Question

In a given R-L A.C circuit, if peak voltage of the source is 10 V, and peak voltage across resistance is 6 V, then what is the voltage across inductor when voltage the across resistance is 3V? ($\sqrt{(}3)=1.732$)

Answer 1

Say, $V_R=6\sin \omega t$
Voltage across the inductor is $\frac{\pi }{2}$ ahead of that of resistance. Therefore, $V_L=8\sin (\omega t+\frac{\pi }{2})=8\cos \omega t$
Given $V_R=3$
$\Rightarrow 6\sin \omega t=3\sin \omega t=\frac{1}{2}\Rightarrow \cos \omega t=\frac{\sqrt{3}}{2}$
Therefore, $V_L=8\cos \omega t=8\times \frac{\sqrt{3}}{2}=6.928$