Question

A $10 \, \mu F $ capacitor is charged by a battery of emf $200\,V$. The energy drawn from the battery, and the energy stored in the capacitor, are respectively

• A. 0.05 J and 0 m J
• B. 0.05 mJ and 0.05 m J
• C. 0.4 J and 0.2 J
• D. 1.0 m J and 0.5 m J

Answer 1

Answer: (C)

Energy drawn from battery
$=$ potential difference $(V) \times$ charge $(q)$
Also, charge $(q)=$ capacitance $(C)$
$\times$ potential difference $(V)$
$\therefore E=V \times C V=C V^{2}$
Given, $C=10\, \mu F =10 \times 10^{-6} F$,
$V=200$ volt
$\therefore E=\left(10 \times 10^{-6}\right)(200)^{2}=0.4 J$
Work done in charging the conductor is stored as potential energy in the electric field, given by
$U=\frac{1}{2} C V^{2}$
Given, $C=10 \,\mu F =10 \times 10^{-6} F$,
$V =200\,$ volt
$U =\frac{1}{2} \times 10 \times 10^{-6} \times(200)^{2} $
$=0.2 \,J$