Current Electricity

Q: An electric bulb is rated 220 V-100 W. The power consumed by it when operated on 110 V will be

(A) 75 W

(B) 40 W

(C) 25 W

(D) 50 W

Solution :

Resistance of electric bulb $\large R = \frac{V^2}{P}$

$\large R = \frac{(220)^2}{100}$

Power consumed at 110 V,

$\large P_{consumed} = \frac{V^2}{R}$

$\large P_{consumed} = \frac{V^2}{\frac{(220)^2}{100}}$

= 25 W

Q: In the circuit, the galvanometer G shows zero deflection. If the batteries A and B have negligible internal resistance, the value of the resistor R will be

(A) 200 ohm

(B) 100 ohm

(C) 500 ohm

(D) 1000 ohm

Solution:

The galvanometer shows G zero deflection ie, current through XY is zero .

As a result potential drop across R is 2 V.

$\large I = \frac{12}{500 + R} $

Voltage across R, V = IR

$\large 2 = \frac{12}{500 + R} \times R$

or , 1000 + 2R = 12 R

or , R = 100 ohm

Q: Two sources of equal emf are connected to an external resistance R. The internal resistances of the two source are R1 and R2 (R2 > R1). If the potential difference across the source having internal resistance R2 , is zero, then

(A) $\large R = \frac{R_2 (R_1 + R_2)}{R_2 – R_1}$

(B) $\large R = R_2 – R_1$

(C) $\large R = \frac{R_1 R_2}{R_1 + R_2}$

(D) $\large R = \frac{R_1 R_2}{R_2 – R_1}$

Solution :

$\large R_{eq} = R_1 + R_2 + R$

$\large I = \frac{2 E}{R_1 + R_2 + R}$

According to the question,

$\large – (V_A – V_B) = E – I R_2 $

$\large 0 = E – I R_2 $

$\large E = I R_2 $

$\large E = (\frac{2 E}{R_1 + R_2 + R}) R_2 $

$\large 2 R_2 = R_1 + R_2 + R$

$\large R = R_2 – R_1 $

Correct Option is (B)

Q: A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now be

(A) doubled

(B) four times

(C) one-fourth

(D) halved

Solution :

$\large H_1 = \frac{V^2}{R} t $

$\large H_2 = \frac{V^2}{R/2} t $

$\large \frac{H_2}{H_1} = 2 $

$\large H_2 = 2 H_1 $

Correct Option is (A)

Q: An energy source will supply a constant current into the load, if its internal resistance is

(A) equal to the resistance of the load

(B) very large as compared to the load resistance

(C) zero

(D) non-zero but less than the resistance of the load

Solution :

$\large I = \frac{E}{R+ r}$

$\large I = \frac{E}{R} = Constant $

Where R = External Resistance

r = Internal resistance = 0

Q: In a Potentiometer experiment the balancing with a cell is at length 240 cm. On shunting the cell with a resistance of 2 Ω , the balancing length becomes 120 cm. The internal resistance of the cell is

(A) 1 Ω

(B) 0.5 Ω

(C) 4 Ω

(D) 2 Ω

Solution : Internal resistance of the Cell is

$\large r = (\frac{l_1}{l_2} – 1)R$

$\large r = (\frac{240}{120} – 1)2 $

r = 2 ohm

Correct Option is (D)

Q: The resistance of hot tungsten filament is about 10 times the cold resistance. What will be the resistance of 100 W and 200 V lamp, when not in use ?

(A) 40 Ω

(B) 20 Ω

(C) 400 Ω

(D) 200 Ω

Solution :

$\large R = \frac{V^2}{P}$

$\large R_{hot} = \frac{V^2}{P} = \frac{200 \times 200}{100}$

= 400 ohm

The resistance of hot tungsten filament is about 10 times the cold resistance.

$\large R_{cold} = \frac{400}{10}$

= 40 ohm

Q: The resistance of a bulb filament is 100 ohm at a temperature of 100°C. If its temperature coefficient of resistance be 0.005 per °C, its resistance will become 200 ohm at temperature of

(A) 300°C

(B) 400°C

(C) 500°C

(D) 200°C

Solution :

Let resistance of bulb filament is R₀ at 0 °C then from expression

R_θ = R₀ (1 + α Δθ)

we have, 100 = R₀[1 + 0.005 x 100]

and 200 = R₀[1 + 0.05 x t ]

where t is temperature in °C at which resistance become 200 ohm.

Dividing the above two equations

$\large \frac{200}{100} = \frac{1 + 0.05 \times t}{1 + 0.005 \times 100}$

t = 400 °C

Q: In a Wheatstone’s bridge, three resistances P, Q and R are connected in the three arms and the fourth arm is formed by two resistances S1 and S2 connected in parallel. The condition for the bridge to be balanced will be

(A) $\large \frac{P}{Q} = \frac{2R}{S_1 + S_2}$

(B) $\large \frac{P}{Q} = \frac{R(S_1 + S_2)}{S_1 S_2}$

(C) $\large \frac{P}{Q} = \frac{R(S_1 + S_2)}{2 S_1 S_2}$

(D) $\large \frac{P}{Q} = \frac{R}{S_1 + S_2}$

Solution : For balanced Wheatstone’s bridge ,

$\large \frac{P}{Q} = \frac{R}{S}$

Here , S1 and S2 are Parallel ,

$\large S = \frac{S_1 S_2}{S_1 + S_2}$

$\large \frac{P}{Q} = \frac{R}{\frac{S_1 S_2}{S_1 + S_2}}$

Correct Option is (B) $\large \frac{P}{Q} = \frac{R(S_1 + S_2)}{S_1 S_2}$

Q: The Kirchholff’s first law ( ∑i = 0)  and second law ( ∑iR = ∑E) . Where the symbols have their usual meanings, are
respectively based on-

(A) Conservation of charge, conservation of momentum

(B) Conservation of energy, conservation of charge

(C) Conservation of momentum, conservation of charge

(D) Conservation of charge, conservation of energy

Solution : Kirchhoff’s 1st law or KCL states that the algebraic sum of current meeting at any junction is equal to zero. In other words, we can say that “the sum of all the currents directed towards a junction in a circuit is equal to the sum of all the currents directed away from that junction.” Thus, no charge has been accumulated at any junction ie,charge is conserved, and hence, we can say that KCL(∑i = 0) is based on conservation of charge.

Kirchhoff’s IInd law or KVL states that algebraic sum of changes in potential around any closed resistor loop must be zero. In
other words, ” around any closed loop, voltage drops are equal to voltages rises “.No energy is gained or lost in circulating a
charge around a loop, thus, we can say that KVL is based on conservation of energy .

Correct Option is (D)