CONTENTS CONTENTS
2 D. C. Circuits CHAPTER OVERVIEW 1. DIRECT CURRENT The current that always flows in one direction is called direct current (d.c.). The current supplied by a cell/battery or d.c. generator is direct current. Thus in Fig. 2.1, the battery supplies direct current to the bulb. The direction of current is along ABCDA and it always flows in this direction. Note that direct current means steady direct current (i.e. one of constant BULB B C magnitude) unless stated otherwise.
2. D.C. CIRCUIT The closed path followed by direct current is called a d.c. circuit. I A d.c. circuit essentially consists of a source of direct voltage (e.g. battery), the conductors used to carry current and the load. Fig. 2.1 shows a torch bulb (i.e. load) connected to a battery through conducting wires. The direct current* starts from the A D positive terminal of the battery and comes back to the starting Fig. 2.1 point via the load. The direct current follows the closed path ABCDA and hence ABCDA is a d.c. circuit. The load for a d.c. circuit is usually a **resistance. In a d.c. circuit, loads (i.e. resistances) may be connected in series or parallel or series-parallel.
3. RESISTORS IN SERIES A number of resistors are said to be connected in series if the same current flows through each resistor and there is only one path for the current flow throughout. Consider three resistors of resistances R1, R2 and R3 connected in series across a battery of E volts as shown in Fig. 2.2 (i). Then total resistance RT is given by, RT = R1 + R2 + R3 A
R1
R2
V1
V2
R3
A
B
RT
B
V3
I
I
E
E
(ii)
(i) Fig. 2.2 * **
This is the direction of conventional current. However, electron flow will be in opposite direction. Other passive elements viz inductance and capacitance are relevant only in a.c. circuits. 23
CONTENTS CONTENTS
24
Objective Electrical Technology
Hence when a number of resistances are connected in series, the total or equivalent resistance is equal to the sum of the individual resistances. Thus we can replace the series connected resistors shown in Fig. 2.2 (i) by a single resistor RT (= R1 + R2 + R3) as shown in Fig. 2.2 (ii). This will enable us to calculate the circuit current easily (I = E/RT). (i) When resistors are connected in series, the total circuit resistance increases. (ii) RT = R1 + R2 + R3 RT R1 R2 R3 or 2 + 2 + 2 2 = V V V V 1 1 + 1 + 1 or = P1 P2 P3 PT where PT is the total power dissipated by the series circuit and P1, P2 and P3 are the powers dissipated by individual resistors.
4. RESISTORS IN PARALLEL A number of resistors are said to be connected in parallel if voltage across each resistor is the same and there are as many paths for current as the number of resistors. Consider three resistors of resistances R1, R2 and R3 connected in parallel across a battery of E volts as shown in Fig. 2.3 (i). Then total resistance RT is given by ;
1 RT
A
I1
R1
I2
R2
I3
R3
=
1 + 1 + 1 R1 R2 R3
A
B
I
RT
B
I
E
E
(i)
(ii) Fig. 2.3
Hence when a number of resistances are connected in parallel, the reciprocal of the total resistance is equal to the sum of reciprocals of individual resistances. Again, we can replace the parallel connected resistors shown in Fig. 2.3 (i) by a single resistor RT shown in Fig. 2.3 (ii). (i) When resistors are connected in parallel, the total circuit resistance decreases. (ii) The total resistance of a parallel circuit is always less than the smallest of the resistances. For example, if three resistors of 1 Ω, 3 Ω and 4 Ω are connected in parallel, the total resistance will be less than 1 Ω. (iii) If n resistors, each of resistance R, are connected in parallel, then total resistance RT = R/n. 1 + 1 + 1 1 (iv) = R1 R2 R3 RT 2
2
or
V RT
=
2
2
V +V +V R1 R2 R3
D. C. Circuits
25
or PT = P1 + P2 + P3 where PT is the total power dissipated by the parallel circuit and P1, P2 and P3 are the powers dissipated by individual resistors.
5. TWO RESISTORS IN PARALLEL A frequent special case of parallel resistors is a circuit that contains two resistors in parallel as shown in Fig. 2.4. The total circuit current I divides into two parts; I1 flowing through R1 and I2 flowing through R2. 1 = 1 + 1 = R2 + R1 (i) Total Resistance : R1 R2 R1 R2 RT I R1R2 Product or RT = i. e., . R1 + R2 Sum Thus, if two resistances of 3 Ω and 6 Ω are connected in parallel, then their total or equivalent resistance R is 3 × 6 18 = =2Ω R = 3+6 9 RR (ii) Branch Currents : E = I RT = I 1 2 R1 + R2 RR 1 = I R2 Now I1 = E = I 1 2 R1 + R2 R1 + R2 R1 R1
FG H
I1
R1
I2
R2
E
Fig. 2.4
IJ K
R2 R1 + R2 R1 = I× R1 + R2
∴
I1 = I ×
Similarly,
I2
i.e., current in any of the two branches = Total current × Thus referring to Fig. 2.5, the currents in the two branches are : I1 = 9 ×
6 =6A 3+6
I2 = 9 ×
3 =3A 3+ 6
Other resistance Sum of the two resistances I1
3Ω
I2
6Ω
9A
Fig. 2.5
6. ADVANTAGES OF PARALLEL CIRCUITS The most useful property of a parallel circuit is the fact that potential difference has the same value between the terminals of each branch of parallel circuit. This feature of the parallel circuit offers the following advantages : (i) The appliances rated for the same voltage but different powers can be connected in parallel without disturbing each other’s performance. Thus a 230 V, 230 W TV receiver can be operated independently in parallel with a 230 V, 40 W lamp. (ii) If a break occurs in any one of the branch circuits, it will have no effect on other branch circuits. Due to above advantages, electrical appliances in homes are connected in parallel. We can switch on or off any light or appliance without affecting other lights or appliances.
26
Objective Electrical Technology
7. INTERNAL RESISTANCE OF A CELL The resistance offered by a cell to the current flow is called internal resistance r of the cell. The internal resistance of a fresh cell is generally low. However, as the cell is used, its internal resistance goes on increasing. The internal resistance of a cell depends upon the following factors : (i) distance between the plates – increases with the increase in distance between the plates (ii) nature of the electrolyte. (iii) concentration of the electrolyte – increases with the increase in concentration of electrolyte. (iv) nature of the electrodes (v) area of the plates – decreases with the increase in plate area. It is a usual practice to show internal resistance of a cell as a series E r resistor external to the cell as shown in Fig. 2.6. Fig. 2.6
8. E.M.F. AND TERMINAL VOLTAGE OF A CELL
(i) When the cell is delivering no current (i.e., on open-circuit), the p.d. across the terminals of the cell is equal to e.m.f. E of the cell as shown in Fig. 2.7 (i). 1
1
I
r
r E
R
E
V
E
(i)
2
2
(ii) Fig. 2.7
(ii) When a resistance R is connected across the cell [See Fig. 2.7 (ii)], current I starts flowing in the circuit. This current causes a voltage drop (= I r) across internal resistance of the cell so that terminal voltage V is less than the e.m.f. E of the cell. E I = R+r or IR+Ir = E But I R = V = Terminal p.d. of the cell ∴ V+Ir = E or E = V+Ir E−V E−V ∵I = V R = Internal resistance of the cell, r = R I V
FH
IK
FH
IK
9. WHEATSTONE BRIDGE This bridge was first proposed by Wheatstone (an English telegraph engineer) for measuring accurately the value of an unknown resistance. It consists of four resistors (two fixed known resistances P and Q, a known variable resistance R and the unknown resistance X whose value is to be found) connected to form a diamond shaped circuit ABCDA as shown in Fig. 2.8 (i). Across one pair of opposite junctions (A and C), battery is connected and across the other opposite pair of junctions (B and D), a galvanometer is connected through the key K. The circuit is called a bridge because galvanometer
D. C. Circuits
27
bridges the opposite junctions B and D. Fig. 2.8 (ii) shows another* way of drawing the Wheatstone bridge. B BATTERY P I1 A
I2
G
C
P
I
B
Q
K
R I
Q
A
D
BATTERY
G
I1
X
C
I2
K R
D
X
(i)
(ii) Fig. 2.8 Let I1 and I2 be the currents through P and R respectively when the bridge is balanced. Since there is no current through the galvanometer, the currents in Q and X are also I1 and I2 respectively. As the galvanometer reads zero, points B and D are at the same potential. This means that voltage drops from A to B and A to D must be equal. Also voltage drops from B to C and D to C must be equal. ∴ I1P = I2R and I1Q = I2X R P ∴ = X Q or PX = QR ...(i) i.e. Product of opposite arms = Product of opposite arms Q ×R Unknown resistance, X = P Hence when the Wheatstone bridge is balanced, the product of resistances of the opposite arms of the bridge are equal. Note that exp (i) is true only under the balanced conditions of Wheatstone bridge.
10. KIRCHHOFF’S LAWS Sometimes we encounter circuits where simplification by series and parallel combinations is impossible. Consequently, Ohm’s law cannot be applied to solve such circuits. Kirchhoff gave two laws to solve such complex circuits, namely : (i) Kirchhoff’s Current Law (KCL) (ii) Kirchhoff’s Voltage Law (KVL) (i) Kirchhoff’s Current Law. This law relates to the currents at the **junctions of an electric circuit and may be stated as under : The algebraic sum of the currents meeting at a junction in an electrical circuit is zero. An algebraic sum is one in which the sign of the quantity is taken into account. For example, consider four conductors carrying currents I1, I2, I3 and I4 and meeting at point O as shown in Fig. 2.9. If we take the signs of currents flowing towards point O as positive, then currents flowing away from * **
Note the four points A, B, C and D, each lying at the junction between two resistors. A galvanometer should bridge a pair of opposite points such as B and D and the battery to the other pair A and C. A junction is that point in an electrical circuit where three or more circuit elements meet.
28
Objective Electrical Technology
point O will be assigned negative sign. Thus, applying Kirchhoff’s I1 current law to the junction O in Fig. 2.9, we have, (I1) + (I4) + (−I2) + (−I3) = 0 O or I1 + I4 = I2 + I3 I2 I3 i.e., Sum of incoming currents = Sum of outgoing I4 currents Hence, Kirchhoff’s current law may also be stated as under : Fig. 2.9 The sum of currents flowing towards any junction in an electrical circuit is equal to the sum of currents flowing away from that junction. Kirchhoff’s current law is rightly called the junction rule. Kirchhoff’s current law is true because electric current is merely the flow of free electrons and they cannot accumulate at any point in the circuit. This is in accordance with the law of conservation of charge. Hence, Kirchhoff’s current law is based on the law of conservation of charge. (ii) Kirchhoff’s Voltage Law (KVL). This law relates to e.m.fs and voltage drops in a closed circuit or loop and may be stated as under : In any closed electrical circuit or mesh, the algebraic sum of all the electromotive forces (e.m.fs) and voltage drops in resistors is equal to zero, i.e., In any closed circuit or mesh, Algebraic sum of e.m.fs + Algebraic sum of voltage drops = 0 The validity of Kirchhoff’s voltage law can be easily established by referring to the loop ABCDA shown in Fig. 2.10. If we start from any point (say point A) R1 R2 in this closed circuit and go back to this point (i.e., point A) B C after going round the circuit, then there is no increase or decrease in potential. This means that algebraic sum of the e.m.fs of all the sources (here only one e.m.f. source is I considered) met on the way plus the algebraic sum of the E voltage drops in the resistances must be zero. Kirchhoff’s D voltage law is based on the law of *conservation of energy, A i.e., net change in the energy of a charge after completing Fig. 2.10 the closed path is zero. Note. Kirchhoff’s voltage law is also called loop rule.
Sign convention. While applying Kirchhoff’s voltage law to a closed circuit, algebraic sums are considered. Therefore, it is very important to assign proper signs to e.m.fs and voltage drops in the closed circuit. The following sign convention may be followed : A rise in potential should be considered positive and fall in potential should be considered negative. (i) Thus in Fig. 2.11 (i), as we go from A B A B A to B (i.e., from negative terminal of the cell to the positive terminal), there is a rise (i) (ii) in potential. In Fig. 2.11 (ii), as we go from Fig. 2.11 A to B, there is also a rise in potential. D D C (ii) In Fig. 2.12 (i), as we go from C C to D, there is a fall in potential. In Fig. (i) (ii) Fig. 2.12 *
As a charge traverses a loop and returns to the starting point, the sum of rises of potential energy associated with e.m.fs in the loop must be equal to the sum of the drops of potential energy associated with resistors.
D. C. Circuits
29
2.12 (ii), as we go from C to D, there is again a fall in potential.
11. BATTERY AND ITS NEED The e.m.f. and current obtained from a single cell is generally small. For instance, an ordinary dry cell has an e.m.f. of 1.5 V and can deliver about 1/8 ampere continuously. Such a cell can, therefore, supply electrical energy to a circuit requiring 1.5 V and not more than 1/8 A. Many occasions arise when higher voltage or higher current or both are required. To meet these needs, a number of cells are suitably connected or grouped. The combination of cells thus obtained is called a battery. Depending upon voltage and current requirements, cells may be connected in three ways to form a battery viz., (i) series grouping (ii) parallel grouping (iii) series-parallel grouping.
12. CELLS IN SERIES When voltage required is more than that of the e.m.f. of a single cell, a number of cells are connected in series to meet the requirement. The cells are said to be connected in series if the negative terminal of one cell is connected to the positive terminal of the next cell and so on. Consider n cells, each of e.m.f. E and internal resistance r, connected in series across an external resistance R as shown in Fig. 2.13. Total battery e.m.f. = n E Internal resistance of the battery = n r Total circuit resistance = R + n r nE ∴ Circuit current, I = R + nr Special Cases (i) If R >> n r, then n r can be neglected as compared to R.
R
I n cells BATTERY
Fig. 2.13
E = n × current due to one cell R (ii) If R << n r, then R can be neglected as compared to nr. ∴
I = n
∴
nE E I = n r = r = current due to one cell
Hence in order to get maximum current in a series grouping of cells, the external resistance (R) should be very high as compared to the internal resistance of the battery (n r).
13. CELLS IN PARALLEL When current required is more than that delivered by a single cell, a number of cells are connected in parallel to meet the requirement. In parallel grouping, positive terminals of all the cells are joined together and in a like manner all the negative terminals are connected together. Consider m rows of cells in parallel, each row containing one cell. Let E and r be the e.m.f. and internal resistance respectively of each cell. Further, let this battery be connected across an external resistance R as shown in Fig. 2.14. E.M.F. of the battery = E Since the cells are connected in parallel, their internal
E m row E E
R
Fig. 2.14
s
30
Objective Electrical Technology
resistances are also in parallel. If rp is the total resistance of the battery, then, 1 = 1 1 1 + + + ... m terms rp r r r = ∴
rp = Total circuit resistance =
∴
Circuit current, I =
or
I =
1 + 1 + 1 + ... m terms m = r r r/m r R + rp = R + m mE E = R + (r / m) m R + r mE mR+ r
Special Cases (i) If R << r, then mR may be neglected as compared to r. ∴ I = m E = m × current due to one cell. r (ii) If r << R, then r may be neglected as compared to m R. mE E = = current due to one cell. ∴ I = mR R Hence in order to get maximum current in parallel grouping of cells, the external resistance (R) should be very low as compared to the internal resistance of each cell.
14. SERIES-PARALLEL GROUPING OF CELLS (MIXED GROUPING) To produce a large voltage, a number of cells are connected in series and to produce a large current, a number of sets of such series-connected cells are connected n cells in parallel. Such a connection is called series-parallel grouping m rows or mixed grouping. Clearly, series-parallel grouping embraces the advantages of both series and parallel groupings. Fig. 2.15 shows series-parallel grouping of cells. Let there be n cells connected in series and m such rows connected in parallel across an external resistance R. Further, let E and r be the e.m.f. and internal resistance of each cell respectively. Resistance of one row of cells = n r R There are m rows in parallel, each having a resistance of n r. If rT is the total internal resistance of the battery, then, Fig. 2.15 1 = 1 + 1 + 1 + ... m terms rT nr nr nr 1 + 1 + 1 + ... m terms m = = nr nr nr ∴ rT = m Total circuit resistance = R + (n r/m) Total battery e.m.f. = E.M.F due to one row = n E nE mnE ∴ Circuit current, I = = R + (n r / m) m R + n r mnE or I = m R + nr
D. C. Circuits
31
Condition for maximum current I = or
I =
mnE m R + nr mnE
d
m R − nr
i
2
+ 2 m R nr
...(i)
As m, n and E are fixed quantities, therefore, numerator of eq. (i) is constant. Hence, current will be maximum when the denominator of eq. (i) is minimum. The denominator will be minimum when quantity
2
d m R − n r i is minimum. Now the minimum value of a squared quantity is zero, i.e., d
m R − nr
2
i
= 0
or
mR = n r or R = nr m i.e., External resistance = Total internal resistance of the battery Hence, in order to get maximum current in series-parallel grouping of cells, the external resistance (R) should be equal to the total internal resistance of the battery (n r/m).
SHORT ANSWER QUESTIONS Q. 1. Why are domestic appliances connected in parallel ? Ans. The domestic appliances are connected in parallel due to the following reasons : (i) The domestic appliances are rated for the same voltage (i.e., 230 V) but different powers. When connected in parallel, the performance of each appliance becomes independent of the other. Thus a 230 V, 250 W TV receiver can be operated independently in parallel with a 230 V, 100 W lamp. (ii) If a break occurs in any one of the branch circuits, it will have no effect on the other branch circuits. Q. 2. Why does a cell possess internal resistance ? Ans. When current flows through a cell, it meets opposition from the electrodes and the electrolyte. This opposition to current offered by the cell is called its internal resistance. Q. 3. Why is Wheatstone bridge method preferred to voltmeter-ammeter method for measuring resistance ? Ans. Wheatstone bridge method is preferred to voltmeter-ammeter method for measuring resistance due to the following reasons : (i) The Wheatstone bridge method is independent of the fluctuations and variations in the supply voltage. (ii) It removes the objectionable feature of voltmeter-ammeter method where the accuracy of measurement is limited by the accuracy of calibration of instruments. Q. 4. Why is Wheatstone circuit called a bridge ? Ans. The circuit is called a bridge because the galvanometer bridges the opposite junctions of the circuit. Q. 5. What is the meaning of algebraic sum ? Ans. The algebraic sum is one in which the sign of the quantity is taken into account. For example, if a current flowing towards a point O is assigned ‘positive’ sign, then current flowing away from point O will be assigned negative sign. Similarly, if a rise in voltage is assigned positive sign, then fall in voltage will be assigned negative sign.
32
Objective Electrical Technology
Q. 6. Which type of grouping of cells will be suitable to obtain higher current and voltage ? Ans. Series-parallel grouping of cells. To produce a large voltage, a number of cells are connected in series and to produce a large current, a number of sets of such series-connected cells are connected in parallel. Such a connection is called series-parallel grouping. Q. 7. When we require a d.c. supply we look for a d.c. source other than a battery. Why ? Ans. Because batteries are costly and require frequent replacement. Q. 8. What is the objection to have lamps in a house-lighting circuit connected in series ? Ans. The lamps in a house-lighting circuit are not connected in series due to the following reasons: (i) If break occurs in any part of the circuit, no current will flow and the entire circuit becomes useless. (ii) A high supply voltage is required if the lamps (or other devices) are to be connected in series. For example, if five 230 V lamps are to be connected in series, then supply voltage would have to be 5 × 230 = 1150 V. Therefore, series connection is not practicable for lighting circuits. (iii) For efficient operation, only those lamps or devices should be connected in series that have the same current rating. However, electrical devices (e.g., heater, toaster, lamp etc.) have different current ratings. Obviously, they cannot be connected in series for efficient operation. Q. 9. Can you measure the e.m.f. of a cell with a voltmeter ? Ans. Not accurately. It is because when voltmeter is connected across the cell, current starts flowing through it. This causes a voltage drop across the internal resistance of the cell. Consequently, the voltmeter will not indicate the e.m.f. of the cell. Remember e.m.f. of a cell is the voltage across the open-circuit terminals of the cell i.e., voltage across the terminals of the cell when it carries no current. Q. 10. When the resistance connected in series with a cell is halved, the current is not exactly doubled but is slightly less. Why ? Ans. This is owing to the internal resistance r of the cell. The current delivered by the cell is I = E R+r When external resistance R is made R/2, the current will be slightly less than 2 I. Q. 11. Why is it easier to start a car engine on a warm day that on a chilly day ? Ans. It is because the internal resistance of a car battery on a chilly day is much less than on a warm day. Q. 12. A low voltage supply should have low internal resistance. Why ? E Ans. I = R+r The maximum current that can be drawn is Imax = E/r ...when R = 0 We can obtain a large maximum current from a low voltage supply if its internal resistance is small. 6Ω Q. 13. How can three resistances of values 2 Ω, 3 Ω and 2Ω 6 Ω be connected to produce an effective B A resistance of 4 Ω ? Ans. Fig 2.16 shows the solution. RAB =
6×3 +2=2+2=4Ω 6+3
3Ω
Fig. 2.16
D. C. Circuits
33
Q. 14. A high voltage supply should have high internal resistance. Why ? Ans. If the internal resistance of a high voltage supply is small, then on accidental short-circuit, a damaging large current will flow through the source of e.m.f. This may damage the high voltage source. For this reason, a high voltage supply should have high internal resistance. Q. 15. The resistance of human body is quite high (a few kΩ). Why does one experience a strong shock from 230 V supply ? Ans. The human body is very sensitive to even minute current. A current as low as a few mA can greatly affect the nervous system. Q. 16. You are given n wires, each of resistance R. What is the ratio of maximum to minimum resistance available from these wires ? Ans. The resistance obtained will be maximum when the wires are connected in series. However, when the wires are connected in parallel, the resistance will be minimum. Rp = R/n ∴ R s = n R and Rs nR 2 ∴ = =n Rp R/n Q. 17. Two cells with same e.m.f. E and different internal resistances r1 and r2 are connected in series to an external resistance R as shown in Fig. 2.17. Find the value of R such that the potential difference across the terminals of first cell is zero. Ans. Circuit current,
P.D. across first cell,
2E I = r1 + r2 + R V1 = E − Ir1 = E −
Q. 18. Ans.
Q. 19. Ans.
Q. 20. Ans.
0 = E−
E
r1
r2 I
I R
FG 2E IJ r H r + r + RK
Fig. 2.17
1
1
or
E
2
2 Er1 r1 + r2 + R
or r1 − r2 = R ∴ R = r1 − r2 It is general belief that a person touching a high power line gets stuck to the line. Is it true? This is not true. The fact is that a current of even a few mA is enough to make the nervous system ineffective for a moment. As a result, the affected person has difficulty to remove his hand off the live wire. We think that the person has got stuck to the line. Why does the glow of lamps become weaker when a heavy current appliance is switched on in the house ? Domestic appliances are connected in parallel across the supply. When heavy current appliance is switched, the total resistance of the system in the room is decreased as the appliance is connected in parallel with other loads. This increases the line current. This in turn causes greater voltage drop in the line. Therefore, the voltage available across the lamps decreases and so is their glow. Is current a scalar or vector ? Current is a scalar. It is because currents at a junction in an electric circuit do not obey the laws of vector algebra.
34
Objective Electrical Technology
OBJECTIVE QUESTIONS 1. A d.c. circuit usually has .......... as the load. (i) resistance (ii) inductance (iii) capacitance (iv) both inductance and capacitance 2. The purpose of load in an electric circuit is to (i) increase the circuit current (ii) utilise electrical energy (iii) decrease the circuit current (iv) none of the above 3. Electrical appliances are not connected in series because (i) series circuit is complicated (ii) power loss is more (iii) appliances have different current ratings (iv) none of the above 4. Electrical appliances are connected in parallel because it (i) is a simple circuit (ii) draws less current (iii) results in reduced power loss (iv) makes the operation of appliances independent of each other 5. Voltage drop across 14·5 Ω resistor in Fig. 2.18 is 14.5 Ω
25.5 Ω
60 Ω
I
+
–
100 W, 200 V
40 W, 200 V
Lamp A
Lamp B
I
+
–
200 V
Fig. 2.19 7. In Fig. 2.19, (i) the lamp A will be brighter than lamp B (ii) the lamp B will be brighter than lamp A (iii) the two lamps will be equally bright (iv) none of the above 8. When cells are arranged in parallel (i) the current capacity increases (ii) the current capacity decreases (iii) the e.m.f. increases (iv) the e.m.f. decreases 9. When a number of resistances are connected in parallel, the total resistance is (i) greater than the smallest resistance (ii) between the smallest and greatest resistance (iii) less than the smallest resistance (iv) none of the above 10. If a d.c. supply of 180 V is connected across terminals AB in Fig. 2.20, then current in 6 Ω resistor will be A
200 V
Fig. 2.18 (i) 29 V (ii) 30·5 V (iii) 14 V (iv) 18 V 6. Referring to Fig. 2.19, the total circuit resistance will be (i) 1000 Ω (ii) 400 Ω (iii) 1400 Ω (iv) 135 Ω
18 Ω
12 Ω 6Ω
B
(i) 10 A (iii) 12 A
Fig. 2.20 (ii) 5 A (iv) 6 A
D. C. Circuits
35
11. A battery of 24 V is applied across terminals AB of the circuit shown in Fig. 2.21. The current in 2 Ω resistor will be A
C
5Ω
8Ω
B
2Ω
E
6Ω
D
2·0 V is 0·1 Ω. It is connected to a resistance of 3·9 Ω. The voltage across the cell is (i) 0·5 V (ii) 1·95 V (iii) 1·9 V (iv) 2 V 18. The total conductance of the circuit shown in Fig. 2.22 is 10 Ω
4Ω
2Ω
F
Fig. 2.21 (i) 3 A (ii) 6 A (iii) 2·5 A (iv) 1·5 A 12. A cell of e.m.f. E is connected across a resistance r. The potential difference between the terminals of the cell is found to be V. The internal resistance of the cell is 2( E − V ) r 2( E − V ) V (i) (ii) E r E −V (iii) (E − V)r (iv) r V 13. An external resistance R is connected to a cell of internal resistance r. The maximum current flows in the external resistance when (i) R < r (ii) R > r (iii) R = r (iv) any other value of R 14. Five cells each of e.m.f. E and internal resistance r are connected in series. If due to oversight, one cell is connected wrongly, then equivalent e.m.f. and internal resistance of the combination are (i) 3E and 5r (ii) 5E and 5r (iii) 3E and 3r (iv) 5E and 4r 15. A wire has a resistance of 12 Ω. It is bent in the form of a circle. The effective resistance between the two points on any diameter of the circle is (i) 6 Ω (ii) 3 Ω (iii) 12 Ω (iv) 24 Ω 16. The smallest resistance obtained by connecting 50 resistances of 1/4 ohm each is (i) 50/4 Ω (ii) 4/50 Ω (iii) 200 Ω (iv) 1/200 Ω 17. The internal resistance of a cell of e.m.f.
F H
I K
1Ω
Fig. 2.22 (i) 13 S (ii) 1·6 S (iii) 6 S (iv) 2·5 S 19. If 10 Ω resistance is removed in Fig. 2.22, then total conductance of the circuit will be (i) 3 S (ii) 6 S (iii) 2 S (iv) 1·5 S 20. The voltage across the parallel circuit shown in Fig. 2.23 is 10 Ω
15 A
2·5 Ω 2Ω
Fig. 2.23 (i) 15 V (ii) 10 V (iii) 30 V (iv) 12·5 V 21. The current in 10 Ω resistor in Fig. 2.23 is (i) 3 A (ii) 2·5 A (iii) 1·5 A (iv) 3·5 A 22. Three 2 ohm resistors are connected to form a triangle. The resistance between any two corners is (i) 6 Ω (ii) 2 Ω (iii) (3/4) Ω (iv) (4/3) Ω 23. A cell of negligible internal resistance and e.m.f. 2 volts is connected to series combination of 2, 3 and 5 ohms. The potential difference across the terminals of 3 Ω resistor will be (i) (2/3) V (ii) 0·6 V (iii) 3 V (iv) 6 V
36
Objective Electrical Technology
R
R
A Ω 10
10 Ω Ω 10
B
C
Fig. 2.25 (i) 60 Ω (ii) 40 Ω (iii) 160/9 Ω (iv) 80/3 Ω 33. Two resistances are in parallel and give equivalent resistance of 6/5 Ω. One of the resistances is broken and the effective resistance is 2 Ω. The resistance of the broken resistor is (i) 3/5 Ω (ii) 2 Ω (iii) 3 Ω (iv) 6 Ω 34. In the circuit shown in Fig. 2.26, the final voltage drop across the capacitor C is V r3 C
P
Q
F
E
Ω 10
A
31. Identical wires of nichrome and copper are connected in series in a circuit. This results in (i) greater current in nichrome (ii) greater current in copper (iii) greater heat in nichrome (iv) greater heat in copper 32. The effective resistance between B and C of letter A containing resistances as shown in Fig. 2.25 is Ω 10
24. A 200 W and 100 W bulb both meant for operation at 220 V are connected in series. When connected to a 220 V supply, the power consumed by them is (i) 66 W (ii) 33 W (iii) 100 W (iv) 300 W 25. An electric fan and a heater are marked as 100 W, 220 V and 1000 W, 220 V respectively. The resistance of heater is (i) zero (ii) greater than that of fan (iii) less than that of fan (iv) equal to that of fan 26. Three cells each of e.m.f. 1·5 V and internal resistance 1 Ω are connected in parallel. The e.m.f. of the combination is (i) 4·5 V (ii) 5 V (iii) 3 V (iv) 1·5 V 27. When the internal resistance of a cell is large compared to the external resistance, then high current is obtained by grouping the cells in (i) series (ii) parallel (iii) series-parallel (iv) all three can be used 28. The e.m.f. of a cell depends upon (i) internal resistance (ii) external resistance (iii) electrolyte and electrodes of the cell (iv) none of these factors 29. Three equal resistors are connected as shown in Fig. 2.24. Find the equivalent resistance between points A and B.
r2
Fig. 2.26 R
B
Vr1 Vr2 (ii) r1 + r2 r1 + r2 V (r1 + r2 ) V (r1 + r2 ) (iii) (iv) r2 r1 + r2 + r3 35. Two identical cells connected in series send 10 A through 5 Ω resistor. When they are connected in parallel, they send 8 A through the same resistor. The internal resistance of each cell is (i) zero (ii) 2·5 Ω (i)
Fig. 2.24 (i) 3R (ii) R/3 (iii) 3R/2 (iv) 2R/3 30. Given three equal resistances. How many combinations of these three resistances can be made (i) three (ii) four (iii) five (iv) two
r1
D. C. Circuits
37
(iii) 10 Ω (iv) 1 Ω 36. Fig. 2.27 shows currents in the part of an electric circuit. Then current i is
(i) 1·7 A (ii) 3·7 A (iii) 1·3 A (iv) 1 A 37. Fig. 2.28 shows a part of a closed circuit. What is the potential difference between points A and B ? 3A A
6Ω
− +
B
3V 1Ω
will be zero in the circuit shown in Fig. 2.31
(i) 4 Ω (ii) 2 Ω (iii) 3 Ω (iv) 6 Ω 41. For what value of unknown resistance X, the potential difference between B and D will be zero in the circuit shown in Fig. 2.32
Fig. 2.28 (i) 6 V (ii) 12 V (iii) 24 V (iv) 18 V 38. Fig. 2.29 shows a part of a closed electrical circuit. Then VA − VB is 2A A
3Ω
2V
1Ω B
Fig. 2.29 (i) − 8 V (ii) 6 V (iii) 10 V (iv) 3 V 39. A current of 2 A flows in a circuit shown in Fig. 2.30. The potential difference VA − VB is
(i) −1 V (ii) +1 V (iii) 4 V (iv) 2 V 40. For what value of unknown resistance X, the potential difference between points B and D
(i) 4 Ω (ii) 2 Ω (iii) 3 Ω (iv) 6 Ω 42. A cell having an e.m.f. of 2·2 V and internal resistance 0·2 Ω is connected to a circuit comprising an ammeter and a resistance of 4 Ω in series with a combination of two resistances of 0·4 Ω each in parallel. What will be the reading of the ammeter ?
38
Objective Electrical Technology
B
4 Ω
6
Ω
2Ω
3V
4Ω
A
Fig. 2.36 (i) 3 Ω (ii) 6 Ω (iii) 9 Ω (iv) 12 Ω 52. The resistance between P and Q in the circuit shown in Fig. 2.37 is R
1Ω
Fig. 2.35 (i) 12 V (ii) 9 V (iii) 18 V (iv) 6 V 48. Kirchhoff’s voltage law deals with (i) conservation of energy (ii) conservation of charge
B
Ω
2A A
4Ω
4Ω
20
Fig. 2.34 (i) 10/3 Ω (ii) 20/3 Ω (iii) 15 Ω (iv) 6 Ω 45. n similar resistors each of resistance r when connected in parallel have the total resistance R. When these resistances are connected in series, the total resistance is 2 (i) nR (ii) R/n 2 (iii) n R (iv) R/n 46. Kirchhoff’s current law at a junction deals with (i) conservation of energy (ii) conservation of momentum (iii) conservation of angular momentum (iv) conservation of charge 47. Fig. 2.35 shows part of a closed circuit. What is the value of VA − VB ?
20 Ω
Ω 7Ω
A
Ω
3
2Ω
(iii) conservation of momentum (iv) conservation of angular momentum 49. A wire of resistance 0·1 Ω/cm is bent to form a square ABCD of side 10 cm. A similar wire is connected between B and D to form the diagonal BD. If a 2 V battery of negligible internal resistance is connected between A and C, then total power dissipated is (i) 2 W (ii) 3 W (iii) 4 W (iv) 6 W 50. A cell supplies a current of 0·9 A through 2 Ω external resistance and 0·3 A through 7 Ω external resistance. The internal resistance of the cell is (i) 0·5 Ω (ii) 1·2 Ω (iii) 1 Ω (iv) 1·5 Ω 51. Four resistances each of value 4 Ω are connected as shown in Fig. 2.36. The equivalent resistance between points A and B is 4Ω
20
(i) 0·5 A (ii) 2·5 A (iii) 1·5 A (iv) 1 A 43. Two batteries of different e.m.fs and same internal resistances are connected in series with an external resistance and current is 3 A. When polarity of one is reversed, the current is 1 A. The ratio of e.m.fs is (i) 2·5 (ii) 2 (iii) 1·5 (iv) 1 44. Five resistances are connected as shown in Fig. 2.34. The effective resistance between points A and B is
B 20 Ω P
S
20 Ω Q
30 Ω
Fig. 2.37 (i) 6 Ω (iii) 18 Ω
(ii) 12 Ω (iv) 24 Ω
D. C. Circuits
39
53. Eight resistors each of resistance 10 Ω are connected as shown in Fig. 2.38. The resistance between points A and B is 10 Ω
10 Ω
10 Ω
A
10 Ω
10 Ω 10 Ω
10 Ω
10 Ω
B
(i) 3 A (ii) 6 A (iii) 9 A (iv) 12 A 57. In Fig. 2.40, the potential difference across 8 Ω resistor is (i) 48 V (ii) 24 V (iii) 960 V (iv) 96 V 58. Eight resistances each of resistance 10 Ω are connected as shown in Fig. 2.41. The resistance between points A and B in the circuit is 10 Ω
Fig. 2.38
10 Ω
10 Ω
A
(i) 30 Ω (ii) 60 Ω (iii) 45 Ω (iv) 90 Ω 54. A 50 V battery is connected across 10 Ω resistor. The current is 4·5 A. The internal resistance of the battery is (i) zero (ii) 5 Ω (iii) 0·5 Ω (iv) 1·1 Ω 55. What is the equivalent resistance between the terminals A and B in Fig. 2.39 C
R
10 Ω
10 Ω
10 Ω
3Ω
3Ω
R
3Ω
A
B
3Ω
D
6Ω 6Ω
R
Fig. 2.39 (i) 2R (ii) 3R (iii) R (iv) R/3 56. It is known that the potential difference across 6 Ω resistor in Fig. 2.40 is 48 V. The entering current I is a+
I 6Ω
12 Ω
48 V
A
0·5A
6Ω 6Ω
8Ω
–
Fig. 2.40
6Ω 6Ω
6Ω
10 Ω
30 Ω
15 Ω
d
B
3Ω
Fig. 2.42 (i) 3 Ω (ii) 2 Ω (iii) 5 Ω (iv) 6 Ω 60. In the circuit shown in Fig. 2.43, find the potential difference VP − VQ
P
c
3Ω
6Ω
6Ω
b
B
Fig. 2.41 (i) 30 Ω (ii) 60 Ω (iii) 45 Ω (iv) 90 Ω 59. Fig. 2.42 shows a network of resistances. The effective resistance between points A and B of the network is
R
R
10 Ω
10 Ω
Fig. 2.43 (i) 3·6 V (iii) 3·0 V
(ii) 6·0 V (iv) 7·2 V
Q
40
61. In the circuit shown in Fig. 2.44, the galvanometer reads zero. The value of resistance R is
(i) 42 Ω (ii) 28 Ω (iii) 21 Ω (iv) 14 Ω 62. Five resistances are connected as shown in Fig. 2.45. The equivalent resistance between points A and B is
(i) 10 Ω (ii) 20 Ω (iii) 5 Ω (iv) 30 Ω 63. The equivalent resistance of the arrangement of resistances shown in Fig. 2.46 between points A and B is
(i) 6 Ω (ii) 8 Ω (iii) 16 Ω (iv) 24 Ω 64. An ordinary dry cell can deliver a current of about (i) 3 A (ii) 2 A (iii) 1/8 A (iv) none of above
Objective Electrical Technology
65. Four cells, each of internal resistance 1 Ω, are connected in parallel. The battery resistance will be (i) 4 Ω (ii) 0·25 Ω (iii) 2 Ω (iv) 1 Ω 66. In the circuit shown in Fig. 2.47, the reading of ammeter is (internal resistance of battery is zero)
(i) 40/29 A (ii) 10/9 A (iii) 5/3 A (iv) 2 A 67. In the circuit shown in Fig. 2.48, find the potential difference across cells E1 and E2
(i) 3·75 V, 7·5 V (ii) 4·25 V, 7·5 V (iii) 3·75 V, 3·75 V (iv) 4·25 V, 4·25 V 68. Three resistances of 4 Ω, 6 Ω and 10 Ω are connected in parallel in a circuit with a battery e.m.f. of 4·53 V. [See Fig. 2.49] If current through 6 Ω resistance is 0· 6 A, the internal resistance of the battery is
(i) 0·2 Ω (iii) 0·4 Ω
(ii) 0·3 Ω (iv) 0·5 Ω
D. C. Circuits
41
69. Referring to Fig. 2.50, the resistance across terminals AE is 13 Ω
A
22
11 Ω
B
Ω
44V
18 Ω
F
14
Ω
9Ω
E
C
5Ω
D
Fig. 2.50 (i) 9 Ω (ii) 18 Ω (iii) 11 Ω (iv) none of above 70. A battery of e.m.f. 2 V and internal resistance 0·5 Ω is connected to a circuit consisting of a resistance of 5 Ω and a galvanometer of resistance 15 Ω connected in series (See Fig. 2.51). When a resistance of 3 Ω is connected across the terminals of the galvanometer, what is the current through the galvanometer ?
(iii) it will increase (iv) data is insufficient 72. In the above question, if the battery has finite value of internal resistance, then what will be the effect on galvanometer reading ? (i) it will not change (ii) it will increase (iii) it will decrease (iv) data is insufficient 73. The current in a coil of resistance 90 Ω is to be reduced by 90%. What value of resistance should be connected in parallel with it ? (i) 9 Ω (ii) 100 Ω (ii) 90 Ω (iv) 10 Ω 74. If a battery of 6 V is applied across terminals 1 and 2 in Fig. 2.53, then current in the horizontal 2 Ω resistor will be 1Ω
2
2V
Ω
1Ω
1Ω
0.5 Ω I 5Ω
Ig
1
15 Ω
Fig. 2.53
G 3Ω
Fig. 2.51 (i) 0·04 A (ii) 1·40 A (iii) 0·40 A (iv) 4·1 A 71. In the circuit shown in Fig. 2.52, the resistance R is increased. What will be the effect on the galvanometer reading if the internal resistance of the battery is zero ?
2
2Ω
(i) 1 A (ii) 2 A (iii) 3 A (iv) 0·5 A 75. The potential difference between points A and B in Fig. 2.54 will be A
5Ω
5Ω
2V + –
5Ω
B 5Ω
G R
(i) it will not change (ii) it will decrease
5Ω
5Ω
Fig. 2.54 (i) 2/3 V (ii) 3 V (iii) 8/9 V (iv) 4/3 V 76. In order to get maximum current in seriesparallel grouping of cells, the external resistance should be ......... the total internal resistance of the battery.
42
Objective Electrical Technology
(i) less than (ii) more than (iii) equal to (iv) none of above 77. Two equal resistances are connected in series across a certain supply. If the resistances are now connected in parallel, the power produced will become (i) two times (ii) four times (iii) one-half (iv) none of above 78. In Fig. 2.55, the switches S1 and S2 are closed. Then total circuit resistance will be 100 W, 200V 40 W, 200V
L2 S1 L3
L1 S2
+
200 V
100 W, 200V
–
e.m.f. 10 V and negligible internal resistance is connected across two resistances of 500 Ω in series. A voltmeter of 1000 Ω resistance is connected across one resistance. What is the reading of the voltmeter ? (i) 4 V (ii) 2 V (iii) 3 V (iv) 1 V 82. Voltmeters V1 and V2 are connected in series across a d.c. line. The voltmeter V1 reads 80 V and has per volt resistance of 200 Ω while V2 has a total resistance of 32 kΩ. The line voltage is (i) 120 V (ii) 160 V (iii) 220 V (iv) 240 V 83. Fig. 2.57 shows a part of a closed circuit. The potential difference between points A and B is 8V 6V 3A
Fig. 2.55
6Ω
3Ω
B
A
(i) 400 Ω (ii) 1200 Ω (iii) 1000 Ω (iv) 2400 Ω 79. In Fig. 2.55, both switches S1 and S2 are closed. Then (i) L1 will be brighter than L2 or L3 (ii) L1 will be dimmer than L2 or L3 (iii) L1 will be as bright as L2 or L3 (iv) none of the above 80. In Fig. 2.55, switches S1 and S2 are closed and the supply voltage is increased to 400 V. Then, (i) lamp L1 will burn out (ii) lamp L2 will burn out (iii) both lamps L2 and L3 will burn out (iv) all the lamps will be safe 81. In the circuit shown in Fig. 2.56, a battery of
Fig. 2.57 (i) 12 V (ii) 24 V (iii) 18 V (iv) 29 V 84. In the Wheatstone bridge shown in Fig. 2.58, P = 9 Ω ; Q = 11 Ω, R = 4 Ω and S = 6 Ω. How much resistance must be put in parallel to resistance S to balance the bridge ? B P A
Q C
G
S
R D
10 V
Fig. 2.58 I 500 Ω
500 Ω
I1 V 1000 Ω
Fig. 2.56
(i) 24 Ω (ii) 26·4 Ω (iii) 15 Ω (iv) 11 Ω 85. A galvanometer together with an unknown resistance in series is connected across two identical batteries each of 1·5 V. When the batteries are connected in series, the galvanometer records a current of 1 A and
D. C. Circuits
43
when the batteries are connected in parallel, the current is 0·6 A. What is the internal resistance of each battery ? (i) 1 Ω (ii) 0·5 Ω (iii) 1/3 Ω (iv) 1·5 Ω 86. Six equal resistances, each of 2 Ω, are connected as shown in Fig. 2.59. The resistance between any two corners is C
Ω
2Ω
2Ω
2
2
2Ω
A
Ω
B
2Ω
Fig. 2.59 (i) 1 Ω (ii) 4 Ω (iii) 2 Ω (iv) 8 Ω 87. Resistances 1 Ω, 2 Ω and 3 Ω are connected in the form of a triangle. If a cell of e.m.f. 1·5 V and negligible internal resistance is connected across 3 Ω, the current through this resistor is (i) 0·25 A (ii) 0·5 A (iii) 1·5 A (iv) 2·5 A 88. The current in 2 Ω resistor shown in Fig. 2.60 is
R
R
R
R
R
A
Ω
1.4 A
25 Ω
220 V
2
Ω 10
section 2·5 mm2 normally in 2 seconds. The current is (i) 25 mA (ii) 5 mA (iii) 20 mA (iv) 4 mA 91. Two similar cells whether joined in series or in parallel have the same current through an external resistanc of 2 Ω. The internal resistance of each cell is (i) 2 Ω (ii) 1 Ω (iii) 0·5 Ω (iv) 1·5 Ω 92. When resistances are connected in parallel, the current divides itself in (i) direct ratio of resistances (ii) inverse ratio of resistances (iii) inverse ratio of potentials (iv) none of above 93. When a d.c. battery of e.m.f. E and internal resistance r delivers maximum power to an external resistance R, the ratio r/R is (i) 1 : 1 (ii) 2 : 1 (iii) 1 : 2 (iv) 1 : 1/2 94. Five identical lamps each of resistance R = 1100 Ω are connected to a 220 V supply as shown in Fig. 2.61. The reading of ideal ammeter is
5Ω
Fig. 2.60 (i) 1·2 A (ii) 0·4 A (iii) 1·4 A (iv) 1 A 89. A wire has a resistance of 6 ohms. It is bent in the form of an equilateral triangle. The effective resistance between any two corners of the triangle is (i) 4·7 Ω (ii) 6 Ω (iii) 4/3 Ω (iv) 3 Ω 90. A charge of 10 mC flows through a cross-
Fig. 2.61 (i) 1/5 A (ii) 2/5 A (iii) 3/5 A (iv) 1 A 95. A torch bulb rated at 4·5W, 1·5V is connected as shown in Fig. 2.62. The e.m.f. of the cell 4.5 W, needed to make the 1.5 V bulb glow at full intensity is I1 1Ω (i) 4·5 V (ii) 1·5 V (iii) 2·56 V I E, r = 2.67 Ω (iv) 13·5 V Fig. 2.62
44
Objective Electrical Technology
96. Two bulbs rated at 25 W, 110 V and 100 W, 99. A standard 40 W tube light is in parallel with 110 V are connected in series to a 220 V a room heater and both are connected to main supply. What will happen to the circuit ? supply line. What will happen when light is switched off ? (i) 100 W bulb will burn out (i) the heater output will increase (ii) 25 W bulb will burn out (ii) the heater output will decrease (iii) both bulbs will burn out (iii) the heater output will remain same (iv) no bulb will burn out (iv) none of above 97. Three equal resistors connected in series across a source of e.m.f. together dissipate 100. Twelve wires, each of resistance R, are 10 W. What would be the power dissipated connected to form a cube. The effective if the same resistors are connected in parallel resistance between two diagonal ends across the same source of e.m.f. ? (points 1 and 7 in Fig. 2.63) will be 6 7 (i) 90 W (ii) 30 W (iii) 180 W (iv) 27 W 3 2 98. In a circuit, two cells of 1·5 V and 2 V e.m.f. having internal resistances of 1 Ω and 2 Ω respectively are connected in parallel so as to send current in the same direction through 8 5 an external resistance of 5 Ω. The circuit current is 4 1 (i) 17 A (ii) 12 A Fig. 2.63 5 5 5R 6R (i) (ii) 5 6 5 (iii) (iv) 5 A A 17 12 (iii) 3R (iv) 12R
ANSWERS TO OBJECTIVE QUESTIONS 1. 6. 11. 16. 21. 26. 31. 36. 41. 46. 51. 56. 61. 66. 71. 76. 81. 86.
(i) (iii) (iv) (iv) (iii) (iv) (iii) (i) (iv) (iv) (i) (iv) (ii) (iv) (i) (iii) (i) (i)
2. 7. 12. 17. 22. 27. 32. 37. 42. 47. 52. 57. 62. 67. 72. 77. 82. 87.
(ii) (ii) (iv) (ii) (iv) (ii) (iv) (iv) (i) (ii) (ii) (iv) (iii) (ii) (ii) (ii) (iv) (ii)
3. 8. 13. 18. 23. 28. 33. 38. 43. 48. 53. 58. 63. 68. 73. 78. 83. 88.
(iii) (i) (iii) (ii) (ii) (iii) (iii) (iii) (ii) (i) (i) (i) (ii) (iv) (iv) (ii) (iv) (iv)
4. 9. 14. 19. 24. 29. 34. 39. 44. 49. 54. 59. 64. 69. 74. 79. 84. 89.
(iv) (iii) (i) (iv) (i) (ii) (ii) (ii) (i) (iii) (iv) (ii) (iii) (iii) (iii) (i) (ii) (iii)
5. 10. 15. 20. 25. 30. 35. 40. 45. 50. 55. 60. 65. 70. 75. 80. 85. 90.
(i) (i) (ii) (i) (iii) (ii) (ii) (ii) (iii) (i) (iii) (iii) (ii) (i) (iv) (i) (iii) (ii)
D. C. Circuits
91. (i) 96. (ii)
45
92. (ii) 97. (i)
93. (i) 98. (iii)
94. (iii) 99. (iii)
95. (iv) 100. (i)
HINTS TO SELECTED OBJECTIVE QUESTIONS 7. The resistance of 40 W lamp is much more (2·5 times) than that of 100 W lamp. Consequently, a greater part of supply voltage will appear across 40 W lamp. 10. So far as terminals AB are concerned, 12 Ω and 6 Ω are in series and this series combination is in parallel with 18 Ω resistor. ∴ RAB = (12 + 6) × 18 = 18 × 18 = 9 Ω (12 + 6) + 18 36 V = 180 = 20 A Circuit current, I = 9 RAB Since there are two parallel paths of equal resistance, current in each path is 10 A. 8×8 11. REF = 6 × 6 = 3 Ω ; RCD = RAB = =4Ω 8+8 6+6 24 = 24 ∴ Circuit current = =6A RAB 4 Current in 5 Ω = 6/2 = 3 A Current in 2 Ω = 3/2 = 1·5 A 6Ω 15. Fig. 2.64 shows the conditions of the problem. Each parallel path has a resistance of 6 Ω. ∴ Effective resistance between two points on the diameter
6 × 6 36 = 6 || 6 = 6 + 6 = 12 = 3 Ω 17.
Circuit current,
I =
E = 2 ⋅0 = 2 ⋅ 0 = 0·5 A R + r 3 ⋅ 9 + 0 ⋅1 4
∴ Voltage across cell, V = E − Ir = 2·0 − 0·5 × 0·1 = 1·95 V 18. Conductance G is the reciprocal of resistance R i.e. G = 1/R
1 RT or 20. ∴ 24. or
∴
1 1 1 = R + R + R 1 2 3
G = G1 + G2 + G3 =
1 +1+1 = 1·6 S 10 2 1
G = G1 + G2 + G3 =
1 + 1 +1 =1S 10 2 ⋅ 5 2
15 A V = IRT = I = = 15 V G 1S RT = R1 + R2 RT R R = 12 + 22 2 V V V 1 = 1 + 1 P1 P2 PT P1P2 200 × 100 = = 66 W PT = P1 + P2 200 + 100
6Ω
Fig. 2.64
46
Objective Electrical Technology
29. It may be seen that one end of each resistor is connected to point A and the other end of each resistor is connected to point B. Hence the three resistors are in parallel. 1 1 + 1 + 1 = 3 ∴ = RAB R R R R or RAB = R/3 32. The effective resistance between points E and F is (10 + 10) × 10 200 20 REF = (10 + 10) + 10 = 30 = 3 Ω 20 80 + 10 = Ω ∴ RBC = 10 + 3 3 34. In the steady state, the capacitor offers infinite resistance to direct current. Therefore no current flows in the branch containing r3 and C. As a result, the potential difference across C is that across r2. V Current through r2 = r1 + r2 P.D. across r2 =
FG V IJ × r = V r Hr + r K r + r 2
2
1
2
1
2
35. Let E and r be the e.m.f. and internal resistance of each cell. For series grouping, 2E = (5 + 2r) × 10 ...(i) For parallel grouping, E = 5+ r ×8 ...(ii) 2 Solving eqs (i) and (ii), we get, r = 2·5 Ω 37. Applying Kirchhoff’s voltage law to the circuit, we have, VA + 3 − 3 × 1 − 3 × 6 = VB or VA + 3 − 3 − 18 = VB ∴ VA − VB = 18 V 39. Current in branch DAC as well as in branch DBC is 1 A. ∴ VD − 2 × 1 = VA and VD − 3 × 1 = VB ∴ VA − VB = (VD − 2) − (VD − 3) = + 1 V 40. When points B and D are at the same potential, this Wheatstone bridge is balanced. Under balanced conditions of the bridge, the products of resistances of opposite arms are equal.
FH
IK
FG H
IJ K
(3 + 1)X = (12 + 4) × 1 × 1 1+1 or X = 2Ω 42. We consider the ammeter to be ideal so that its resistance is zero. ∴ Total circuit resistance is 0⋅4 × 0⋅4 RT = 0·2 + 4 + = 0·2 + 4 + 0·2 = 4·4 Ω 0⋅4 + 0⋅4 E = 2⋅2 = 0⋅5 A Circuit current, I = RT 4 ⋅ 4 Therefore the reading of ammeter will be 0·5 A. 43. Let E1, E2 be the e.m.fs of the two cells and r be the internal resistance of each cell. If R is the external resistance then, E1 + E2 E1 − E2 = 3 and =1 R+r R+r ∴
D. C. Circuits
47
E1 + E2 = 3 E1 − E2 On solving, E1/E2 = 2 R = r/n ∴ r = n R When n resistors, each of resistance r (= n R) are connected in series, then total resistance is 2 R′ = n × n R = n R Let E and r be the e.m.f. and internal resistance of the cell respectively. For the first case, 0·9 = E ...(i) 2+r E For the second case, 0·3 = ...(ii) 7+r Solving eqs (i) and (ii), we get, r = 0·5 Ω Here PRQSP is a balanced Wheatstone bridge. Therefore branch RS is ineffective and can be considered as removed. 20 × 30 = 12 Ω ∴ RPQ = 20 Ω | | 30 Ω = 20 + 30 The network shown in Fig. 2.39 can be redrawn as shown in Fig. 2.65 (i). It is a balanced Wheatstone bridge. Therefore, point C and D are at the same potential. Since no current flows in the branch CD, this branch is ineffective in determining the equivalent resistance between terminals A and B and can be removed. The circuit then reduces to that shown in Fig. 2.65 (ii). ∴
45.
50.
52.
55.
C
C
R
R
R
A
R
R
B
R
A
B
R
R
R
D
D
(i)
(ii)
Fig. 2.65 The branch ACB (= R + R = 2R) is in parallel with branch ADB (= R + R = 2R), ∴
RAB =
(2 R) × (2 R) =R 2R + 2R
56. Current through 6 Ω resistor, I1 = 48/6 = 8 A Current through 12 Ω resistor, I2 = 48/12 = 4 A ∴ Entering current, I = I1 + I2 = 8 + 4 = 12 A 61. It is a balanced Wheatstone bridge. Therefore product of resistances of opposite arms is equal i.e. 40 × 7 = 10 × R 40 × 7 = 28 Ω or R = 10 66. We consider the ammeter to be ideal so that its resistance is zero. 4 × 5 20 Total circuit resistance = = Ω 4+5 9 10 90 9 = = A Circuit current = 20 / 9 20 2
48
Objective Electrical Technology
Current through 5 Ω resistor = 9 × 4 = 9 × 4 = 2 A 2 4+5 2 9 Therefore, ammeter reading will be 2 A. 70. The equivalent resistance of parallel combination of 15 Ω and 3 Ω is 15 × 3 45 Rp = 15 + 3 = 18 = 2 ⋅ 5 Ω Total circuit resistance, RT = 0·5 + 5 + 2·5 = 8 Ω E = 2 = 0 ⋅ 25 A Total circuit current, I = RT 8 Current through galvanometer, Ig = I × 3 = 0 ⋅ 25 × 3 = 0 ⋅ 04 A 15 + 3 18 S 73. The conditions of the problem are represented in Fig. 2.66. The voltage across the required resistance S and 90 Ω is 0.9 I the same i.e. 0·1 I
0·9 I × S = 0·1I × 90 or S = 90/9 = 10 Ω 77. Suppose each resistance is R.
I
90 Ω
Fig. 2.66
2
For series connection,
V 2R 2 V = R/2
P1 =
For parallel connection, P2
(∵ RT = R + R = 2R)
FH∵ R
T
= R 2
IK
P2 = 4 or P2 = 4P1 P1 80. The resistance of each 100 W lamp is 400 Ω. The total resistance of the parallel circuit will be 200 Ω. The resistance of 40 W lamp is 1000 Ω. The applied voltage of 400 V will appear across 40 W lamp and the parallel circuit in the ratio of 1000 : 200 = 5 : 1. Consequently the voltage across 40 W lamp will far exceed 200 V and it will burn. 81. Resistance of parallel resistances of 500 Ω and 1000 Ω is 1000 × 500 1000 Rp = 1000 + 500 = 3 Ω Total circuit resistance is 1000 + 2500 Ω 500 = RT = 3 3 E = 10 = 3 A Circuit current, I = R 2500 / 3 250 T Current in 500 Ω resistor across which voltmeter is connected is 1000 = 3 × 1000 = 1 A I1 = I × 500 + 1000 250 1500 125 P.D. across 500 Ω resistor = I1 × 500 = 1 × 500 = 4 V 125 Therefore, the voltmeter will read 4 V. 82. Resistance of voltmeter V1 is R1 = 80 × 200 = 16000 Ω = 16 kΩ V 80 V −3 = 5 mA = 5 × 10 A Circuit current, I = 1 = R1 16 kΩ Reading of V2 = IR2 = 5 mA × 32 kΩ = 160 V ∴ Line voltage = V1 + V2 = 80 + 160 = 240 V ∴
D. C. Circuits
49
87. Fig. 2.67 shows the conditions of the problem. Total circuit resistance is (2 + 1) × 3 RT = (2 + 1) + 3 = 1 ⋅ 5 Ω E 1⋅ 5 Circuit current, I = R = 1⋅ 5 = 1 A T Current in 3 Ω resistor is
3 = 1 × 3 = 0⋅5 A 3+3 6 91. Let E and r be the e.m.f. and internal resistance of each cell respectively. 2E E ∴ = 2 + 2r 2+r/2 2 1 or = 2 + 2r 2+r/2 On solving, we get, r = 2Ω I1 = I ×
(1 ⋅ 5) Rb = V = = 0 ⋅5 Ω 4 ⋅5 P 4 ⋅5 Rated current of bulb, I1 = =3A 1⋅ 5 In order that bulb glows at full intensity, 3 A must pass through it. 0⋅5 × 1 Total circuit resistance, RT = 2 ⋅ 67 + = 2·67 + 0·33 = 3 Ω 0⋅5 + 1 E =E ∴ Current supplied by battery, I = RT 3 I1 = I × 1 = E× 1 1 + 0 ⋅ 5 3 1⋅ 5 or 3 = E× 1 3 1⋅ 5 or E = 3 × 3 × 1·5 = 13·5 V 98. Apply Kirchhoff’s voltage law to solve the problem. 100. Symmetry about entrance point 1 and exit point 7 shows that points 2, 4 and 5 are at the same potential. Similarly points 3, 6 and 8 are at the same potential. Therefore the circuit reduces to the one shown in Fig. 2.68. 2
2
95. Resistance of bulb,
R R
R
R
R
R
R
2,4,5
1 R
R R
3,6,8
7 R
R
Fig. 2.68 ∴ Resistance between points 1 and 7 = R + R + R = 5R 3 6 3 6
GO To FIRST
