Q: A wire of resistance R is cut into 8 equal pieces. From these pieces two equivalent resistances are made by adding four of these together in parallel. Then these two sets are added in series. The net

aligned & Resistance of each part = R / 8 \\ & R _ eq = 1 4 ( R / 8)+ 1 4 ( R / 8)= R 16 aligned

Question 1

A galvanometer of resistance 100\ gives full scale deflection for a current of 1 mA. It is converted into an ammeter of range 0 - 10 A. The shunt required is :

Accepted Answer

Given G = 100\ , I_g = 1 mA = 10^ -3 A , and I = 10 A . The shunt resistance S required to convert a galvanometer into an ammeter is given by: S = I_g G I - I_g Substituting the values: S = 10^ -3 100 10 - 10^ -3 Since 10 10^ -3 , we can approximate 10 - 10^ -3 10. S 0.1 10 = 0.01\ Answer: 0.01\

Question 2

A room heater is rated 400 W, 220 V. If the supply voltage drops to 200 V, what will be the power consumed (approximately) ?

Accepted Answer

The resistance of the room heater is given by R = V^2 P Substituting the rated values, R = 220^2 400 = 48400 400 = 121 When the supply voltage drops to V' = 200 V, the power consumed is P' = V'^2 R P' = 200^2 121 = 40000 121 330.57 W Therefore, the power consumed is approximately 331 W. Answer: 331 W

Question 3

A resistor is connected to a battery of 12 V emf and internal resistance 2\ . If the current in the circuit is 0.6 A, the terminal voltage of the battery is :

Accepted Answer

The terminal voltage V of a battery discharging in a circuit is given by the relation: V = E - Ir Substituting the given values E = 12 V, r = 2\ , and I = 0.6 A: V = 12 - (0.6 2) V = 12 - 1.2 V = 10.8 V Answer: 10.8 V

Question 4

Consider two circuits, (A) and (B), each having two resistors. One of them has a positive temperature coefficient of resistance, + , while the other one has a negative temperature of coefficient, - ,

Accepted Answer

For circuit (A), the two resistors are connected in series. The equivalent resistance is given by: R_A = R_0(1 - T) + R_0(1 + T) = 2R_0 The current in circuit (A) is I_A = V R_A = V 2R_0 . Since R_A is independent of the temperature change T, the current I_A remains constant. For circuit (B), the two resistors are connected in parallel. The equivalent resistance is given by: R_B = R_0(1 - T) R_0(1 + T) R_0(1 - T) + R_0(1 + T) R_B = R_0^2(1 - ^2 T^2) 2R_0 = R_0 2 (1 - ^2 T^2) As the temperature i

Question 5

A uniform metallic wire having resistance 4\ is bent to form a square loop (ABCD) (see figure). A resistance of 2\ is connected between points B and D and a battery of 2 V is connected across points A

Accepted Answer

The total resistance of the uniform wire is 4\ . Since it is bent to form a square loop ABCD, the resistance of each of its four sides is equal. Resistance of each side = 4 4 = 1\ . Thus, R_ AB = R_ BC = R_ CD = R_ DA = 1\ . The given circuit can be redrawn as a Wheatstone bridge where the arms are AB, BC, AD, and DC, and the central arm is BD. The battery is connected across the opposite corners A and C. Let's check the balance condition of the Wheatstone bridge: R_ AB R_ AD = 1 1 = 1 R_ BC R_

Question 6

In a metre bridge experiment (see figure), the positions of the cell, E, and galvanometer, G, are interchanged. We shall observe in the galvanometer :

Accepted Answer

Let the nodes of the circuit be identified as follows: Node A: Left L-shaped metallic strip. Node B: Right L-shaped metallic strip. Node C: Central metallic strip. Node D: The point on the wire where the jockey touches. The resistances in the four arms of the circuit are: 1. Resistance between A and C is R_1. 2. Resistance between B and C is R_2. 3. Resistance between A and D is R_ AD (left part of the metre bridge wire). 4. Resistance between B and D is R_ DB (right part of the metre bridge wir

Question 7

A wire of resistance R is cut into 8 equal pieces. From these pieces two equivalent resistances are made by adding four of these together in parallel. Then these two sets are added in series. The net

Accepted Answer

aligned & Resistance of each part = R / 8 \ & R _ eq = 1 4 ( R / 8)+ 1 4 ( R / 8)= R 16 aligned

Question 8

The current passing through the battery in the given circuit, is :

Accepted Answer

Simplified Diagram: i = V R _ eq = 5 1 3 + 8 3 +1.5+5.5 = 5 10 =0.5 ~A

Question 9

A constant voltage of 50 V is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is :-

Accepted Answer

by KCL ;Current in branch CD =18-16=2 ~A

Question 10

A uniform wire of diameter d carries a current of 100 mA when the mean drift velocity of electrons in the wire is v. For a wire of diameter d 2 of the same material to carry a current of 200 mA , the

Accepted Answer

current i=n A v_d e ( v_d= mean drift velocity ) =n ( D^2 4 ) v_d e i D^2 v_d 100 200 = (d)^2 ( d 2 )^2 v v^ v^ =2 2^2 v=8 v

Question 11

In an electrical circuit, the voltage is measured as V=(200 4) volt and the current is measured as I=(20 0.2) A . The value of the resistance is:

Accepted Answer

R= V l = 200 20 =10 R R = V V + l l = 4 200 + 0.2 20 = 4+2 200 = 6 200 R= 6 200 R= 6 20 = 3 10 =0.3 Resistance =R R=(10 0.3)

Question 12

The given circuit shows a uniform straight wire A B of 40 cm length fixed at both ends. In order to get zero reading in the galvanometer G, the free end of J is to be placed from B at:

Accepted Answer

8 x = 12 40-x 2 x = 3 40-x 80-2 x=3 x 16=x from B =40-16=24 ~cm

Question 13

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Houses made of concrete roofs overlaid with foam keep the room hotter during summer.

Accepted Answer

Layer of foam create an insulation which prohibits heat transfer, and hence roofs of houses containing foam keep room cooler during summer.

Question 14

A uniform metal wire of length / has 10 resistance. Now this wire is stretched to a length 2 / and then bent to form a perfect circle. The equivalent resistance across any arbitrary diameter of that c

Accepted Answer

R_0=10 After stretching its length upto 2l R_1=n^2 R_0 =4 R_0 =40 R_ A B = 20 20 20+20 =10

Question 15

The terminal voltage of the battery, whose emf is 10 V and internal resistance 1 , when connected through an external resistance of 4 as shown in the figure is:

Accepted Answer

Current in circuit i= 10 4+1 =2 ~A aligned Terminal voltage & =E-i R \ & =10-2 1=8 ~V aligned

Question 16

A wire of length ' 1 and resistance 100 is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again conne

Accepted Answer

Divided into 10 parts aligned R & = l A \ R^ & = l 10 A = R 10 \ R_S & =5 R 10 [series] \ R_S & =50 \ R_P & = R 50 [parallel] \ R_ eq & =R_S+R_P \ & =52 aligned

Question 17

Choose the correct circuit which can achieve the bridge balance.

Accepted Answer

In option (4), 10 15 = 10 5+R_D The diode can conduct and have resistance R_D=10 because diode have dynamic resistance. In that case bridge will be balanced

Current Electricity — Previous Year Questions & Solutions