Explanation

(a) The rate of dissipation of electric energy is called electric power
$\ \ \ \ \ \ W = Vit$
The electric power dissipated will be is given by
$\ \ \ \ \ \ \ \ P = \frac{W}{t} = \frac{Vit}{t} = Vi$
$\ \ \ \ \ \ \ \ \ \ \ \ = \frac{V^{2}}{R}$ …(i)
When resistance is doubled, then let electric power is $P^{'}.$
$\therefore\ \ P^{'} = \frac{V^{2}}{2R}$ …(ii)
From Eqs, (i) and (ii), we get
$\ \ \ \ \ P^{'} = \frac{1}{2}P$
So, power becomes $\frac{1}{2}$ of initial value.