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A capacitor of 4 mu F is connected as sh...

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

A

0

B

`4 muC`

C

`16 muC`

D

`8 muC`

Text Solution

Verified by Experts

The correct Answer is:
D
Current in the lower arm of the circuit , `I = (2.5 V)/(2 Omega + 0.5 Omega) = 1A`
Potential difference across the internal resistance of cell ` = (0.5 Omega ) (1A) = 0.5V`
and potenital differeece across the `4 muF` capacitor `= 2.5 V - 0.5 = 2V`
Charge on the capacitor plates, `Q = CV = (4 muF)(2V)`
` = 8 muC`.
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