Problems Section 3-2 Kirchhoff’s Laws
P3.2-1
Apply KCL at node a to get
2 + 1 = i + 4 ⇒ i = -1 A
The current and voltage of element B adhere to the passive convention so (12)(-1) = -12 W is power received by element B. The power supplied by element B is 12 W. Apply KVL to the loop consisting of elements D, F , E , and C to get 4 + v + (-5) – 12 = 0 ⇒ v = 13 V The current and voltage of element F do not adhere to the passive convention so (13)(1) = 13 W is the power supplied by element F . Check: The sum of the power supplied by all branches is -(2)(-12) + 12 – (4)(12) + (1)(4) + 13 – (-1)(-5) = 24 +12 – 48 + 4 +13 –5 = 0
P3.2-2
Apply KCL at node a to get
2 = i2 + 6 = 0 ⇒ i2 = −4 A
3 = i4 + 6 ⇒ i4 = -3 A
Apply KCL at node b to get
Apply KVL to the loop consisting of elements A and B to get -v2 – 6 = 0 ⇒ v2 = -6 V Apply KVL to the loop consisting of elements C , D, and A to get -v3 – (-2) – 6 = 0 ⇒ v4 = -4 V Apply KVL to the loop consisting of elements E , F and D to get 4 – v6 + (-2) = 0 ⇒ v6 = 2 V Check: The sum of the power supplied by all branches is
−(6)(2) – (-6)(-4) – (-4)(6) + (-2)(-3) + (4)(3) + (2)(-3) = -12 - 24 + 24 + 6 + 12 – 6 = 0 P3.2-3
KVL : −12 − R 2 (3) + v = 0 (outside loop) v = 12 + 3R 2 or R 2 =
KCL
i+
12 R1
(c)
12 R1
or R1 =
12 3−i
v = 12 + 3 ( 3) = 21 V i = 3−
(b)
3
− 3 = 0 (top node)
i = 3−
(a)
v − 12
R 2 =
12 6
=1 A
2 − 12 3
=−
10 3
Ω ; R1 =
12 3 − 1.5
=8Ω
(checked using LNAP 8/16/02) 24 = − 12 i, because 12 and i adhere to the passive convention.
∴ i = − 2 A and and R1 = 9 = 3v,
∴ v = 3V
12
= 2.4 2.4 Ω 3+ 2 becaus becausee 3 and v do not adhere adhere to the passiv passivee conventi convention on
and R 2 =
3 − 12
= −3 Ω 3 The situations described in (b) and (c) cannot occur if R1 and R2 are required to be nonnegative.
P3.2-4
12 i = =2A 1 6 20 i = = 5A 2 4 i = 3− i = − 2 A 3 2 i = i +i = 3A 4 2 3
Powe Powerr abso absorb rbed ed by the the 4 Ω resi resist stor or = 4 ⋅ i 2 = 100 100 W 2 Powe Powerr abso absorb rbed ed by the the 6 Ω resi resist stor or = 6 ⋅ i 2 = 24 W 1 Powe Powerr abso absorb rbeed by the the 8 Ω resi resist stor or = 8 ⋅ i 2 = 72 W 4
(checked using LNAP 8/16/02)
P3.2-5
v1 = 8 V v2 = −8 + 8 +12 = 12 V v3 = 2⋅ 4 = 8 V v32
4Ω : P =
4
v22
6Ω : P =
6 2
v1
8Ω : P =
(checked using LNAP 8/16/02)
= 16 W
8
= 24 W
=8 W
P3.2-6
P2 mA = − ⎡3 × 2 ×10
⎣
(
−3
) ⎤⎦ = −6 ×10 −
3
= −6 mW
P1 mA = − ⎡ −7 × 1 ×10−3 ⎤ = 7 ×10 10 −3 = 7 mW
⎣
(
)⎦
(checked using LNAP 8/16/02)
P3.2-7
P2 V = + ⎡ 2 × 1×10 −3 ⎤ = 2 × 10 10 −3 = 2 mW P3 V
)⎦ ⎣ ( = + ⎡⎣3 × ( −2 ×10−3 ) ⎤⎦ = −6 ×10 −3 = −6 mW (checked using LNAP 8/16/02)
P3.2-8
KCL: i R = 2 +1 ⇒ iR = 3 A KVL: v R + 0 − 12 = 0 ⇒ vR = 12 V
∴ R =
v R i R
=
12 3
=4Ω
(checked using LNAP 8/16/02)
P3.2-9
KVL: v R + 56 + 24 = 0 ⇒ vR = −80 V KCL: i R + 8 = 0 ⇒ iR = −8 A
∴ R =
v R i R
=
−80 = 10 Ω −8
(checked using LNAP 8/16/02)
P3.2-10
KCL at node b:
5.61 7
=
3.71 − 5.61 12 − 5.61
+
R1
5
⇒ 0.801 = ⇒ R1 =
KCL at node a:
3.71 2
+
3.71 − 5. 5.61 4
+
3.71 −12 R2
−1.9 R1
+1.278
1.9 1.278 − 0.801
= 3.983 ≈ 4 Ω
= 0 ⇒ 1.855 + ( −0.475 ) + ⇒ R 2 =
8.29 1.855 − 0.475
−8.29 R2
=0
= 6.007 ≈ 6 Ω
(checked using LNAP 8/16/02)
P3.2-11
The subscripts suggest a numbering of the sources. Apply KVL to get v1 = v 2 + v 5 + v 9 − v 6 i1 and v1 do not adhere to the passive convention, so
(
p1 = i1 v1 = i1 v 2 + v 5 + v 9 − v 6
is the power supplied by source 1. Next, apply KCL to get
(
i 2 = − i1 + i 4 i 2 and v 2 do not adhere to the passive convention, so
)
)
(
)
p2 = i 2 v 2 = − i1 + i4 v2
is the power supplied by source 2. Next, apply KVL to get
(
v3 = v6 − v5 + v9
)
i 3 and v 3 adhere to the passive convention, so
(
(
p3 = − i3 v3 = − i3 v6 − v5 + v9
))
is the power supplied by source 3. Next, apply KVL to get v 4 = v 2 + v5 + v8 i 4 and v 4 do not adhere to the passive convention, so
(
p 4 = i4 v 4 = i 4 v 2 + v 5 + v 8
)
is the power supplied by source 4. Next, apply KCL to get
( (
i 5 = i 3 − i 2 = i 3 − − i1 + i 4
)) = i
1
+i 3 +i 4
i 5 and v 5 adhere to the passive convention, so
(
)
p5 = − i5 v5 = − i1 + i3 + i 4 v 5
is the power supplied by source 5. Next, apply KCL to get
(
i 6 = i 7 − i1 + i 3
)
i 6 and v 6 adhere to the passive convention, so
(
(
p6 = − i6 v6 = − i7 − i1 + i3
is the power supplied by source 6. Next, apply KVL to get v 7 = −v 6 i 7 and v 7 adhere to the passive convention, so
)) v
6
(
)
p7 = − i7 v7 = − i7 − v6 = i7 v 6
is the power supplied by source 7. Next, apply KCL to get i8 = −i 4 i 8 and v 8 do not adhere to the passive convention, so
(
)
p8 = i8 v8 = − i 4 v8 = − i 4 v8
is the power supplied by source 8. Finally, apply KCL to get i 9 = i1 + i 3 i 9 and v 9 adhere to the passive convention, so
(
)
p9 = − i9 v9 = − i1 + i3 v 9
is the power supplied by source 9. 9
(Check:
∑ p
n
= 0 .)
n =1
P3.2-12
The subscripts suggest a numbering of the circuit elements. Apply KCL to get i 2 + 0. 2 + 0 . 3 = 0
⇒ i 2 = −0.5 A
The power received by the 6 Ω resistor is 2
p 2 = 6 i 2 = 6 ( −0.5 ) = 1.5 W 2
Next, apply KCL to get i 5 = 0.2 + 0.3 + 0.5 = 1.0 A
The power received by the 8 Ω resistor is 2
p 5 = 8 i 5 = 8 (1) = 8 W 2
Next, apply KVL to get v 7 = 15 V
The power received by the 20 Ω resistor is p 7 =
v7
2
20
=
15 2 20
= 11.25 W
is the power supplied by source 7. Finally, apply KCL to get i 9 = 0.2 + 0.5 = 0.7 A
The power received by the 5 Ω resistor is 2
p 9 = 5 i 9 2 = 5 ( 0.7 ) = 2.45 W P3.2-13
We can label the circuit as follows:
The subscripts suggest a numbering of the circuit elements. Apply KCL at node b to get i 4 + 0.25 + 0.75 = 0
⇒ i 4 = −1.0 A
Next, apply KCL at node d to get i 3 = i 4 + 0.25 = −1.0 + 0.25 = −0.75 A
Next, apply KVL to the loop consisting of the voltage source and the 60 Ω resistor to get v 2 − 15 = 0
⇒ v 2 = 15 V
Apply Ohm’s law to each of the resistors to get i2 =
v2
60
=
15 60
= 0.25 A ,
v 3 = 10 i 3 = 10 ( −0.75 ) = −7.5 V
and v 4 = 20 i 4 = 20 ( −1) = −20 V
Next, apply KCL at node c to get i1 + i 2 = i 3
⇒ i1 = i 3 − i 2 = −0.75 − 0.25 = −1.0 A
Next, apply KVL to the loop consisting of the 0.75 A current source and three resistors to get v6 − v4 − v3 − v2 = 0
⇒ v 6 = v 4 + v 3 + v 2 = −20 + ( −7.5) + 15 = −12.5 V
Finally, apply KVL to the loop consisting of the 0.25 A current source and the 20 Ω resistor to get v 5 + v 4 = 0 ⇒ v 5 = −v 4 = − ( − 20 ) = 20 V (Checked: LNAPDC 8/28/04)
P3.2-14
We can label the circuit as follows: The subscripts suggest a numbering of the circuit elements. Apply KCL at node b to get i1 + 1.5 = 0
⇒ i1 = −1.5 A
Apply KCL at node d to get i 5 + 0.5 = 1.5
⇒ i 5 = 1.0 A
Apply KCL at node f to get i 8 + 0.5 = 0
⇒ i 8 = −0.5 A
Apply Ohm’s law to each of the 10 Ω resistors to get
v 1 = 10 i1 = 10 ( −1.5 ) = −15 V , v 5 = 10 i 5 = 10 (1) = 10 V and v 8 = 10 i 8 = 10 ( −0.5 ) = −5 V
Apply KVL to the loop consisting of the voltage sources and the 25 Ω resistor to get
−5 + 15 + v 4 = 0 ⇒ v 4 = −10 V Apply Ohm’s law to the 25 Ω resistor to get i4 =
v4
25
=
−10 25
= −0.4 A
Apply KCL at node a to get i1 + i 2 = i 4
⇒ i 2 = i 4 − i1 = −0.4 − ( −1.5) = 1.1 A
Apply KCL at node e to get i6 + i8 = i 4
⇒ i 6 = i 4 − i 8 = −0.4 − ( −0.5) = 0.1 A
Apply KVL to the loop consisting of the 1.5 A current source, the 5 V voltage source and two 10 Ω resistors to get v1 + v 3 − v 5 + 5 = 0 ⇒
v 3 = − 5 + v 5 − v1 = − 5+ 10− (− 15) = 20 V
Finally, apply KVL to the loop consisting of the 0.5 A current source, the 15 V voltage source and two 10 Ω resistors to get v 7 + v 8 − 15 + v 5 = 0 ⇒
(
)
v 7 = 15 − v 5 + v 8 = 15 − (10 + ( − 5) ) = 10 V
(Checked: LNAPDC 8/28/04)
P3.2-15
We can label the circuit as shown. The subscripts suggest a numbering of the circuit elements. Apply KVL to node the left mesh to get 15 i1 + 25 i1 − 20 = 0 ⇒
i1 =
20 40
= 0.5 A
Apply KVL to node the left mesh to get
v 2 − 25 i1 = 0 ⇒
v 2 = 25 i1 = 25 ( 0.5) = 12.5 V
Apply KCL to get i m = i 2 . Finally, apply Ohm’s law to the 50 Ω resistor to get im = i2 =
v2
50
=
12.5 50
= 0.25 A (Checked: LNAPDC 9/1/04)
P3.2-16
We can label the circuit as shown. The subscripts suggest a numbering of the circuit elements. Ohm’s law to the 8 Ω resistor to get v1 i1 = 8 Apply KCL at the top node of the CCCS to get i1 + 0.25 v 1 = i 2
⇒ i 2 = i1 + 0.25 v 1 =
v1
8
+ 0.25 v 1 = 0.375 v1
Ohm’s law to the 12 Ω resistor to get
(
)
v 2 = 12 i 2 = 12 0.375 v 1 = 4.5 v1
Apply KVL to the outside loop to get v1 + v 2 − 20 = 0
⇒ v 1 + 4.5 v1 = 20 ⇒ v1 =
20 5.5
= 3.636 V
Apply KCL to get i m = i 2 . Finally, i m = i 2 = 0.375 v1 = 0.375 ( 3.636) = 1.634 A
(Checked: LNAPDC 9/1/04)
P3.2-17
We can label the circuit as shown. The subscripts suggest a numbering num bering of the circuit elements. Ohm’s law to the 48 Ω resistor to get v1 = 48 i 1 Apply KCL at the top node of the CCCS to get i1 + 5 i 1 = i 2
⇒ i 2 = 6 i1
Ohm’s law to the 4 Ω resistor to get
( )
v m = 4 i 2 = 4 6 i 1 = 24 i1
Apply KVL to the outside loop to get v1 + v m − 24 = 0 ⇒
48 i1 + 24 24 i 1 = 24 ⇒
i1 =
24 72
=
1 3
A
Finally,
⎛1⎞ v m = 24 i1 = 24 ⎜ ⎟ = 8 V ⎝ 3⎠ (Checked: LNAPDC 9/1/04)
P3.2-18
We can label the circuit as shown. The subscripts suggest a numbering num bering of the circuit elements. Apply KCL at the top node of the current source to get i1 = i 2 + 0.25
Apply Ohm’s law to the resistors to get
(
)
v1 = 20 i1 and v 2 = 60 i 2 = 60 i 1 − 0.25 = 60 i 1 − 15
Apply KVL to the outside to get
v 2 + 80 i1 + v 1 = 0
⇒
( 60 i
1
− 15) + 80 i 1 + 20 i 1 = 0 ⇒ i 1 =
15 160
= 0.09375 A
Finally, v m = 80 i1 = 80 ( 0.09375) = 7.5 V
(Checked: LNAPDC 9/1/04) P3.2-19
i=
R1 =
4.8 12
= 0.4 A and v =
12 − 7.2 0.4
3.6 0.5
= 12 Ω and R 2 =
= 7.2 V 7.2
0 . 4 + 0. 5
=8 Ω
(Checked: LNAPDC 9/28/04)
P3.2-20
Apply KCL at node a to determine the current in the horizontal resistor as shown. Apply KVL to the loop consisting of the voltages source and the two resistors to get -4(2-i) + 4(i) - 24 = 0 ⇒ i = 4 A P3.2-21
−18 + 0 − 12 − va = 0 ⇒ va = − 30 V and im =
2 5
va + 3 ⇒
im = 9 A
P3.2-22
−va − 10 + 4va − 8 = 0 ⇒ va =
18 3
= 6 V and vm = 4 va = 24 V
Section 3-3 Series Resistors and Voltage Division P3.3-1
6 6 v = 12 = 12 = 4 V 1 6+ 3+5+ 4 18 3 5 10 v = 12 = 2 V ; v = 12 = V 2 18 3 18 3 4 8 v = 12 = V 4 18 3
(checked using LNAP 8/16/02)
P3.3-2
(a ) R = 6 + 3 + 2 + 4 = 15 Ω (b) i =
(c)
28
28
= 1.867 A 15 p = 28 ⋅ i =28( =28(1. 1.86 867) 7)=5 =52. 2.27 27 W R
=
(28 V and i do not adhere to the passive convention.)
(checked using LNAP 8/16/02)
P3.3-3 i R2 = v = 8 V
12 = i R1 + v = i R1 + 8
⇒ 4 = i R1
(a) i = (b) i =
8 R 2
4 R1
= =
8 100 4 100
; R1 =
4
; R2 =
8
=
i i
=
( c ) 1.2 = 12 i ⇒ i = 0.1 A ;
4 ⋅ 100 8 8 ⋅ 100 4
R1 =
4 i
= 50 Ω = 200 Ω
= 40 Ω; R2 =
8 i
= 80 Ω (checked using LNAP 8/16/02)
P3.3-4
Voltage division v1 =
16
12 = 8 V 16 + 8 4 v3 = 12 = 4 V 4+8 KVL: v3 − v − v1 = 0 v = −4 V
(checked using LNAP 8/16/02)
P3.3-5
⎛v ⎞ s ⎜ ⇒ R = 50 − 1⎟ ⎜v ⎟ ⎝ o ⎠ with v = 20 V and v > 9 V, R < 61.1 Ω ⎫ 0 ⎪ s ⎬ R = 60 Ω with v = 28 V and v < 13 V, R > 57.7 Ω ⎪ 0 s ⎭
⎛ 100 ⎞ using voltage divider: v = ⎜ v 0 ⎝ 100 + 2 R ⎟⎠ s
P3.3-6
⎛ 240 ⎞ ⎟ 18 = 12 V 1 2 0 2 4 0 + ⎝ ⎠ 18 ⎛ ⎞ b.) 18 ⎜ ⎟ = 0.9 W + 1 2 0 2 4 0 ⎝ ⎠ ⎛ R ⎞ c.) ⎜ ⎟ 18 = 2 ⇒ 18 R= 2 R+ 2 (120 ) ⇒ ⎝ R + 120 ⎠ a.) ⎜
d.) 0.2 =
R R + 120
R= 15 Ω
⇒ ( 0.2 ) (120 ) = 0.8 R ⇒ R = 30 Ω (checked using LNAP 8/16/02)
P3.3-7 All of the elements are connected in series.
Replace the series voltage sources with a single equivalent voltage having voltage 12 + 20 – 18 = 14 V. Replace the series 15 Ω, 5 Ω and 20 Ω resistors by a single equivalent resistance of 15 + 5 + 20 = 40 Ω.
By voltage division 14 ⎛ 10 ⎞ ⎟14 = = 2.8 V 40 ⎠ 5 ⎝ 10 + 40
v=⎜
(checked: LNAP 6/9/04)
P3.3-8
Use voltage division to get
⎛ 10 ⎞ ⎟ (120 ) = 20 V 1 0 5 0 + ⎝ ⎠
va = ⎜
Then i a = 0.2 ( 20 ) = 4 A
The power supplied by the dependent source is given by p = (120 ) i a = 480 W
(checked: LNAP 6/21/04)
P3.3-9 (a) Use voltage division to get vm =
aR p
(1 − a ) R p + R p
v s = av s
therefore
⎛ vs ⎞ ⎟ θ ⎝ 360 ⎠
vm = ⎜
So the input is proportional to the input.
⎛1⎞ ⎟ θ . When 15 ⎝ ⎠
(b) When vs = 24 V then v m = ⎜ θ
θ
= 45o then vm = 3 V. When vm = 10 V then
= 150o . (checked: LNAP 6/12/04)
P3.3-10 Replace the (ideal) voltmeter with the equivalent open circuit. Label the voltage measured by the meter. Label some other element voltages and currents.
Apply KVL the left mesh to get 8 i a + 4 i a − 24 = 0
Use voltage division to get
⇒ ia = 2 A
vm =
5 5+3
4ia =
5 5+3
4 ( 2) = 5 V
(checked using LNAP 9/11/04) P3.3-11
⎛ 3 ⎞ ⎟= 3V ⎝ 3+9 ⎠
From voltage division v3 = 12 ⎜
then i =
v3 =1A 3
The power absorbed by the resistors is: (12 ) ( 6 ) + (12 ) ( 3) + (12 ) ( 3) = 12 W The power supplied by the source is (12)(1) = 12 W.
P3.3-12 P = 6 W and R1 = 6 Ω i2 =
P R1
=
6 6
= 1 or i =1 A
v0 = i R1 =(1) (6)=6V
from KVL: − v
s
+ i (2 + 4 + 6 + 2) = 0
⇒ v = 14 i = 14 V s
Section 3-4 Parallel Resistors and Current Division P3.4-1
1 6
1 1 i = 4= 4= A 1 1 + 1 + 1 +1 1+ 2 + 3 + 6 3 6 3 2 1 1 2 3 i = 4 = A; 2 1 + 1 + 1 +1 3 6 3 2 1 1 2 i = 4 =1 A 3 1 + 1 + 1 +1 6 3 2 1 1 i = 4=2A 4 1 + 1 + 1 +1 6 3 2
P3.4-2
(a) (b) (c)
1 R
=
1
+
1
+
1
6 12 4 v = 6 ⋅ 2 = 12 V
=
1 2
⇒ R = 2Ω
p = 6 ⋅12 = 72 W
P3.4-3
i=
8 R1
or R1 =
8 i
8 = R2 (2 − )i ⇒ i= 2 −
(a) i = 2 − (b) i =
8 12
8 12
=
2 3
=
4 3
8 =6Ω 4 3 8 =6Ω 2
A ; R1 =
A ; R2 =
2−
3
8 R 2
or
R2 =
8 2−i
1 will caus causee i= 2 = 1 A. The The curr curren entt in both both R1 and and R 2 will will be 1 A. R1 = R 2 will 2 R1 R 2 1 2 ⋅ = 8 ; R1 = R2 ⇒ 2 ⋅ R1 = 8 ⇒ R1 = 8 ∴ R1 = R2 = 8 Ω 2 R1 + R 2
(c)
P3.4-4
Current division: 8 i = ( −6 ) = −2 A 1 16 + 8 8 i = ( −6 ) = −3 A 2 8+8 i = i −i 1 2
= +1 A
P3.4-5
⎛ R ⎞ 1 ⎟ i and current division: i = ⎜ 2 ⎜ R + R ⎟ s 2⎠ ⎝ 1 Ohm's Law: v = i R yields o 2 2
⎛ v ⎞⎛ R + R ⎞ 2⎟ i = ⎜ o ⎟⎜ 1 s ⎜ R ⎟⎜ R ⎟ 1 ⎠ ⎝ 2 ⎠⎝ plug pluggi ging ng in R = 4Ω, v> 9 V giv gives
i > 3.15 3.15 A o s 1 and R= 6Ω, v < 13 V gives i < 3.47 A 1 o s So any 3.15 A < i < 3.47 A keeps 9 V < v < 13 V. s o
P3.4-6
⎛ 24 ⎞ ⎟ 1.8 = 1.2 A ⎝ 12 + 24 ⎠ ⎛ R ⎞ b) ⎜ 1 .6 ⇒ 2 R = 1.6 R + 1.6 (12 ) ⇒ R = 48 Ω ⎟ 2 = 1. R + 12 ⎝ ⎠ a) ⎜
c ) 0.4 =
R R + 12
⇒ ( 0.4 ) (12 ) = 0.6 R ⇒ R = 8 Ω
P3.4-7
(a) To insure that ib is negligible we require i1 =
15 R1 + R 2
10 ×10 −6 ) = 10 −3 ≥ 10 (10
So R1 + R 2 ≤ 150 k Ω
To insure that the total power absorbed by R1 and R2 is no more than 5 mW we require 2
15
R1 + R 2
≤ 5 ×10−3
⇒
R1 + R 2 ≥ 45 k Ω
Next to cause vb = 5 V we require 5 = vb =
R 2 R1 + R 2
15
⇒
R1 = 2 R 2
For example, R1 = 40 kΩ, R 2 = 80 kΩ , satisfy all three requirements.
(b)
KVL gives
(80 ×10 ) i
+ v b − 15 = 0
3
1
KCL gives vb
i1 =
40 ×10
3
+ 15 ×10−6
Therefore
⎛
vb
(80 ×10 ) ⎜ 40 ×10 3
⎝
3
⎞ + 15 ×10 −6 ⎟ + v b = 15 ⎠
Finally 3v b + 1.2 = 15
⇒
vb =
13.8 3
= 4. 6 V
P3.4-8
All of the elements of this circuit are connected in parallel. Replace the parallel current sources by a single equivalent 2 – 0.5 + 1.5 = 3 A current source. Replace the parallel 12 Ω and 6 Ω 12 × 6 resistors by a single = 4 Ω resistor. 12 + 6
By current division 12 ⎛ 4 ⎞ 3 = = 1.714 A ⎟ 7 ⎝ 3+ 4⎠
i=⎜
(checked: LNAP 6/9/04) P3.4-9
Each of the resistors is connected between nodes a and b. The resistors are connected in parallel and the circuit can be redrawn like this:
Then
40 20 40 = 10 Ω
So v = 10 ( 0.003) = 0.03 = 30 mV
(checked: LNAP 6/21/04)
P3.4-10
2
R L =
5 × 10
−3
=
0.025
= 80 Ω
12 + R L
( 30 ×10 − ) 3
(
R1 + 12 + R L
)
so 1 6
=
92 R1 + 92
⇒ R1 = 410 Ω (checked: LNAP 6/21/04)
P3.4-11
Use current division to get ia = −
75
30 × 10− ) = −22.5 mA ( 25 + 75 3
So v b = 50 −22.5 × 10−
(
3
) = −1.125 V
The power supplied by the dependent source is given by −3 p = − 30 × 10
(
) ( −1.125) = 33.75 mW (checked: LNAP 6/12/04)
P3.4-12
(a) Using current division
⎛ 30 ⎞ =⎜ ⎟1 R ⎝ R + 30 ⎠
20
⇒
20 ( R + 30 ) =
(R30 )
(b) The power supplied by the current source is p= iv= (1) ⎡⎣ (1) (10) + 20⎤⎦ = 30 W P3.4-13
Using voltage division
⇒
= 60 Ω R
R1
8= R1 +
40 R 2
× 24
⇒
1 3
=
R 2 + 40
(
R1 R 2 + 40
(
R1R 2 + 40 R1 + R 2
(
⇒
) )
)
R1 R2 + 40 R1 + R2 = 3 R1 R2 + 120 R1
⇒
R1 =
40 R 2 2 R 2 + 80
Using KVL
24 = 8 + R 2 (1.6 )
⇒
R 2 = 10 10 Ω
Then R1 =
40 (10 ) 2 (10 10 ) + 80
=4Ω
P3.4-14
Using KCL .024 = 0.0192 +
0.384 R 2
⇒
R 2 =
0.384 0.0048
= 80 Ω
Using current division 0.384
=
2R
R
1
(
R1 + R 2 + 80
)
× 0.024
⇒
16 =
R 2R
1
+ 1R
2R+ 80
=
80 1R
+ 160 1R
⇒
R1 = 40 Ω
P3.4-15
Replace the (ideal) ammeter with the equivalent short circuit. Label the current measured by the meter. Apply KCL at the left node of the VCCS to get 1.2 =
va
10
+ 0.2 v a = 0.3 v a
⇒ va =
1.2 0.3
=4V
Use current division to get im =
30 30 + 10 10
0. 2 v a =
30 30 + 10 10
0.2 ( 4 ) = 0.6 A
(checked using LNAP 9/11/04)
Section 3-5 Series Voltage Sources and Parallel Current Sources P3.5-1 The voltage sources are connected in series and can be replaced by a single equivalent voltage source. Similarly, the parallel current sources can be replaced by an equivalent current source.
After doing so, and labeling the resistor currents, we have the circuit shown. Apply KCL at the top node of the current source to get i1 + 1.75 = i 2
Apply KVL to the outside loop to get 5 + 2i2
+ 2 i1 = 0
so 5 + 2 ( i1 + 1.75 ) + 2 i1
=
0
⇒
i1 = −
8.5 4
= −2.125
A
and i 2 = −2.125 + 1.75 = −0.375
A
The power supplied by each sources is: Source 8-V voltage source
Power delivered −8 i1 = 17 W
3-V voltage source
3 i1
3-A current source
3× 2 i 2
1.25-A current source
= −6.375
W
= −2.25
W
−1.25 × 2 i 2 = 0.9375
W
(Checked using LNAP, 9/14/04)
1
P3.5-2 The 20-Ω and 5-Ω resistors are connected in parallel. The 20 × 5 = 4 Ω . The 7-Ω resistor is equivalent resistance is 20 + 5 connected in parallel with a short circuit, a 0- Ω resistor. 0×7 The equivalent resistance is = 0 Ω , a short circuit. 0+7
The voltage sources are connected in series and can be replaced by a single equivalent voltage source. After doing so, and labeling the resistor currents, we have the circuit shown. The parallel current sources can be replaced by an equivalent current source. Apply KVL to get −5 + v1 − 4
( 3.5) = 0
⇒
v 1 = 19
V
The power supplied by each sources is: Source 8-V voltage source 3-V voltage source 3-A current source 0.5-A current source
Power delivered 3.5 ) = −7 W −2 ( 3. −3
( 3.5) = −10.5 W
3 ×19 = 57 W 0.5 ×19 = 9.5 W (Checked using LNAP, 9/15/04)
2
P3.5-3 The voltage sources are connected in series and can be replaced by a single equivalent voltage source. Similarly, the parallel current sources can be replaced by an equivalent current source.
After doing so, and labeling the resistor currents, we have the circuit shown. Apply KCL at the top node of the current source to get i1 + 1.75 = i 2
Apply KVL to the outside loop to get 5 + 2 i2
+ i1 = 0
so 5 + 2 ( i1 + 1.75 ) + 2 i1
=
0
⇒
i1 = −
8.5 4
= −2.125
A
and i 2 = −2.125 + 1.75 = −0.375
A
The power supplied by each sources is: Source 8-V voltage source
Power delivered −8 i1 = 17 W
3-V voltage source
3 i1
3-A current source
3× 2 i 2
1.25-A current source
= −6.375
W
= −2.25
W
−1.25 × 2 i 2 = 0.9375
W
(Checked using LNAP, 9/14/04)
3
Section 3-6 Circuit Analysis P3.6-1
(a)
R = 16 +
(b)
v =
48 ⋅ 24
= 32 Ω
48 + 24 32 ⋅ 32
32 + 32 24 = 16 V ; 32 ⋅ 32
8+
32 + 32 16 1 i= = A 32 2
(c)
i2 =
48
(b)
1 R p
=
3⋅ 6 3+ 6 1 1 12
=6Ω
+ + 6
1 6
⇒ R p = 2.4 Ω then
1
48 + 24 2
P3.6-2
(a) R1 = 4 +
⋅
R 2 = 8 + R p = 10.4 Ω
=
1 3
A
(c) KCL: i2 + 2 = i1
and
− 24 + 6 i2 + R 2i1 = 0
⇒ − 24 + 6 (i1 − 2) + 10.4i1 = 0 ⇒
i1 =
36 16.4
= 2.195 A
⇒ v1 =i1 R 2 =2.2 (10.4)= 22.83 V
1 ( d ) i2 =
1
6 1
+ +
1
( 2.195) = 0.878 A,
6 6 12 v2 = ( 0.878) (6) = 5.3 V (e) i3 =
6 3+ 6
i2 = 0.585 A
⇒ P = 3 i32 = 1.03 W
P3.6-3 Reduce the circuit from the right side by repeatedly replacing series 1 Ω resistors in parallel with a 2 Ω resistor by the equivalent 1 Ω resistor
This circuit has become small enough to be easily analyzed. The vertical 1 Ω resistor is equivalent to a 2 Ω resistor connected in parallel with series 1 Ω resistors:
i1 =
1+1 2 + (1 + 1)
(1.5 ) = 0.75
A
P3.6-4
(a)
1 R2
=
1 24
+
1 12
+
1 8
⇒ R2 = 4 Ω
and
R1 =
(10 + 8) ⋅ 9
b10 + 8g + 9
= 6Ω
(b)
First, apply KVL to the left mesh to get −27 + 6 ia + 3 ia = 0 ⇒ ia = 3 A . Next, apply KVL to the left mesh to get 4 ib − 3 ia = 0 ⇒ ib = 2.25 A . (c)
1 i2 =
1 24
8 1
1
8
12
+ +
2.25 25 = 1.125 A
and
L
v1 = − b10gM
9
O
3P = −10 V
N b10 + 8g + 9 Q
P3.6-5
30 10 + 30
R2 R2 + 10
20 =
v1 = 6 ⇒
12 = 8 ⇒ R2 = 20 Ω
R1 b10 + 30g R1 + b10 + 30g
⇒ R1 = 40 Ω
Alternate values that can be used to change the numbers in this problem: meter reading, V 6 4 4 4.8
Right-most resistor, Ω 30 30 20 20
R1, Ω 40 10 15 30
v1 = 8 V
P3.6-6
P3.6-7
1×10−3 =
24 12 ×10 + R p 3
⇒ R p = 12 ×103 = 12 kΩ
21×10 ) R ( = ( 21×10 ) + R 3
12 ×10 = R p 3
3
⇒ R = 28 kΩ
P3.6-8
⎛
⎞ ⎟⎟ = 15.963 V 1 3 0 5 0 0 2 0 0 2 0 + + ⎝ ⎠ ⎛ 100 ⎞ ⎛ 10 ⎞ .963) ⎜ ⎟ = 12.2 2.279 V ∴v = v ⎜ = (15.963 ⎟ h ⎝ 100 + 30 ⎠ ⎝ 13 ⎠
Voltage division ⇒ v = 50 ⎜ ⎜
130 500
v ∴ i = h = .12279 A h 100
P3.6-9
P3.6-10
Req =
15 ( 20 + 10 ) 15 + ( 20 + 10 )
= 10 Ω
60 ⎛ 30 ⎞ ⎛ 60 = −6 A, ib = ⎜ ia = − ⎟⎜ Req ⎝ 30 + 15 ⎠ ⎜⎝ Req
P3.6-11
⎞ ⎛ 20 ⎞ 40 V ⎟⎟ = 4 A, vc = ⎜ ⎟ ( −60 ) = − 40 + 2 0 1 0 ⎝ ⎠ ⎠
a)
Req = 24 12 =
b)
(24)(12) 24 + 12
=8Ω
from voltage division: 100 5 ⎛ 20 ⎞ 100 v = 40 ⎜ V∴ i = 3 = A = ⎟ x x 20 3 ⎝ 20 + 4 ⎠ 3 5 ⎛ 8 ⎞ from current division: i = i ⎜ A = x ⎝ 8 + 8 ⎟⎠ 6
P3.6-12
9 + 10 + 17 = 36 Ω a.)
b.)
36 R 36+R
36 (18 ) 36+18
= 12 Ω
36 ) ⇒ R = 36 Ω = 18 ⇒ 18 R = (18) ( 36
P3.6-13 Req =
2 R( R ) 2 R + R
=
2 3
R
2
240 Pdeliv. = = =1920 W Req 2 R to ckt 3 Thus R=45 Ω v
P3.6-14
R = 2 + 1 + eq
( 6 12 ) + ( 2 2 ) = ∴i =
40
=
40
Req
8
A
and
3+ 4+1= 8 Ω
=5A
Using current division
⎛ 6 ⎞ ⎟ = ( 5) ⎝ 6 + 12 ⎠
i1 = i ⎜
( 13 )
=5
3
⎛ 2 ⎞ 5 1 ⎟ = ( 5) 2 = 2 A ⎝ 2+ 2⎠
i2 = i ⎜
( )
P3.6-15
( R 4 R) + ( 2 R 3 R) =
(
4
R+
5
6
R = 2R
5
)
R + 2 R ( R + ( 2 R 2 R) ) = R + ( 2 R 2 R) = 2 R
So the circuit is equivalent to
Then
12 = 0.1( R+ ( 2 R 2 R) ) = 0.1( 2 R)
⇒
R= 60 Ω
(checked: ELAB 5/31/04) P3.6-16 The circuit can be redrawn as
va =
R ( R + ( R 2 R) )
2 R + R ( R + ( R 2 R) )
vc =
R 2 R R+ ( R 2 R)
vb =
R R + R
vs =
vc =
1 2
2 5
vs =
va =
vc =
1 21
5
vs
21
2 21
vs
vs
(Checked using LNAP 5/23/04)
P3.6-17 vo =
vs 5 (10 10 ) vs = vs = 10 + (10 10 ) 15 3
vR + vo − vs = 0 iR =
vR
10
⇒
=
2 30
vR =
⇒
3
vs
vs
2
4 2 1 ⎛ 2 ⎞ P=⎜ v s ⎟ (10 ) = vs ≤ 90 4 ⎝ 30 ⎠
2
vs ≤
90 16
=
3 10 4
= 2.37 V
(checked: LNAP 5/31/04)
P3.6-18
The voltage across each strain gauge is 0.2 × 10
−3
≥
vs
vs
2
so the current in each strain gauge is
vs
240
.
2
480
⇒
− v s ≤ 96 × 10 = 0.31 V 3
(checked: LNAP 6/9/04)
P3.6-19 (a) R1 = 10 ( 30 + 10 ) = 8 Ω R 2 = 4 + (18 9 ) = 10 Ω R 3 = 6 ( 6 + 6 ) = 4 Ω
(b)
i =1 A v1 = 8 V, v 2 = 4 V
(c) v4 = −
10
8 = −2 V 10 + 30 9 1 i5 = − 1= − A 9 + 18 3 ⎛ 1⎞ v 7 = −18 ⎜ − ⎟ = +6 V ⎝ 3⎠ i6 =
4 12
=
1 3
A (checked: LNAP 6/6/04)
P3.6-20
Replace series and parallel combinations of resistances by equivalent resistances. Then KVL gives
16 ) i = 48 ⇒ ( 20 + 4 + 8 + 16
i = 0.5 A
v a = 20 i = 10 V , v b = 16 i = 8 V and v c = 8 i = 4 V
Compare the original circuit to the equivalent circuit to get
⎛ 10 || (10 + 30) ⎞ ⎛ 8 ⎞ ⎟va = −⎜ ⎟10 = −4 V + + + 1 2 1 0 | | ( 1 0 3 0 ) 1 2 8 ⎝ ⎠ ⎝ ⎠
v1 = − ⎜
v 2 = −v c = −4 V
⎛ 20 ⎞ ⎛ 1⎞ = − i ⎟ ⎜ ⎟ ( 0.5 ) = −0.1 A ⎝ 20 + 80 ⎠ ⎝ 5⎠
i3 = − ⎜
⎛ 30 ⎞ ⎛1⎞ ⎟ v 1 = − ⎜ ⎟ ( −4 ) = 1 V ⎝ 10 + 30 ⎠ ⎝ 4⎠
v4 = −⎜
4 ⎛ ⎞ ⎛ 1⎞ v c = ⎜ ⎟ ( 4) = 1 V ⎟ ⎝ 5+6+6⎠ ⎝ 4⎠
v5 = ⎜
⎛
⎞ ⎛1⎞ ⎟⎟ i = − ⎜ ⎟ ( 0.5 ) = −0.25 A ⎝ 2⎠ ⎝ 16 + ( 4 + 6 + 6 ) ⎠
i6 = −⎜ ⎜
16
(checked: LNAP 6/10/04) P3.6-21
Replace parallel resistors by equivalent resistors: 6 || 30 = 5 Ω and 72 || 9 = 8 Ω
A short circuit in parallel with a resistor is equivalent to a short circuit. R eq = 36 || ( 8 + 10 ) = 12 Ω v=
8 8 + 10
v ab =
i=
v
8
4 9
(18 ) = 8 V
=1 A (checked: LNAP 6/21/04)
P3.6-22
Replace parallel resistors by an equivalent resistor: 8 || 24 = 6 Ω
A short circuit in parallel with a resistor is equivalent to a short circuit. Replace series resistors by an equivalent
resistor: 4+6 = 10 Ω Now
9 = R eq = 5 + (12 || R || 10 ) so 4=
R × R +
60 11 60
⇒ R = 15 Ω
11 (checked: LNAP 6/21/04)
P3.6-23 Req = ( R || ( R + R) || R) || ( R || ( R + R) || R ) R || ( R + R) || R = 2 R ||
Req =
2 5
R||
2 5
R=
R
⇒
5
R
2
=
2 5
R
R= 5 Req = 250 Ω
(checked: LNAP 6/21/04) P3.6-24 9.74
ia =
8
⎛ 9.74 ⎞ ⎟ ⎝ 8 ⎠
9.74 − 6.09 = r i a = r ⎜
= 1.2175 A
⇒
V ⎛ 9.74-6.09 ⎞ r = ⎜ ⎟8 = 3 A ⎝ 9. 7 4 ⎠
v b = 12 − 9.74 = 2.26 V
gv b +
6. 0 9 8
+
9.74 8 g=
−
2.26
gv b vb
8
=
=0
−6.696 2.26
⇒
gv b = −1.696 A
= −0.75
(checked: LNAP 6/21/04)
P3.6-25 va =
20 20 20 + ( 20 20 20 )
vs =
1 3
vs
3 1 ⎛ 12 ⎞ 10v a ) = × 10 × v s = 2v s ( ⎟ 5 3 ⎝ 12 + 8 ⎠
vo = ⎜
So vo is proportional to vs and the constant of proportionality is 2
V V
.
P3.6-26 vs ⎛ 40 ⎞ ⎛ 4 ⎞ ⎛ vs ⎞ 4 = ⎟ ⎜ ⎟ ⎜ ⎟ = vs 1 0 ) ⎝ 5 ⎠ ⎝ 10 ⎠ 50 ⎝ 40 + 10 ⎠ 2 + ( 40 10
ia = ⎜
100 ⎛ 4 ⎞ 8 ⎛ 40 ⎞ 50i a ) = − vs = − vs ( ⎟ ⎜ ⎟ 3 ⎝ 50 ⎠ 3 ⎝ 20 + 40 ⎠
io = − ⎜
The output is proportional to the input and the constant of proportionality is −
P3.6-27 Replace the voltmeter by the equivalent open circuit and label the voltage measured by the meter as vm.
The 10-Ω resistor at the right of the circuit is in series with the open circuit that replaced the voltmeter so it’s current is zero as shown. Ohm’s law indicates that the voltage across that 10- Ω resistor is also zero. Applying KVL to the mesh consisting of the dependent voltage source, 10-Ω resistor and open circuit shows that vm = 8 ia
The 10-Ω resistor and 40- Ω resistor are connected in parallel. The parallel combination of these resistors is equivalent to a single resistor with a resistance equal to 40 × 10 40 + 10
=8 Ω
8 A 3 V
.
Figure a shows part of the circuit. In Figure b, an equivalent resistor has replaced the parallel resistors. Now the 4- Ω resistor and 8- Ω resistor are connected in series. The series combination of these resistors is equivalent to a single resistor with a resistance equal to 4 + 8 = 12 Ω. In Figure c, an equivalent resistor has replaced the series resistors. Here the same three circuits with the order reversed. The earlier sequence of figures illustrates the process of simplifying the circuit by repeatedly replacing series or parallel resistors by an equivalent resistor. This sequence of figures illustrates an analysis that starts with the simplified circuit and works toward the original circuit.
Consider Figure a. Using Ohm’s law, we see that the current in the 12- Ω resistor is 2 A. The current in the voltage source is also 2 A. Replacing series resistors by an equivalent resistor does not change the current or voltage of any other element of the circuit, so the current in the voltage source must also be 2 A in Figure b. The currents in resistors in Figure b are equal to the current in the voltage source. Next, Ohm’s law is used to calculate the resistor voltages as shown in Figure b. Replacing parallel resistors by an equivalent resistor does not change the current or voltage of any other element of the circuit, so the current in the 4- Ω resistor in Figure c must be equal to the current in the 4- Ω resistor in Figure b. Using current division in Figure c are yields
⎛ 40 ⎞ ⎟ 2 = 1.6 A 4 0 1 0 + ⎝ ⎠
ia = ⎜
Finally, v m = 8 i a = 8 × 1.6 = 12.8 V
P3.6-28 Replace the ammeter by the equivalent short circuit and label the current measured by the meter as im.
The 10-Ω resistor at the right of the circuit is in parallel with the short circuit that replaced the ammeter so it’s voltage is zero as shown. Ohm’s law indicates that the current in that 10- Ω resistor is also zero. Applying KCL at the top node of that 10- Ω resistor shows that i m = 0.8 v a
Figure a shows part of the circuit. The 2- Ω resistor and 4- Ω resistor are connected in series. The series combination of these resistors is equivalent to a single resistor with a resistance equal to
2+4 = 6 Ω
P3.6-29 Use current division in the top part of the circuit to get
⎛ 40 ⎞ ⎟ ( −3) = −2.4 A ⎝ 40 + 10 ⎠
ia = ⎜
Next, denote the voltage measured by the voltmeter as vm and use voltage division in the bottom part of the circuit to get ⎛ R ⎞ ⎛ −5 R ⎞ 5 − = vm = ⎜ i ( ) a ⎟ ⎜ ⎟ ia ⎝ 18 + R ⎠ ⎝ 18 + R ⎠ Combining these equations gives: 12 R ⎛ −5 R ⎞ ⎟ ( −2.4 ) = 18 + R ⎝ 18 + R ⎠
vm = ⎜
When vm = 4 V, 4=
12 R 18 + R
⇒ R =
4 × 18 12 − 4
=9 Ω
P3.6-30 Use voltage division in the top part of the circuit to get 2 ⎛ 12 ⎞ − v = − vs ( ) s ⎟ 5 ⎝ 12 + 18 ⎠
va = ⎜
Next, use current division in the bottom part of the circuit to get 80 ⎞ ⎛ 16 ⎞ ⎛ 5 v = − ( ) a ⎟ ⎜ ⎟ va ⎝ 16 + R ⎠ ⎝ 16 + R ⎠
im = − ⎜
Combining these equations gives: 80 ⎞ ⎛ 2 ⎞ ⎛ 32 ⎞ ⎛ ⎟ ⎜ − vs ⎟ = ⎜ ⎟ vs 1 6 5 1 6 + R + R ⎝ ⎠⎝ ⎠ ⎝ ⎠
im = ⎜ −
a. When vs = 15 V and im = 5 A 400 ⎛ 32 ⎞ = 80 Ω ⎟ 15 ⇒ 80 + 5 R = 480 ⇒ R = 16 5 + R ⎝ ⎠
5=⎜
b. When vs = 15 V and R = 24 Ω
⎛ 32 ⎞ ⎟ 15 = 12 A ⎝ 16 + 24 ⎠
im = ⎜
c. When im = 3 A and R = 24 Ω 4 ⎛ 32 ⎞ vs = vs ⎟ 5 ⎝ 16 + 24 ⎠
3=⎜
⇒ vs =
15 4
= 3.75 V
P3.6-31 R eq = ( ( R + 4 ) || 20 ) + 2 =
12 =
a.
20 R + 80 R + 24
+ 2 ⇒ 10 =
R eq =
b.
( R + 4 ) × 20 20 R + 80 +2 = +2 R + 24 ( R + 4 ) + 20
20 R + 80
⇒
R + 24
20 (14 ) + 80 14 + 24
R+ 24 = 2 R+ 8 ⇒
R= 16 Ω
+ 2 = 11.5 Ω (Checked: LNAPDC 9/28/04)
P3.6-32
Replace the ideal voltmeter with the equivalent open circuit and label the voltage measured by the meter. Label the element voltages and currents as shown in ( b). Using units of V, A,
and W:
Using units of V, mA, k
and mW:
a.) Determine the value of the voltage measured by the meter.
a.) Determine the value of the voltage measured by the meter.
Kirchhoff’s laws give
Kirchhoff’s laws give
12 + v R = v m and −i R = −i s = 2 ×10−3 A
12 + v R = v m and −i R = −i s = 2 mA Ohm’s law gives
Ohm’s law gives
(
v R = −25 i R
)
v R = − 25 ×103 i R
Then
Then
(
)
(
v R = − 25× 103 i R = − 25× 103
v R = −25 i R = −25 ( −2 ) = 50 V
) ( −2 ×10−3 )
= 50 V
v m = 12 + v R = 12 + 50 = 62 V
v m = 12 + v R = 12 + 50 = 62 V
b.) Determine the power supplied by each element. voltage source current source resistor
12 i s = −12 −2 × 10−3
(
( )
b.) Determine the power supplied by each element.
)
voltage source
( )
12 i s = −12 ( −2 )
= − 24 ×10−3 W
(
62 2 ×10
−3
= −24 mW
) = 124 ×10−
3
v R i R = 5 0 −2 × 10 −
(
3
W
)
current source
62 ( 2 ) = 124 mW
resistor
v R i R = 50 ( − 2 )
= − 1 0 0 × 1 0 −3 W total
= −100 mW total
0
P3.6-33 12 +
40 × 10 40 + 10
+ 4 = 12 Ω
P3.6-34
( 60 + 60 + 60 ) × 60 = 45 Ω ( 60 + 60 + 60 ) + 60
0
Section 3-8 How Can We Check …
P3.8-1
(a) 7 + ( −3) = 4
(node a )
4 + ( −2 ) = 2
(node b)
− 5 = −2 + ( − 3 )
(node c )
(b)
−1 − ( − 6 ) + ( − 8 ) + 3 = 0 −1 − 2 − ( − 8 ) − 5 = 0
(loop a - b - d - c - a ) (loop a - b - c - d - a )
The given currents and voltages satisfy these five Kirchhoff’s laws equations.
*P3.8-2
(a)
i
vs
= R1
2.4 =
from row 1
1.2 =
from row 2
+ R2 vs R1 vs
+ 10
R1
so
2.4
1.2 R 1 = v s =
(
10 R 1+
)
⇒
10 R 1 =
then vs
(b)
i
24
When R2 = 20 Ω then i = When R2 = 30 Ω then
v
=
When R2 = 40 Ω the i =
24
40 24
and
10 + R 2
= 0.8 A and
30 720
50
=
= 2.4 (10 ) = 24 V
= 18 V .
= 0.48 A .
v
=
480 30
v
=
24 R 2 10 + R 2
= 16 V .
Ω
(c) When R2 = 30 Ω then
i
v
=
When R2 = 40 W then
24
=
40 960 50
= 0.6 A . = 19.2 V . (checked: LNAP 6/21/04)
P3.8-3
(a)
i
=
R1 R1
+ R2
is
From row 1 4 3
R1
=
R1
+ 10
⇒
4R1 + 40 = 3 R1i s
⇒
6R1 + 120 = 7 R1i s
is
From row 2 6 7
=
R1
+ 20
R1
is
So 4 R1 + 40 3 R1
= is =
6 R1 + 120
⇒
7 R1
28R1 + 280 = 18 R1 + 360
⇒
R1
=8Ω
Then
4 3 (b)
When R2 = 40 Ω then i =
i
=
24
48 tabulated data is consistent.
(c) When R2 = 80 Ω then
i
=
8 8+
8 8 + 10
( 3) =
2
R
= 0.5 A and
=
24 88
=
3 11
⇒
is
24 8+
v
=
A and
2
R
and
960 48
v
=
is
v
=3A
= R2 i =
24 R 2 8+
2
R
= 20 V . These are the values in the table so
24 (80 ) 88
=
240 11
V. (checked: LNAP 6/21/04)
P3.8-4
KVL bottom loop: − 14 + 0.1i A + 1.2iH = 0
− 12 + 0.05i B + 1.2iH = 0 KCL at left node: i A+ i B = i H KVL right loop:
This alone shows the reported results were incorrect. Solving the three above equations yields: i A
= 16.8 A
i B
= −6.49 A
i H
= 10.3 A
∴ Reported values were incorrect.
P3.8-5
⎛ ⎝
Top mesh: 0 = 4 i a + 4 i a + 2 ⎜ ia +
(
⎞ − i b ⎟ = 10 ( −0.5) + 1 − 2 ( −2) 2 ⎠
1
)
Lower left mesh: vs = 10 + 2 i a + 0.5 − i b = 10 + 2 ( 2) = 14 V Lower right mesh: vs + 4 i a = 12 ⇒ vs = 12 − 4 ( −0.5) = 14 V The KVL equations are satisfied so the analysis is correct.
P3.8-6
Apply KCL at nodes b and c to get: KCL equations: Node e: −1 + 6 = 0.5 + 4.5 Node a:
0.5 + i c = −1 ⇒
Node d:
ic
+ 4 = 4.5 ⇒
ic ic
= −1.5 mA
= 0.5 mA
That's a contradiction. The given values of i a and ib are not correct.
P3.8-7
KCL at node a:
i
= i +i
3 1 2 − 1.167 = − 0.833 + ( −0.333)
− 1.167= − 1.166 OK KVL loop consisting of the vertical 6 Ω resistor, the 3 Ω and4Ω resisto istorrs, and the voltage source: 6i + 3i + v + 12 = 0 3 2 yields yields v = −4.0 V not v = −2.0 V