Arun
Last Activity: 4 Years ago
When steady state is reached, the current I from the battery is
9=I(3+6)orI=IA
Potential difference across 3Ω resistance is 3 V and potential difference across
6Omega resistance is 6 V.
Potential difference across
3muF capacitor is 3V and pd
across 6muF capacitor is 6 V.
Therefore, charge on 3muF
capacitor is
Q_1 = 3 xx 3 = 9muC
Charge on 6muF capacitor is
Q_2 = 6 xx 6 = 36 muC
Charge (-Q_1) is shifted from the positive plate of 6muF capacitor.
The remaining charge on the positive plate of 6muF capacitor is
shifted through the switch. Therefore, charge passing through
the switch is