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
.

`0muC`

`4muC`

`16muC`

`8muC`

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The correct Answer is:

As capacitor offer infinite resistance for DC circuit. So current from cell will not flow across branch of `4muF and 10Omega`. So current will flow across 2 ohm branch.

So, Potential Difference (PD) across `2Omega` resistance `V=RI=2xx1=2` Volt. As battery, capacitor and 2 branches are in parallel. So PD will remain same across all three branches. As current does not flow through capacitor branch, so no potential drop will be across `10Omega`. So PD across `4muF` capacitor = 2 Volt Q = CV = `2muFxx2V=8muC`

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 .

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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
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