Two capacitors of $3\,pF$ and $6\,pF$ are connected in series and a potential difference of $5000\,V$ is applied across the combination. They are then disconnected and reconnected in parallel. The potential between the plates is
Medium
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(b) $\frac{1}{C} = \frac{1}{3} + \frac{1}{6}$ $==>$ $C = 2\,pF$
Total charge$ = 2 \times {10^{ - 12}} \times 5000 = {10^{ - 8}}\,C$
The new potential when the capacitors are connected in parallel is
$V = \frac{{2 \times {{10}^{ - 8}}}}{{(3 + 6) \times {{10}^{ - 12}}}} = 2222\,V$
Total charge$ = 2 \times {10^{ - 12}} \times 5000 = {10^{ - 8}}\,C$
The new potential when the capacitors are connected in parallel is
$V = \frac{{2 \times {{10}^{ - 8}}}}{{(3 + 6) \times {{10}^{ - 12}}}} = 2222\,V$
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($1$) In Process $1$, the energy stored in the capacitor $E_C$ and heat dissipated across resistance $E_D$ are released by:
$[A]$ $E_C=E_D$ $[B]$ $E_C=E_D \ln 2$ $[C]$ $\mathrm{E}_{\mathrm{C}}=\frac{1}{2} \mathrm{E}_{\mathrm{D}}$ $[D]$ $E_C=2 E_D$
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Given the answer quetion ($1$) and ($2$)
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