Question

Two capacitors of $3\mu F$ and $6\mu F$ are connected in series and a potential difference of 900 V is applied across the combination. They are then disconnected and reconnected in parallel. The potential difference across the combination is

• A. Zero
• B. 100 V
• C. 200 V
• D. 400 V

Answer 1

Answer: (D)

$Q=CV$, Here Q is a constant
$\therefore C\propto \frac{1}{V}$
$\therefore \frac{C_1}{C_2}=\frac{V_2}{V_1}$
$\Rightarrow \frac{3}{6}=\frac{V_2}{V_1}$
$\Rightarrow V_1=2V_2$
Also $V_1+V_2=900V$
$\therefore 2V_2+V_2=900V$
$V_2=300V$ and $V_1=600V$
Common potential $V=\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}$
$=\frac{3\times 10^{-6}\times 600+6\times 10^{-6}\times 300}{3\times 10^{-6}+6\times 10^{-6}}$
$=\frac{1800+1800\times 10^{-6}}{9\times 10^{-6}}=\frac{3600}{9}$
$=400V$