A $5 \mu F$ capacitor is connected in series with $10 \mathrm{\mu F}$ capacitor. When a $300 \mathrm{v}$ potentid dillerence is applied across this combination, the total energy stored in the capacitors is

Question

Aicharya · Accepted Answer

According to question the figure can be drawn as below

The equivalent capacitance,
$
\begin{aligned}
\frac{1}{C_{e q}} &=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{1}{5}+\frac{1}{10} \\
&=\frac{2+1}{10}=\frac{3}{10} \\
\Rightarrow \quad C_{e q} &=\frac{10}{3} \mu \mathrm{F}
\end{aligned}
$
Now, the energy stored in the capacitor is
$
\begin{aligned}
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 \mathrm{J}
\end{aligned}
$

A $5 \mu F$ capacitor is connected in series with $10 \mathrm{\mu F}$ capacitor. When a $300 \mathrm{v}$…

Solution