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

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

Tardigrade Admin · Accepted Answer

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

5 V

to

10 V,

W = \frac{1}{2} C (V_{2}^{2} - V_{1}^{2})

∴ W = \frac{1}{2} C (1 0^{2} - 5^{2}) = \frac{1}{2} C (75)

Work done in changing the potential from

10 V

to

15 V,

W^{'} = \frac{1}{2} C (1 5^{2} - 1 0^{2})

∴ W^{'} = \frac{1}{2} C (125)

\Rightarrow \frac{W ^{'}}{W} = \frac{125}{75} = 1.67

We get

W^{'} = 1.67 W

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

$W$

$0.6 W$

$1.25 W$

$1.67 W$

Q. 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

W

0.6W

1.25W

1.67W

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

$W$

$0.6 W$

$1.25 W$

$1.67 W$