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

The Correct Option is 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} = 2 V_{2} $ Also $ V_{1} + V_{2} = 900 \,V $ $ \therefore 2V_{2} + V_{2} = 900\, V $ $ V_{2} = 300 \,V$ and $V_{1} = 600\, V $ Common potential $ V = \frac{C_{1}V_{1} + C_{2}V_{2}}{C_{1} + C_{2}} $ $ = \frac{3\times10^{-6} \times600 + 6 \times10^{-6} \times300}{3 \times10^{-6} + 6 \times10^{-6} } $ $= \frac{1800 +1800 \times 10^{-6}}{9\times 10^{-6}} = \frac{3600}{9} $ $ = 400 \,V $