Explanation

(b) The instantaneous voltage through the given device
$e = 80\sin{100\pi t}$
Comparing the given instantaneous voltage with standard instantaneous voltage
$e = e_{0}\sin{\omega t.}$
We get $e_{0} = 80\ V$
Where $e_{0}$ is the peak value of voltage
Impedance $(Z) = 20\Omega$
Peak value of current $I_{0} = \frac{e_{0}}{Z}$
$= \frac{80}{20} = 4A$
Effective value of current (root mean square value of current).
$I_{rms} = \frac{I_{0}}{\sqrt{2}}$
$= \frac{4}{\sqrt{2}} = 2\sqrt{2} = 2.828\ A$