Question

An alternating voltage $E=200\sin 300t$ is applied across a series combination of $R=10ohm$ and an inductor of $800mH$ calculate.
(i) impedance of the circuit.
(ii) Peak value of current in the circuit.

Answer 1

$\left(i\right)$ Impedance $Z=\sqrt{R^2+X_L^2}$

$=\sqrt{R^2+(\omega {L)}^2}$

$=\sqrt{{10}^2+({300\times 800\times {10}^{-3})}^2}$

$Z=240.2\Omega$

$\left(ii\right)$ $I_0=\frac{V_0}{Z}=\frac{200}{240.2}=0.832A$