Answer
a) $I=0.79A$
b) $R=120.33\Omega$
c) If the resistance is halved then the power will be doubled.
Work Step by Step
(a) We know that
$P=VI$
This can be rearranged as:
$I=\frac{P}{V}$
We plug in the known values to obtain:
$I=\frac{75}{95}$
$I=0.79A$
(b) As $P=\frac{V^2}{R}$
This can be rearranged as:
$R=\frac{V^2}{P}$
We plug in the known values to obtain:
$R=\frac{(95)^2}{75}$
$R=120.33\Omega$
(c) We know that
$P=\frac{V^2}{R}$
According to given condition $R^{\prime}=\frac{R}{2}$
$\implies P=\frac{V^2}{R\prime}$
$\implies P=\frac{V^2}{\frac{R}{2}}=2\frac{V^2}{R}=2P$
Thus, if the resistance is halved then the power will be doubled.
