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2. Electric Potential and Capacitance
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Two condensers, one of capacity $C$ and other of capacity $C/2$ are connected to a $V-$ volt battery, as shown in the figure. The work done in charging fully both the condensers is
A
$\frac{1}{4}\,C{V^2}$
B
$\;\frac{3}{4}\,C{V^2}$
C
$\;\frac{1}{2}\,C{V^2}$
D
$\;3\,C{V^2}$
(AIPMT-2007)
Solution
As the capacitors are connected in parallel, therefore potential difference across both the condensors remains the same.
$\therefore Q_{1}=C V ;$
$Q_{2} =\frac{C}{2} V $
$\text { Also }, Q =Q_{1}+Q_{2}$
$ = CV+\frac{C}{2} V=\frac{3}{2} C V$
Work done in charging fully both the condensors is given by
$W=\frac{1}{2} Q V=\frac{1}{2} \times\left(\frac{3}{2}\, C V\right) V=\frac{3}{4} \,C V^{2}$
Standard 12
Physics
