Tiêu đề 1. Current Electricity (S.C.Q.) E.pdf ✅

PHYSICS Q.1 When a galvanometer is shunted with a 4 resistance, the deflection is reduced to one-fifth. If the galvanometer is further shunted with a 2 wire. the further reduction (find the ratio of decrease in current to the previous current) in the deflection will be (the main current remains the same) (A) (8/13) of the deflection when shunted with 4  only (B) (5/13) of the deflection when shunted with 4  only (C) (3/4) of the deflection when shunted with 4  only (D)(3/13) of the deflection when shunted with 4  only [A] Sol. Case I Rg × 5 I =       − 5 I I × 4  Rg = 16  Case II G 2 I I1 Rg = 16 4 16 I1 = 6 42 (I–I1)  I1 = I/13 So decrease in current to previous current 13 8 I/5 I/5 I/13 = − = Q.2 Three batteries of emf 1V and internal resistance 1  each are connected as shown. Effective emf of combination between the points PQ is – 1  P 1  1  1 V 1 V 1 V Q R (A) Zero (B) 1V (C) 2V (D) 3 2 V [A] Sol. 2  2V 1  1V Q P Enet = 1 2 1 2 2 1 r r E r – E r + = 2 1 2 – 2 + = 0 Q.3 In an experiment according to set up, when E1 = 12 volt and internal resistance zero, E = 2 volt. The galvanometer reads zero, then X would be- G E D C E1 500  A B X  (A) 200 (B) 500 (C) 100 (D) 10 [C] Sol. Voltage across X is 2V So         500+X E1 X = 2          500+X 12 X = 2  12 X = 1000 + 2X  10X = 1000 X = 100 Q.4 The resistances in wheat stone's bridge circuit shown in the fig. have different values. The current through, the galvanometer is zero. If all the thermal effects are negligible. The current through the galvanometer will not be zero. When – R1 G R2 R3 R4 (A) the battery emf is doubled (B) the battery and galvanometer are interchanged (C) all resistance in the circuits are doubled (D) resistance R1 and R2 are interchanged [D]
Q.5 Equivalent resistance between the points A and B is – 4  4  4  4  4  A B (A) 1 (B) 2 (C) 3 (D) 4  Sol. 4 A B 4 4 4 4 Q. 6 The space between two coaxial cylinders whose radii are a and b (where a < b) as shown in figure, is filled with a conducting medium. The specific conductivity of the medium is .  = r 0 , where r is radial distance from common axis of both cylinder. Assuming L >> b, where L is the length of cylinder, then resistance of system is - b a L  (A) 4 L 0 b a   − (B) 2 L 0 b a   − (C) L 0 b a   − (D) L 0 6 b a   −  Q. 7 If ammeter has zero resistance then – 2R R R R A  (A) Reading of ammeter is 6R  (B) Reading of ammeter is 7R  (C) Reading of ammeter is 8R  (D) Reading of ammeter is 9R   Q.8 In the circuit shown, the coil has inductance and resistance. When X is joined to Y, the time constant is  during growth of current. When the steady state is reached, rate of production of heat in the coil is "P" joule/sec. X is now joined to Z, and after long time of joining X to Z – Y X Z (A) the total heat produced in the coil is P (B) the total heat produced in the coil is 2 1 P (C) the total heat produced in the coil is 2P (D) the data given is not sufficient to reach a conclusion Sol.1[B] Let L and R be the inductance and the resistance of the coil respectively Let E = e.m.f. of the cell.  = Time constant, I0 = E/R P = 2 0 I R = R E 2 Energy stored in the coil = 2 1 2 0 LI = 2 1 L         2 2 R E = 2 1       R L         R E 2 = 2 1 P = total heat produced in the coil Q. 9 All the bulbs below are identical. Which bulb(s) light(s) most brightly ? 1 2 3 4 5 (A) 1 only (B) 2 only (C) 3 and 4 (D) 1 and 5 [D]
¿Q. 10 In the given circuit, the voltmeter reading is 4.5 V. Assuming that the voltmeter is ideal, current through 12 resistance is - V 3 6 9 10 12 15 2 20 I I (A) 1 A (B) 0.5 A (C) 0.25 A (D) 0.1 A Sol. [B] The voltmeter is ideal, its resistance Rv →  Fig. shows the current distribution in the circuit . Voltmeter will not draw any current. V 3 6 9 10 12 15 2 20 I I I1 I–I1 I–I1 I1 O A I1 ' 1 I ' 1 I ' 1 I '' 2 I '' 2 I '' 1 I '' 1 I ''' 2 I ''' 2 I I1 ' 2 I ' 2 I I1 I1 Potential difference across 9  resistance = 4.5 V (given) Hence, current in 9 resistance = 9 4.5 = 0.5 A (I = R V ) i.e., ' 1 I = 0.5 A The same current ( ' 1 I ) passes through 3. Obviously, 9  and 3  are in series and their equivalent, i.e., 12 is in parallel with 6  between A and B. Dividing the current in the inverse ratio of resistances between A and B, '' 1 ' 1 I I = 12 6 = 2 1 '' 1 I = 2 ' 1 I = 2 × 0.5 = 1 A and I1 = ' 1 I + '' 1 I = 0.5 + 1 = 1.5 A at junction C, I1 divides into three parts. Since the resistances 10, 12, 15  are in parallel between C and D, current will distribute in the inverse ratio of resistances.  ' 2 I : '' 2 I : ''' 2 I = 10 1 : 12 1 : 15 1 = 6 : 5 : 4 ' 2 I = 15 5 × 1.5 = 0.5 amp So ' 2 I = 6k, '' 2 I = 5k, ''' 2 I = 4k (k being a constant of proportionality) and I1 = ' 2 I + '' 2 I + ''' 2 I = 15k but I1 = 1.5 A  15 k = 1.5 or k = 0.1 so '' 2 I = 5k = 0.5 A Thus current through 12 resistance is 0.5 A Q. 11 An accumulator battery B of e.m.f E and internal resistance r being charged from a DC supply whose terminals are T1 and T2 The connecting wires have uniform resistance. D.C. supply T1 T2 B E,r + – Moving from T1 to T2 through B, the potential V is plotted against distance x. The correct curve is - (A) >E x V (B) >E x V (C) >E x V (D) >E x V [A] Q.12 In the circuit shown, each resistance is 2. The potential V1 as indicated in the circuit, is equal to – 12V 5V V1 (A) 11 V (B) – 11V
(C) 9 V (D) – 9 V Sol. [D] i = 7 7V = 1 A Current flows in anticlockwise direction in the loop. Therefore 0 – 1 × 2 – 1 × 2 – 5 = V1 V1 = – 9V Q.13 Two resistances, a 60 ohm and an unknown one are connected to a power source in a series arrangement. This way the power of the unknown resistance is 60 watt. What is the least voltage of the power source ? (A) 60 Volt (B) 120 Volt (C) 140 Volt (D) 180 Volt Sol. [B] 60 R V V1 V2 2 1 V V = R 60 ........(1) V1 + V2 = V .........(2) from (1) & (2)  V2 = 60 R VR + ...........(3) R V 2 2 = 60 watt ........(4) from (3) & (4) V2R = 60 (60 + R)2 V2R = 60(3600 + R2 + 60R) 0 = 60R2 + 3600 R – V2R + 60 × 3600 (3600 – V2 – 60 × 60 × 2) (3600 – V2 + 60 × 60 × 2) > 0 (3 × 3600 – V2 ) (–3600 – V2 ) > 0 3 × 3600 – V2 < 0 60 × 3600 < V2 60 × 3 < V Q. 14 If all meters are ideal and reading of voltmeter 3 is 6V. Power supplied by voltage source is - A V A A 1 5 2 20 4 V 30 10 3 Voltge source (A) 10 Watt (B) 38 Watt (C) 20 Watt (D) 30 Watt Sol. [D] Current through 20W = 20 6 current through 10W = 10 6 Total current supplied by voltage source = 10 6 + 20 6 = 27 Volt Voltage of battery = 6 + 27 = 33 volt Power supplied = 10 9 × 33 = 29.7 or 30 Watt Potential difference across 30 W =       + 20 6 10 6 × 30 Q.15 Eight resistances each of resistance 5 are connected in the circuit as shown in figure. The equivalent resistance between A and B is – A B (A)  3 8 (B)  3 16 (C)  7 15 (D)  2 19 Sol. [A] The given circuit can be redrawn as A B A B 40/3 10 5   RAB =  3 8