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A $20 \mu F$ capacitor and $ 40 \; \mu F$ capacitor are connected in series across $400\; V$ supply line. The charged capacitors are then disconnected from the line and reconnected with their positive plates together and negative plates together. No external voltage is applied. What is potential difference across each capacitor .

$(A)\;\large\frac{1600}{9} \\ (B)\;\large\frac{800}{9} \\ (C)\; 400 \\ (D)\;800 $

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In series,
Equivalent $C=\large\frac{20 \times 40}{20 + 40}$
$\qquad= \large\frac{40}{3}$$ \mu F$
Charge on each capacitor $ =\large\frac{40}{3}$$ \times 400 $
$\qquad= \large\frac{16000}{3}$$\mu c$
In series charge on both capacitor is equal.
=> Total charge $=\large\frac{2 \times 16000}{3} $
$\qquad = \large\frac{32000}{3}$$\mu c$
New capacitors are connected in parallel.
Equivalent $C= 20+40=60 \mu F$
Potential $V'= \large\frac{Q'}{C'}$
$\qquad= \large\frac{32000}{3 \times 60}$
$\qquad= \large\frac{1600}{9}$
Hence A is the correct answer.
answered Jan 6, 2014 by meena.p
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