Question

As shown in the figure below, if a capacitor $C$ is charged by connecting it with resistance $R$ then, the energy stored in the capacitor is

• A. $\frac{1}{2}C{V}^2$
• B. More than $\frac{1}{2}C{V}^2$
• C. Less than $\frac{1}{2}C{V}^2$
• D. Zero

Answer 1

The correct option is A $\frac{1}{2}C{V}^2$

Energy stored in a capacitor is given by
$E=\frac{1}{2}C{V}_c^2...\left(1\right)$
Applying KVL in circuit $V=V_R+V_C$
In steady state current through resistance = 0
$\therefore V_R=IR=0$
Hence, $V=V_C$
From equation (1),
$E=\frac{1}{2}C{V}^2$