Q: Two resistances R 1 = X Ω and R 2 = 1 Ω are connected to a wire A B of uniform resistivity, as shown in the figure. The radius of the wire varies linearly along its

Resistance of frustum shaped conductor shown is R = ρ I π a b For the shown conductor in the diagram. r = a + b 2 = 0 . 2 + 1 2 = 0 . 6 mm Thus, the resistance of left half is P = ρ × 0 . 5 × 10 6 π × 0 . 2 × 0 . 6 And the resistance of right half is Q = ρ × 0 . 5 × 10 6 π × 0 . 6 × 1 For Wheatstone to be balanced R 1 P = R 2 Q ⇒ X π × 0 . 2 × 0 . 6 ρ × 0 . 5 × 10 6 = 1 &#

Q: In Circuit- 1 and Circuit- 2 shown in the figures, R 1 = 1 Ω , R 2 = 2 Ω and R 3 = 3 Ω . P 1 and P 2 are the power dissipations in Circuit- 1 and

When S 1 and S 2 are open For circuit 1 R eq 1 = 1 + 5 × 1 2 5 + 1 2 = 1 + 5 11 = 16 11 Ω P 1 = V 2 R eq = 6 2 16 × 11 = 36 × 11 16 = 24 . 75 W For circuit 2 R eq 2 = 6 11 Ω P 2 = V 2 R eq = 6 2 6 × 11 = 36 × 11 6 = 66 W Therefore, P 2 > P 1 Option A is correct ⇒ If 2 A source is used in both cases. P 1 = i 2 R eq 1 = 2 2 × 16 11 = 64 11 = 5 . 818

Question 1

A metal wire of cross-sectional area 0.5 mm^2 and length 100 m is connected across a battery of e.m.f. 2 V and internal resistance 1\ . The density, atomic mass and electrical conductivity of the meta

Accepted Answer

The number density of conduction electrons n is given by the number of atoms per unit volume, since there is one conduction electron per atom: n = d N_A M Substituting the given values: n = 6.35 10^3 6 10^ 23 63.5 10^ -3 = 6 10^ 28 m ^ -3 The resistance of the wire R is: R = L A R = 100 2 10^8 0.5 10^ -6 = 100 10^2 = 1\ The total resistance of the circuit is R_ total = R + r = 1 + 1 = 2\ . The current in the circuit is: I = E R_ total = 2 2 = 1 A The drift velocity v_d is given by the relation I

Question 2

As shown in the figure, the resistance of a galvanometer G can be found by the half-deflection method. Here the resistance R_2 is adjusted such that when the key K is closed the deflection in the galv

Accepted Answer

Let the resistance of the galvanometer be G and the series resistance be R_1. When the key K is open, the current through the galvanometer is: I_1 = V R_1 + G When the key K is closed, the total resistance of the circuit becomes: R_ eq = R_1 + G R_2 G + R_2 The total current from the battery is I = V R_ eq . The current through the galvanometer is: I_2 = I R_2 G + R_2 = V R_1 + G R_2 G + R_2 R_2 G + R_2 = V R_2 R_1(G + R_2) + G R_2 In the half-deflection condition, I_2 = I_1 2 : V R_2 R_1(G + R_

Question 3

Two resistances R 1 = X &#160; &#937;  and R 2 = 1 &#160; &#937;  are connected to a wire A B of uniform resistivity, as shown in the figure. The radius of the wire varies linearly along its

Accepted Answer

Resistance of frustum shaped conductor shown is R = &#961; I &#960; a b For the shown conductor in the diagram. r = a + b 2 = 0 . 2 + 1 2 = 0 . 6 &#160; mm Thus, the resistance of left half is  P = &#961; &#215; 0 . 5 &#215; 10 6 &#960; &#215; 0 . 2 &#215; 0 . 6 And the resistance of right half is  Q = &#961; &#215; 0 . 5 &#215; 10 6 &#960; &#215; 0 . 6 &#215; 1 For Wheatstone to be balanced R 1 P = R 2 Q &#8658; X &#960; &#215; 0 . 2 &#215; 0 . 6 &#961; &#215; 0 . 5 &#215; 10 6 = 1 &#

Question 4

In Circuit- 1 and Circuit- 2  shown in the figures, R 1 = 1 &Omega; , R 2 = 2 &Omega;  and  R 3 = 3 &Omega; . P 1  and P 2  are the power dissipations in Circuit- 1  and

Accepted Answer

When  S 1  and  S 2  are open For circuit 1 R eq 1 = 1 + 5 &#215; 1 2 5 + 1 2 = 1 + 5 11 = 16 11 &#160; &#937; P 1 = V 2 R eq = 6 2 16 &#215; 11 = 36 &#215; 11 16 = 24 . 75 &#160; W For circuit 2 R eq 2 = 6 11 &#160; &#937; P 2 = V 2 R eq = 6 2 6 &#215; 11 = 36 &#215; 11 6 = 66 &#160; W Therefore,  P 2 &#62; P 1 Option  A  is correct &#8658;   If  2  A source is used in both cases. P 1 = i 2 R eq &#160; 1 = 2 2 &#215; 16 11 = 64 11 = 5 . 818

Question 5

The figure shows a circuit having eight resistances of 1   &Omega; each, labelled R 1 to R 8 , and two ideal batteries with voltages &epsilon; 1 = 12   V and &epsilon; 2 = 6   V . Which

Accepted Answer

Point A  and B  are at same potential so they can be merged/folded. Now, let  x  and  y  be the current flowing from  6 &#160; V  and  12 &#160; V  batteries. The current distribution is shown below. Applying Kirchhoff's voltage law, For loop  1 : - 1 2 x - 1 2 x + y - 1 x + 6 = 0 &#8658; - 2 x - y 2 = - 6 &#8658; - 4 &#160; x - y = - 12 &#8658; 4 x + y = 12 &#160; &#160; &#160; &#160; &#8943; 1 For loop  2 : 1 2 y + 1 y + 12 + 1 2

Question 6

In the balanced condition, the values of the resistances of the four arms of a Wheatstone bridge are shown in the figure below. The resistance R 3 has temperature coefficient 0 . 0004 ° C - 1 . If the

Accepted Answer

R 3 = 300 ( 1 + α Δ T ) = 300 1 + 4 10000 × 100 = 312 V 1 = 41 . 9354 = 41 . 94 V 2 = 41 . 666 = 41 . 67 V 1 - V 2 = 41 . 94 - 41 . 67 = 0 . 27

Question 7

Shown in the figure is a semicircular metallic strip that has thickness t and resistivity ρ . Its inner radius is R 1 and outer radius is R 2 . If a voltage V 0 is applied between its two ends, a curr

Accepted Answer

Resistance of the strip = ρ π r tdr R = net resistance 1 R = ∫ t d r ρ π r t π ρ ∫ R 1 R 2 d r r R = π ρ t ln R 2 R 1 Hence, current i = V 0 R = V 0 t π ρ ln R 2 R 1 for elementary dr, di = V 0 ρ π r tdr = V 0 tdr ρ π r i = nAe v V 0 t d r ρ π r = n t d r e v eE = m v 2 r = m V 0 2 n 2 ℓ 2 ρ 2 π 2 r 2 × 1 r m v 2 r = m V 0 2 n 2 ℓ 2 ρ 2 π 2 r 2 × 1 r E ∝ 1 r 3 E = k 1 r 3 Δ V = ∫ E d r = k ∫ R 1 R 2 d r r 2 ∝ V 0 2 ∴ ∝ i 2

Current Electricity — Previous Year Questions & Solutions