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(B) The instantaneous current is $i = \sin \omega t$,so the peak current $i_0 = 1 \ A$.The instantaneous potential difference is $E = 100 \cos \omega

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}}$

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(B) The instantaneous current is $i = \sin \omega t$,so the peak current $i_0 = 1 \ A$.The instantaneous potential difference is $E = 100 \cos \omega t = 100 \sin(\omega t + \frac{\pi}{2})$.Comparing this with the standard form $E = E_0 \sin(\omega t + \phi)$,the phase difference between voltage and current is $\phi = \frac{\pi}{2}$.The wattless current is the component of the current that is $90^\circ$ out of phase with the voltage,given by $i_w = i \sin \phi$.The $r.m.s.$ value of the wattless current is $I_{w,rms} = \frac{i_0}{\sqrt{2}} \sin \phi$.Substituting the values: $I_{w,rms} = \frac{1}{\sqrt{2}} \sin(\frac{\pi}{2}) = \frac{1}{\sqrt{2}} 	imes 1 = \frac{1}{\sqrt{2}} \ A$.

The instantaneous values of current and potential difference in an alternating circuit are $i = \sin \omega t$ and $E = 100 \cos \omega t$ respectively. The $r.m.s.$ value of the wattless current (in $A$) in the circuit is:

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