 2-37


Compute the power as px =

Seventh Edition

v2 (100)2 = = 3.125W REQ 3200

To find i x , perform a source transformation on the left side and combine the three resistors on the right side as described above. The resulting cicuit is shown in Figure P2–71(a). Apply current division to solve for i x

(a)

(b)

(c)

(d) Figure P2–71

as follows, noting the direction of i x is opposite that of the source 1 2 3 ix = ( 50mA) = ( 50 mA) = 1 1 1 3+2+3 + + 2 3 2

−

−

−12.5 mA

To find v x , start with the original circuit and perform a source transformation as above, but leave the three right resistors intact. After the source transformation, combine the 2-kΩ and 3-kΩ resistors in parallel to get the circuit in Figure P2–71(b). Perform another source transformation to get the circuit in Figure P2–71(c). Combine the 1.2-kΩ and 1.5-kΩ resistors in series and perform one final source transformation to get the circuit in Figure P2–71(d). Combine the three resistors in parallel to get a single equivalent resistance of 421.875 Ω, which still has vx across it. Compute the voltage by multiplying the current by the equivalent resistance to get v x = (0.02222)(421.875) = 9.375 V. Problem 2–72. Use circuit reduction to find v x and i x in Figure P2–72. To find ix , combine the two right resistors in series and perform a source transformation to get an equivalent circuit with a 1.3333-mA current source in parallel with a 18-kΩ and two 12-kΩ resistors. Apply current division to find ix as follows 1 12 ix = (1.3333 mA) = 500 µA 1 1 1 + + 18 12 12 In the reduced circuit, there are two 12-kΩ resistors in parallel, so they share the same voltage drop and current. Therefore, 500 µA flows through the 4-kΩ resistor. We can calculate its voltage directly as vx = (4000)(500 10 6 ) = 2 V.

×

−

Problem 2–73. Use source transformation to find i x in Figure P2–73. Perform the source transformation on the current source in parallel with the 220-Ω resistor to get the circuit in Figure P2–73(a). All of the elements in the circuit are in series, so we can combine the two voltage


Page 2-38


Seventh Edition

(a)

(b) Figure P2–73

sources and the two resistors to get the equivalent circuit in Figure P2–73(b). Throughout the process, we have not disrupted i x , so we can calculate ix = 37/370 = 100 mA. Problem 2–74. Select a value for Rx so that i x = 0 A in Figure P2–74. Perform a source transformation on both voltage sources to get the equivalent circuit shown in Figure P2– 74. Note the negative voltage source influences the direction of the current in the left transformation. In the

Figure P2–74

equivalent circuit, all three resistors are in parallel and, therefore, share the same voltage. If i x = 0 A, then the voltage drop across each resistor must be zero and no current flows through them. Therefore, all of the current from one source flows through the other source. Solve for R x as follows 24 Rx

= 0.4

Rx =

60 Ω

Problem 2–75. Use source transformations in Figure P2–75 to relate v O to v 1 , v 2 , and v 3 . Perform a source transformation on each voltage source to get the equivalent circuit shown in Figure P2– 75. The current sources are in parallel, so they combine by summing to give an equivalent current source of

Figure P2–75

iS = (v1 + v2 + v3 )/R. The three resistors are in parallel, so they combine to yield an equivalent resistance of R EQ = R/3. Apply Ohm’s law to compute the output voltage vO = i S REQ = (v1 + v2 + v3 )/3.


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Seventh Edition

Problem 2–76. The current through RL in Figure P2–76 is 100 mA. Use source transformations to find RL . Validate your answer using OrCAD. Perform a source transformation on the voltage source to get a 1-A current source in parallel with a 100-Ω resistor. Combine the resulting two 100-Ω resistors in parallel to get a 50-Ω resistor. Perform another source transformation to get a 50-V source in series with a 50-Ω resistor, which is also in series with the other 100-Ω resistor and RL . The 50-V source produces 100 mA through the circuit, so the equivalent resistance is REQ = 50/(0.1) = 500 Ω. The equivalent resistance is also the sum of the three resistors in series R EQ = 50 + 100 + RL , so we can solve for R L = 350 Ω. The OrCAD circuit in Figure P2–76 confirms the solution.

Figure P2–76

Problem 2–77. Select R x so that 50 V are across it in Figure P2–77. On the left side of the circuit, perform a source transformation to get a 400-mA current source in parallel with a 500-Ω resistor. Combine the 500-Ω resistor in parallel with the 1-kΩ resistor to get a 333.33-Ω resistor. Perform a second source transformation to get a 133.33-V source in series with the 333.33-Ω resistor. That resistor is in series with R x . On the right side of the circuit combine the 800-Ω and 200-Ω resistors in series to get a 1-kΩ resistor. That resistor is in parallel with the right 1-kΩ resistor, which combine to yield a 500-Ω resistor. Figure P2–77 displays the resulting circuit. Apply voltage division to solve for R x

Figure P2–77


Page 2-40


Seventh Edition

Rx (133 V) 333 + Rx + 500

50 V = 50(Rx + 833)

= 133Rx

83Rx =

41667

Rx =

500 Ω

Problem 2–78. The box in the circuit in Figure P2–78 is a resistor whose value can be anywhere between 8 kΩ and 80 kΩ. Use circuit reduction to find the range of values of v x . Perform a source transformation to get a 5-mA current source in parallel with a 10-kΩ resistor. In the resulting circuit, combine the two 10-kΩ resistors in parallel to get a 5-kΩ resistor in parallel with the current source. Perform another source transformation to get a 25-V voltage source in series with a 5-kΩ resistor, which are also in series with the variable resistor and the right 10-kΩ resistor. Apply voltage division, once with each extreme value of the variable resistor, to find the range of values for v x . vx,Min

=

10 (25) = 2.6316V 5 + 80 + 10

vx,Max

=

10 (25) = 10.8696V 5 + 8 + 10

Problem 2–79. A circuit is found to have the following element and connection equations: v1 = 24 V

−v + v + v = 0 −v + v + v = 0

v2 = 8000 i2

1

2

3

3

4

5

v3 = 5000 i3

i1 + i2 = 0

v4 = 4000 i4

−i + i + i = 0 −i + i = 0 2

v5 = 16000 i5

3

4

4

5

Use MATLAB to solve for all of the unknown voltages and currents associated with this circuit. Sketch one possible schematic that matches the given equations. There are many valid approaches to solve this problem using MATLAB. One way is to write the equations in matrix form and solve by inverting the matrix. Choose a vector of variables as x = [v1 v 2 v 3 v 4 v 5 i 1 i 2 i 3 i 4 i 5 ]T and write the equations in matrix form as follows. A x = B where

A =

    −  

1 0 0 0 0 1 0 0 0 0

0 1 0 0 0 1 0 0 0 0

0 0 1 0 0 1 1 0 0 0

−


0 0 0 1 0 0 1 0 0 0

0 0 0 0 1 0 1 0 0 0

0 0 0 0 0 0 0 1 0 0

0 8000 0 0 0 0 0 1 1 0

−

−

0 0 5000 0 0 0 0 0 1 0

−

0 0 0 4000 0 0 0 0 1 1

−

−

0 0 0 0 16000 0 0 0 0 1

−

     

B =

     

24 0 0 0 0 0 0 0 0 0

      Page 2-41


Seventh Edition

Solve for the unknown values by calculating x = A

1

−

B

The following MATLAB code provides the solution. A = [1 0 0 0 0 0 0 0 0 0; 0 1 0 0 0 0 -8000 0 0 0; 0 0 1 0 0 0 0 -5000 0 0; 0 0 0 1 0 0 0 0 -4000 0; 0 0 0 0 1 0 0 0 0 -16000; -1 1 1 0 0 0 0 0 0 0; 0 0 -1 1 1 0 0 0 0 0; 0 0 0 0 0 1 1 0 0 0; 0 0 0 0 0 0 -1 1 1 0; 0 0 0 0 0 0 0 0 -1 1]; B = [24 0 0 0 0 0 0 0 0 0]'; x = A \B

The corresponding MATLAB output is shown below. x = 24.0000 16.0000 8.0000 1.6000 6.4000 -0.0020 0.0020 0.0016 0.0004 0.0004

Figure P2–79 displays one possible circuit that corresponds to the given equations.

Figure P2–79

In summary, the voltages and currents are as follows


v1 = 24 V

i1 =

−2 mA

v2 = 16 V

i2 = 2mA

v3 = 8 V

i3 = 1.6 mA

v4 = 1.6 V

i4 = 0.4 mA

v5 = 6.4 V

i5 = 0.4 mA

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Seventh Edition

Problem 2–80. Consider the circuit of Figure P2–80. Use MATLAB to find all of the voltages and currents in the circuit and find the power provided by the source. Label the source as v 1 with current i 1 and the resistors from left to right as R 2 to R 7 with corresponding voltages and currents. Write the following element and connection equations by applying Ohm’s law, KVL, and KCL: v1 = 120 V

−v + v + v = 0 −v + v + v = 0 −v + v + v = 0

v2 = 150000 i2 v3 = 220000 i3

1

2

3

3

4

5

5

6

7

v4 = 68000 i4

i1 + i2 = 0

v5 = 56000 i5

−i + i + i = 0 −i + i + i = 0 −i + i = 0

v6 = 47000 i6 v7 = 33000 i7

2

3

4

4

5

6

6

7

Using a matrix approach, define a vector of variables as x = [v1 v 2 v 3 v 4 v 5 v 6 v 7 i 1 i 2 i 3 i 4 i 5 i 6 i 7 ]T and write the equations in matrix form as follows. A x = B The following MATLAB code provides the solution. A = [1 0 0 0 0 0 0 0 0 0 0 0 0 0; 0 1 0 0 0 0 0 0 -150000 0 0 0 0 0; 0 0 1 0 0 0 0 0 0 -220000 0 0 0 0; 0 0 0 1 0 0 0 0 0 0 -68000 0 0 0; 0 0 0 0 1 0 0 0 0 0 0 -56000 0 0; 0 0 0 0 0 1 0 0 0 0 0 0 -47000 0; 0 0 0 0 0 0 1 0 0 0 0 0 0 -33000; -1 1 1 0 0 0 0 0 0 0 0 0 0 0; 0 0 -1 1 1 0 0 0 0 0 0 0 0 0; 0 0 0 0 -1 1 1 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 1 1 0 0 0 0 0; 0 0 0 0 0 0 0 0 -1 1 1 0 0 0; 0 0 0 0 0 0 0 0 0 0 -1 1 1 0; 0 0 0 0 0 0 0 0 0 0 0 0 -1 1; ]; B = [120 0 0 0 0 0 0 0 0 0 0 0 0 0]'; x = A \B

The corresponding MATLAB output is shown below. x = 120.0000e+000 82.1192e+000 37.8808e+000 25.5188e+000 12.3620e+000 7.2627e+000 5.0993e+000 -547.4614e-006 547.4614e-006 172.1854e-006


Page 2-43


Seventh Edition

375.2759e-006 220.7506e-006 154.5254e-006 154.5254e-006

Problem 2–81. Consider the circuit of Figure P2–80 again. Use OrCAD to find all of the voltages, currents and power delivered or absorbed. Verify that the sum of all power in the circuit is zero. The OrCAD circuit in Figure P2–81 presents the solution. The power supplied by the voltage source is 65.70 mW and the power dissipated by the resistors is 44.96+6.523+9.577+2.729+1.122+0.788 = 65.70 mW, so the sum of all power in the circuit is zero.

−

Figure P2–81

Problem 2–82. The circuit of Figure P2–82 is called a “bridge-T ” circuit. Use OrCAD to find all of the voltages and currents in the circuit. The OrCAD circuit in Figure P2–82 presents the solution. Note that resistor R 3 has no current following through it.

Figure P2–82

Problem 2–83. Nonlinear Device Characteristics (A)


Page 2-44


Seventh Edition

The circuit in Figure P2–83 is a parallel combination of a 50-Ω linear resistor and a varistor whose i-v characteristic is i V = 2.6 10 5 v 3 . For a small voltage, the varistor current is quite small compared to the resistor current. For large voltages, the varistor dominates because its current increases more rapidly with voltage.

×

−

(a). Plot the i-v characteristic of the parallel combination. For a given voltage v , the current through the 50-Ω resistor is i1 = v/50 and the current through the varistor is i V = 2.6 10 5 v 3 . The total current is i = i 1 + iV . The following MATLAB code plots the i-v characteristic.

×

−

% Set the range of voltages to plot v = -200:1:200; % Compute the current through the resistor iR = v/50; % Compute the current through the varistor iV = 2.6e-5*v.ˆ3; % Sum the two path currents to get the total current iTotal = iR + iV; % Plot the i-v characteristic plot(v,iTotal, 'b','LineWidth' ,3) xlabel('Voltage (V)') ylabel('Current (A)') grid on

The corresponding MATLAB output is shown below.

250

200

150

100

50 ) A ( t n e r r u C

0

−50

−100

−150

−200

−250 −200

−150

−100

−50

0 Voltage (V)

50

100

150

200

(b). State whether the parallel combination is linear or nonlinear, active or passive, and bilateral or nonbilateral. The parallel combination is nonlinear based on the curved shape of the i-v characteristic. The combination is passive because the power is always positive, which means it is absorbing power. The combination is bilateral because the i-v charateristic has odd symmetry. (c). Find the range of voltages over which the resistor current is at least 10 times as large as the varistor current.


Page 2-45


Seventh Edition

Solve the following expression for a range of voltages. i1

≥ 10i

v 50

V

≥ 10

v



2.6

× 10

−

5 3

v

3

≥ 0.013v v ≤ 76.923 |v| ≤ 8.77 V



2

(d). Find the range of voltages over which the varistor current is at least 10 times as large as the resistor current. Solve the following expression for a range of voltages. iV 2.6

× 10

≥ 10i

5 3

−

v

1

≥ 10

v2

v 50

 

≥ 7692.3 |v| ≥ 87.7 V Problem 2–84. Transistor Biasing (D) The circuit shown in Figure P2–84 is a typical biasing arrangement for a BJT-type transistor. The actual transistor for this problem can be modeled as 0.7-V battery in series with a 200-kΩ resistor. Biasing allows signals that have both a positive and negative variation to be properly amplified by the transistor. Select the two biasing resistors RA and R B so that 3 V 0.1 V appears across R B . Label the voltage across the 220-kΩ resistor as vT . Write a KVL equation with resistor RB and the transitor to get

±

−3 + 0.7 + v

T =

0

Solve for vT = 2.3 V. The current through the 220-kΩ resistor is iT = 2.3/200000 = 11.5 µA. The voltage across RB is 3 V, which makes the voltage across RA = 15 3 = 12 V. As part of the design, choose the current through R B to be approximately equal to the transistor current of 11.5 µA. The required resistance is R B = 3/(11.5 10 6 ) = 260.87 kΩ. A standard resistor value that is close is 270 kΩ. With R B = 270 kΩ, the current through R B is i B = 3/(270000) = 11.11 µA. Applying KCL, the current through R A is the sum of the currents through RB and the transistor, which yields iA = 11.5 + 11.11 = 22.61 µA. The required resistance is RA = 12/(22.61 10 6 ) = 530.71 kΩ. Create RA by combining a 330-kΩ and two 100-kΩ resistors in series.

−

×

−

×

−

Problem 2–85. Center Tapped Voltage Divider (A) Figure P2–85 shows a voltage divider with the center tap connected to ground. Derive equations relating vA and v B to v S , R 1 , and R 2 . Using the passive sign convention and KCL, we have iS = iA = iB . Calculate the magnitude of the current by combining the resistors in series and using Ohm’s law.

−

iA =


vS R1 + R2

Page 2-46


Seventh Edition

Apply Ohm’s law to each resistor to find v A and v B vA = i A R1 =

R1 vS R1 + R2

vB = i B R2 =

−R v

2 S

R1 + R2

Problem 2–86. Active Transducer (A) Figure P2–86 shows an active transducer whose resistance R(V T ) varies with the transducer voltage V T 2 as R(V T ) = 0.5V T + 1. The transducer supplies a current to a 12-Ω load. At what voltage will the load current equal 100 mA? The resistors are in series, so REQ = R(V T ) + 12 Ω. Apply Ohm’s law to find an expression for iL in terms of V T and set i L = 100 mA. iL = 0.1 =

V T V T V T V T = = = 2 2 REQ R(V T ) + 12 0.5V T + 1 + 12 0.5V T + 13

2 (0.1)(0.5V T + 13) = V T 2 0.05V T + 1.3 = V T 2 V T

− 20V + 26 = 0 T

V T = 1.3977V or 18.6023V Problem 2–87. Interface Circuit Choice (E) You have a practical voltage source that can be modeled as a 5-V ideal source in series with a 1-kΩ source resistor. You need to use your source to drive a 1-kΩ load that requires exactly 2 V across it. Two solutions are provided to you as shown in Figure P2–87. Validate that both meet the requirement then select the b est solution and give the reason for your choice. Consider part count, standard parts, accuracy of meeting the specification, power consumed by the source, etc. In the circuit with Interface #1, combine the two right resistors in parallel to get an equivalent resistance of 666.7 Ω. Using voltage division, we can confirm that 2 V appears across the 1-kΩ load resistor. vL =

666.7 (5) = 2 V 1000 + 666.7

In the circuit with Interface #2, apply voltage division directly to confirm that 2 V appears across the 1-kΩ load resistor. 1000 vL = (5) = 2 V 1000 + 500 + 1000 Both circuits exactly meet the specification. In addition, both interface designs use a single, non-standard resistor value when considering resistors at the 10% tolerance level. With Interface #1, the source experiences an equivalent resistance of 1.667 kΩ and provides pS = v S2 /REQ = 25/1667 = 15 mW of power. With Interface #2, the source experiences an equivalent resistance of 5 kΩ and provides ps = 25/5000 = 5 mW. If we want to minimize the power provided by the source, Interface #2 is a better choice, since all other factors are equal.

±

Problem 2–88. Programmable Voltage Divider (A) Figure P2–88 shows a programmable voltage divider in which digital inputs b 0 and b 1 control complementary analog switches connecting a multitap voltage divider to the analog output vO . The switch positions in the figure apply when digital inputs are low. When inputs go high the switch positions reverse. Find the analog output voltage for (b1 , b0 ) = (0, 0), (0, 1), (1, 0), and (1, 1) when v REF = 12 V. There are four equal resistors in series with a voltage source, so each drops one quarter of the total voltage, or 3 V in this case. As we cycle through the four combinations of the digital inputs, the switchs connect the output voltage to be across zero, one, two, or three resistors, in that order. The output voltages are therefore 0 V, 3 V, 6 V, and 9 V. The following table summarizes the results.


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Seventh Edition

b1 0 0 1 1

b0 0 1 0 1

vO (V) 0 3 6 9

Problem 2–89. Analog Voltmeter Design (A, D, E) Figure P2–89(a) shows a voltmeter circuit consisting of a D’Arsonval meter, two series resistors, and a two-position selector switch. A current of iFS = 400 µA produces full-scale deflection of the D’Arsonval meter, whose internal resistance is R M = 25 Ω. (a). (D) Select the series resistances R1 and R 2 so a voltage v x = 100 V produces full-scale deflection when the switch is in position A, and voltage v x = 10 V produces full-scale deflection when the switch is in position B. First, solve for R2 such that a 10-V input at position B causes 400 µA to flow through the two resistors. R2 + RM =

v 10 = i 400 10

×

6

−

= 25kΩ

Solving for R2 , we get R2 = 25000 25 = 24.975 kΩ. Now solve for R1 such that a 100-V input at position A causes 400 µA to flow through all three resistors.

−

R1 + R2 + RM =

v 100 = i 400 10

×

−

6

= 250kΩ

R1 = 225kΩ

(b). (A) What is the voltage across the 20-kΩ resistor in Figure P2–89(b)? What is the voltage when the voltmeter in part (a) is set to position A and connected across the 20-kΩ resistor? What is the percentage error introduced connecting the voltmeter? Using voltage division, the voltage across the 20-kΩ resistor is 20 V when the voltmeter is not connected. When the voltmeter is set in position A and connected in parallel to the 20-kΩ resistor, it is equivalent to placing a 250-kΩ resistor in parallel with the 20-kΩ resistor. The equivalent resistance of the parallel combination is REQ = 20 250 = 18.5185 kΩ. Applying voltage division to this case yields the following result: 18.5185 vM = (50) = 19.084V 30 + 18.5185



The percentage error in this case is 4.58%. (c). (E) A different D’Arsonval meter is available with an internal resistance of 100 Ω and a full-scale deflection current of 100 µA. If the voltmeter in part (a) is redesigned using this D’Arsonval meter, would the error found in part (b) be smaller or larger? Explain. With a full-scale deflection current of 100 µAfor an applied voltage of 100 V, (switch in position A,) the total resistance of the meter must be 1 MΩ. The increased meter resistance will draw less current when it is connected to the 20-kΩ resistor and have a smaller impact on the voltage. The error will decrease. The new equivalent resistance of the meter in parallel with the 20-kΩ resistor is REQ = 19.6078 kΩ, the measured voltage is 19.7628 V, and the error is 1.19%. Problem 2–90. MATLAB Function for Parallel Equivalent Resistors (A) Create a MATLAB function to compute the equivalent resistance of a set of resistors connected in parallel. The function has a single input, which is a vector containing the values of all of the resistors in parallel, and it has a single output, which is the equivalent resistance. Name the function “EQparallel” and test it with at least three different resistor combinations. At least one test should have three or more resistor values.


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the analysis of linear circuits and design ch02

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