Question

An uncharged capacitor of $50\mu \text{F}$ is connected to a battery of $10\text{V}$. What amount of energy supplied by the battery is lost as heat while charging the capacitor ?

• A. $5\text{mJ}$
• B. $10\text{mJ}$
• C. $1.25\text{mJ}$
• D. $2.5\text{mJ}$

Answer 1

The correct option is D $2.5\text{mJ}$

Given:
$C=50\mu \text{F};10\text{V}$

Work done by battery is given by

$W=QV=C{V}^2$

And, energy stored in capacitor will be

$E=\frac{1}{2}C{V}^2$

$\Rightarrow E=\frac{1}{2}\times 50\times {10}^{-6}\times {10}^2=2500\mu \text{J}$

$\Rightarrow$ Heat dissipated $\left(H\right)$ in the connecting wires $=\frac{1}{2}C{V}^2$

$\Rightarrow H=\frac{1}{2}\times 50\times {10}^{-6}\times {10}^2=2500\mu \text{J}=2.5\text{mJ}$

Key concept-Now, work done by battery = Energy stored in the capacitor + Heat dissipated in the connecting wires.