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In the circuit shown in the figure, initially, K_{1} is closed and K_{2} is open. What are the charges on each capacitor? Then K_{1} was opened and K_{2} was closed. What will be the charge on each capacitor now?

Answers (1)

When key K_{1} is closed and key K_{2} is open, then capacitors C_{1} and C_{2} are connected in series with the battery.

Therefore, the charge stored in the capacitors C_{1} and C_{2} will be same as Q_{1}=Q_{2}

Therefore, Q_{1}=Q_{2}=q=\left ( \frac{C_{1}}{\left ( C_{1}+C_{2} \right )} \right )E=18\mu C

Assuming only capacitors C_{2} and C_{3} are placed in parallel,

C_{2}V'+ C_{3}V'=Q_{2}



Q{_{2}}^{'}=3CV'=9\; \mu C

Q{_{3}}=3CV'= \mu C

Q{_{1}}^{'}=18\; \mu C


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