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Q.
A $5 \,\mu \,F$ capacitor is connected in series with a $10 \,\mu \,F$ capacitor. When a $300\, V$ potential difference is applied across this combination, the energy stored in the capacitors is
WBJEEWBJEE 2015Electrostatic Potential and Capacitance
Solution:
According to question the figure can be drawn as below
The equivalent capacitance,
$\frac{1}{C_{\text {eq }}} =\frac{1}{C_{1}}+\frac{1}{C_{2}}$
$=\frac{1}{5}+\frac{1}{10}$
$=\frac{2+1}{10}=\frac{3}{10}$
$\Rightarrow C_{\theta q} =\frac{10}{3} \mu F$
Now, the energy stored in the capacitor is
$U=\frac{1}{2} C V^{2}$
$=\frac{1}{2} \times \frac{10}{3} \times 10^{-6} \times 300 \times 300$
$=\frac{3}{10 \times 2}=\frac{3}{20}=0.15\, J$
