Current Electricity-02-OBJECTIVE SOLVED

OBJECTIVE SOLVED 1. n identical cells, each of emf  and internal resistance r, are joined in series to form a closed circuit. One cell (A) is joined with reversed polarity. The potential difference across each cell, except A, is (a) 2 (b) n 1  n n (c) n  2  (d) 2n  . n Ans. (a) Solution: See the figure i  (n  2) nr VB  VA  ir      (n  2) r   1  n  2   2 . n  2 i A  r B nr  n  n 2. An ammeter and a voltmeter are joined in series to a cell. Their readings are A and V respectively. If a resistance is now joined in parallel with the voltmeter. (a) both A and V will increase (b) both A and V will decrease (c) A will decrease, V will increase (d) A will increase, V will decrease. Ans. (d) Solution: When a resistance is joined in parallel with the voltemeter, the total resistance of the circuit decreases. Current will increase and ammeter reading will increase. Potential difference across the ammeter increases thus potential difference across voltmeter decraeses. 3. An ideal cell is connected to a capacitor and a voltmeter in series. The reading V of the voltmeter (added in parallel with resistor) is plotted against time. Which of the following best represents the resulting curve? (a) V (b) V (c) V (d) V . Ans. (b) Solution: This is basically an RC circuit, charging from a cell. The resistance (R) of the voltmeter is the resistance in the circuit. The voltage across R = circuit current  R = reading of the voltmeter (V). Thus the nature of the V-t curve is the same as the nature of the I-t curve. 4. A capacitor of capacitance C has charge Q. It is connected to an identical capacitor through a resistance. The heat produced in the resistance is (a) Q2 Q2 2C (b) 4C Q2 (c) Ans. (b) 8C (d) dependent on the value of the resistance. Solution: As the capacitors are identical, they will finally have charge Q/ 2 each. Initial energy of the system   2 Ei  2C .  (Q / 2)2  Q2 Final energy of the system Ef  2  2C   4C Heat produced = loss in energy   2  Ei  Ef  4C . 5. The charge on a capacitor decreases  times in time t, when it discharges through a circuit with a time constant  . (a) t   (b) t  ln  (c) t  (ln  1) (d) t  ln1  1  .    Ans. (b) Solution:   Q  Q e t /   Q /  et /   1  or et /    or   ln . 6. A milliammeter of range 10 mA and resistance 9 is joined in a circuit as shown. The metre gives full-scale deflection for current I when A and B are used as its terminals, i.e., current entres at A and leaves at B (C is left isolated). The value of I is 9, 10 mA 0.1 0.9 A B C Solution: or ig  10mA  0.01A VA  VB  (I  ig )0.1  ig  9.9 I  0.1  10i g I I or I  10  0.01  1A . A B 0.1 7. The charge flowing through a resistance R varies with time t as Q = at – bt2. The total heat produced in R by the time current ceases is (a) (c) Ans. (a) Solution: a3R 6b (b) a3R 2b (b) Q  at  bt2 i  dQ  a  2bt dt a3R 3b a3R b . i =0 for t  t0  a / 2b , i.e., current flow from t =0 to t  t0 . t0 a3R The heat produced   i2 R dt . Putting the value of i we get heat produced = 0 6b . 8. The effective resistance between points P and Q of the electri- cal circuit shown in the figure is (a) 2 Rr/(R + r) (b) 8R(R + r) / (3R + r) (c) 2r + 4R (d) 5R/2 + 2r. Ans. (a) P Q 2R 2R Solution: Let us connect P and Q through a battery as shown. About vertical line MN network is symmetric with symmetric elements marked as (I, I) (II, II) and (III, III) current in symmetrical element is same  P Q 2R N 2R As no current goes to MO at M and NO at N the connection between M and O, N and O has no significance thus effective network is P Q 2R N 2R 4R = P Q 4R  RPQ 2Rr = R  r . 9. A 100W bulb B1 , and two 60 W bulbs B2 and B3 , are connected B1 B2 to a 250 V source, as shown in the figure. Now W1 , W2 and W3 are the output powers of the bulbs Then B1 , B2 and B3 respectively. (a) (c) Ans. (d) Solution:  W1  W2  W3 W1  W2  W3 V 2 P  , P  Rated Power R P1  P2 R 2  R1 (b) (d) W1  W2  W3 W1  W2  W3 . 250V B1 and B2 in series current is same.  W2  W1 P2  P3  R 2  R 3 …(i) But voltage across B3 will be 250 V while across B2 less than 250 V hence W2  W3 …(ii) From (i) and (ii) W1  W2  W3 10. In the given circuit, with steady current, the potential drop across the capacitor must be (a) V (b) V/2 (c) V/3 (d) 2V/3. Ans. (c) Solution: In steady state, no current will pass through the capacitor. Moving along periphery of the circuit 2V  2iR  iR  V  0  i  V / 3R For the upper loop, V  Vc  iR  V  0  | VC | iR  V / 3 . 11. A wire of length L and 3 identical cells of negligible internal resistances are connected in series. Due to the current, the temperature of the wire is raised by T in a time t. A number N of similar cells is now connected in series with a wire of the same material and cross-section but of length 2L. the temperature of the wire is raised by the same amount T in the same time t. The value of N is (a) 4 (b) 3 (c) 8 (d) 9. Ans. (b) [(3)2 /(L / A)]  (LA)s T V2 Solution: [(N)2 /( 2L / A)] (2L A)s T (using R = msT )  = resistivity; s = specific heat capacity of material of the wire. A = area of cross section 9  1 N 2  2 2 N 2  9  N  3 . R3 12. In the given circuit, it is observed that the current I is independent of the value of the resistance R 6 . Then the resistance values must R4 satisfy (a) R R R  R R R (b) 1  1  1  1 1 2 5 3 4 6 R 5 R 6 R1  R 2 R 3  R 4 (c) Ans. (c) R1 R 4  R 2 R 3 (d) R1R 3  R 2 R 4  R 5 R 6 . Solution: For the current to be independent of balanced Wheatstone Bride. R 6 . the resistance R1 , R 2 , R 3 and R 4 must form a 13. A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is similarly charged to potential difference 2V. The charging battery is now disconnected and the capacitors are connected to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is (a) zero (b)  3 CV 2 2   (c)  25 CV 2 6 (d)  9 CV 2 . 2     Ans. (b) Solution: + + + –Q1=C1V1=CV – – Let initially charge on capacitors be Q1 and Q2 and final charge Q  – – and Q2 + – Q  Q  Q  Q …(i) + – 1 2 2 1 Finally the two capacitors will have the same potential (let V ). – – – – – Q2=C2V2= + – (2C(2V)=4CV V  Q1  Q2 …(ii) C1 C2 Solving (i) and (ii) V  V The electrostatic energy of the capacitors will be E  1 (C  C )V2  3 CV2 . 2 1 2 2 14. In the given circuit P  R , the reading of the galvanometer is same with switch S open or closed. Then (a) (c) Ans. (a) IR  IG IQ  IG (b) (d) IP  IG IQ  IR . Solution: Since the closing of switch S does not affect the current in the galvanometer, the current passing through G will be the same as that passing through the resistance R. Hence, IR  IG . 15. In the circuit shown in figure, the current through (a) the 8 left resistor is 0.50 A (b) the 8 right resistor is 0.25A 3 2 2 9V (c) the 4 (d) the 4 Ans. (a), (b), (d) resistance is 0.50 A resistor is 0.25A. 2 2 2 Solution: The net resistance of the circuit is 9 . The given circuit can be transformed into 2  2    The current flowing in the circuit I  V  9 A  1.0A    9V  2 R 9 The flow of current in the circuit is as follows: 0.5A 0.25A   16. Capacitor C1 of capacitance 1 microfarad and capacitor C 2 of capacitance 2 microfarad are separately charged fully by a common battery. The two capacitors are then separately allowed to discharge through equal resistors at time t  0 . (a) The current in each of the two discharging circuits is zero at t = 0 (b) The currents in the two discharging circuits at t = 0 are equal but not zero. (c) The currents in the two discharging circuits at t = 0 are unequal (d) Capacitor C1 loses 50% of its initial charge sooner than C 2 Ans. (b) and (d) Solution: Instantaneous charge on a capacitor, q  q et / RC  CVet / RC loses 50% of its initial charge. Instantaneous current i(t)  dq / dt  CV 1 et / RC  V et / RC At t = 0,  RC  R i 0  V / R Since V and R are the same for both capacitors, the initial current in the two is same and is non-zero. During discharge: instantaneous charge q(t)  q 0 et / RC Let q  q0 / 2 at t  t , Then q0  q et / RC 2 0 t  RC ln 2 If 1 and 2 are the times in which the two capacitors lose 50% of their charge. Then 1  RC1 ln 2  C1  1 2 RC2 ln 2 C2 2  1  2 / 2 . 17. Three resistances R, 2R and 3R are connected in parallel to a battery. Then (a) The potential drop across 3R is maximum (b) The current through each resistance is same (c) The heat developed in 3R is maximum (d) The heat developed in R is maximum. Ans. (d) Solution: In parallel combination, potential drop across each resistance is same. V 2 Heat developed R V 2 H1  R V 2 H 2  2R V 2 H3  3R  Heat developed in resistance R is maximum. j 18. A wire of resistance 10 is bent to form a circle. P and Q are the points on the circumference of the circuit dividing it into a quadrant and are connected to a 3V battery having internal resistance 1 as shown in the figure. The currents in the two parts of the circle are (a) 5 26 A and 15 26 A (B) 4 A and 25 12 A 25 (c) Ans. (d) 3 A and 9 A 25 25 (d) 6 23 A and 18 A . 23 Solution: Resistance of smaller section  1 10  2.5 4 Resistance of bigger section  3 10  7.5 4 The two resistances are in parallel. Resultant resistance  7.5  2.5  7.5  2.5  1.875  7.5  2.5 10 i  3  1  1.875 3 2.875  24 A 23 4 1.875 Current in smaller section  23  18 A 2.5 23 24 1.875 Current in bigger section  23  7.5 6 A . 23 19. A uniform wire has electric resistance R. The wire is cut into n equal parts. All wires are put parallel to each other and joined at the ends. The resistance of the combination is (a) R/n (b) R / n 2 (c) R (d) none of these. Ans. (b) Solution:  R   l A  R  l R Hence, resistance of each wire is n For the resistance of the combination, 1  n  n  n  n times R R R R  n  n R R  R . n 2 20. A wire of resistance R is stretched to double its length. Its new resistance is (a) R (b) R/2 (c) 4R (d) R/4. Ans. (c) Solution: Volume of the wire will remain the same. Let initially the length and radius be l1 and r1 respectively and after stretching the length and radius be l2 and r2 respectively. Then r 2l  r 2l 1 1 2 2 Given l2  2l1 Then r 2l  r 2 (2l ) 1 1 2 1 r 2  2r 2 1 2 R   11 A1 or R   l1 r 2 R   l2 A 2   l2 r 2 or R   (2l1 )  r 2   2    4l1 r 2   = 4R.