A capacitor of 4 μ F is connected as shown in the circuit. The internal resistance of the battery is 0.5 Ω The amount of charge on the capacitor plates will be

Question

A capacitor of 4 μ F is connected as shown in the circuit. The internal resistance of the battery is 0.5 Ω The amount of charge on the capacitor plates will be <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/0cd43f8b2ed10c0df9f2bcc80ae37885-.png" /> (A) 0 (B) 4 μ C (C) 16 μ C (D) 8 μ C

Tardigrade Admin · Accepted Answer

Current in the lower arm of the circuit,

I = \frac{2.5 V}{2 Ω + 0.5 Ω} = 1 A

,
Potential difference across the internal resistance of cell

= (0.5 Ω) (1 A) = 0.5 V

and potential difference across the

4 μ F

capacitor

= 2.5 V - 0.5 V = 2 V

Charge on the capacitor plates,

Q = C V = (4 μ F) (2 V)

= 8 μ C

Q. A capacitor of $4 μ F$ is connected as shown in the circuit. The internal resistance of the battery is $0.5 Ω$ The amount of charge on the capacitor plates will be

$0$

$4 μ C$

$16 μ C$

$8 μ C$

Q. A capacitor of 4μF is connected as shown in the circuit. The internal resistance of the battery is 0.5Ω The amount of charge on the capacitor plates will be

0

4μC

16μC

8μC

Current in the lower arm of the circuit, I=2Ω+0.5Ω2.5V​=1A, Potential difference across the internal resistance of cell =(0.5Ω)(1A)=0.5V and potential difference across the 4μF capacitor =2.5V−0.5V=2V Charge on the capacitor plates, Q=CV=(4μF)(2V) =8μC

Q. A capacitor of $4 μ F$ is connected as shown in the circuit. The internal resistance of the battery is $0.5 Ω$ The amount of charge on the capacitor plates will be

$0$

$4 μ C$

$16 μ C$

$8 μ C$