Question

A $4.00\mu F$ capacitor and a $6.00\mu F$ capacitor are connected in parallel across a $660V$ supply line. The charged capacitors are disconnected from the line and from each other, and then reconnected to each other with terminals of unlike sign together. Find the final charge on each and the voltage across each capacitors.

Answer 1

Before disconnection , as the both capacitor are in parallel so potential across them will be same i.e $660V$.

Now using $Q=CV$, the charge on $4\mu F$ is $Q_1=(4\times {10}^{-6})\times 660=2.64\times {10}^{-3}C$ and the charge on $6\mu F$ is $Q_2=(6\times {10}^{-6})\times 660=3.96\times {10}^{-3}C$

When disconnected and again connected each other with terminals of unlike sign together, the total charge is $Q_t=Q_2+(-Q_1)=(3.96-2.64){10}^{-3}=1.32\times {10}^{-3}C$ and total capacitance $C_t=C_1+C_2=4+6=10\mu F$

After re-connection, they are in parallel so the potential across will be same.

Thus, common potential is $V_c=\frac{Q_t}{C_t}=\frac{1.32\times {10}^{-3}}{10\times {10}^{-6}}=132V$

Now charges become , $Q_1^{\prime }=(4\times {10}^{-6})\times 132=5.28\times {10}^{-4}C$ and $Q_2^{\prime }=(6\times {10}^{-6})\times 132=7.92\times {10}^{-4}C$