Question

If P.D across a capacitor is changed from 15V to 30V, work done is W. What will be the work done when P.d is chnaged from 30V to 60V. (a) W (b) 4W (c) 3W (d) 2W

Answer 1

Answer = (b) 4 W

Given,

V1 = change in P.D  

    = 30 V - 15 V

V2 = Change in P.D

    = 60 V - 30 V

Now,

We know,

$W= \frac{1}{2}CV^{2}$

From above,

W ∝ $V^{2}$

Thus , $\frac{W2}{W1} =\frac{(\triangle V2)^{2}}{(\triangle V1)^{2}}$

On putting the values,

=$\frac{(60-30)^{2}}{(30-15)^{2}}$

= $(\frac{30}{15})^{2}$

= 4

now ,

$\frac{W2}{W1} = 4$

W2 = 4 W1

Thus, Work done when potential difference is changed from 30 V to 60 V  is 4W.

Answer 2

Answer

Answer = 4W

Explanations:-

According to the question given here,

V1 is the Change in P.D and V2 is Change in P.D

$= (30 V - 15 V) \\ </p><p> = (60 V - 30 V)$

$W= \frac{1}{2}CV^{2}$

Here,

$W ∝V^{2}$

$Therfore,\\{W2}and{W1} =\frac{(\triangle V2)^{2}}{(\triangle V1)^{2}}$

Adding the values we have:-

Now within few steps we will get our answer!

$=\frac{(60-30)^{2}}{(30-15)^{2}}$

$=(\frac{30}{15})^{2}$

$\frac{W2}{W1} = 4$