Question

For the circuit shown in the figure, the EMF of the generator is $E$. The current through the inductor is $3\text{A}$, while the current through the condenser is $1\text{A}$ then current drawn from the generator is

Answer 1

Current lead EMF by phase $\frac{\pi }{2}$ for capacitor and Current lags behind EMF by phase$\frac{\pi }{2}$ for inductor.

Let $I_C$ be the current passing through capacitor, $I_L$ be the current passing through the inductor and $I$ be the current drawn from the generator.

Thus, $I=I_C+I_L$

$\Rightarrow I=\frac{E_0}{X_C}\sin (\omega t+\frac{\pi }{2})+\frac{E_0}{X_L}\sin (\omega t-\frac{\pi }{2})$

$\Rightarrow I=[\frac{E_0}{X_C}-\frac{E_0}{X_L}]\cos \omega t$

RMS current drawn from the generator is
$I_v=\frac{E_0}{\sqrt{2}}(\frac{1}{X_C}-\frac{1}{X_L})$
$I_v=I_{Crms}-I_{Lrms}$

$\therefore I_v=3-1=2\text{A}$

Accepted answer : $2$