Question

The amount of work done in increasing the voltage across the plates of capacitor from 5V to 10V is 'W'. The work done in increasing it from 10V to 15V will be

• A. W
• B. 0.6 W
• C. 1.25 W
• D. 1.67 W

Answer 1

Answer: (D)

1.67 W

Let the capacitance of the capacitor be $C$.

Energy stored in the capacitor $E=\frac{1}{2}C{V}^2$

Work done in changing the potential from 5V to 10V, $W=\frac{1}{2}C(V_2^2-V_1^2)$

$\therefore$ $W=\frac{1}{2}C({10}^2-5^2)=\frac{1}{2}C\left(75\right)$

Work done in changing the potential from 10V to 15V, $W^{\prime }=\frac{1}{2}C({15}^2-{10}^2)$

$\therefore$ $W^{\prime }=\frac{1}{2}C\left(125\right)$

$⟹$ $\frac{W^{\prime }}{W}=\frac{125}{75}=1.67$

We get $W^{\prime }=1.67W$