An inductor of reactance $1 \Omega$ and a resistor of resistance $3 \Omega$ are connected in series to the…
An inductor of reactance $1 \Omega$ and a resistor of resistance $3 \Omega$ are connected in series to the terminals of $10 \mathrm{~V}$ (rms) AC source. The power dissipated in the circuit is
$33.3 \mathrm{~W}$
$30 \mathrm{~W}$
$31.6 \mathrm{~W}$
$20 \mathrm{~W}$
Solution
Average power dissipated.
$P_{\mathrm{avg}}=\frac{v_{\mathrm{rms}}^2 R}{z^2}$
Here, $Z=$ impedence $=\sqrt{X_L^2+R^2}$
where, $X_L=$ inductive reactance $=L \omega$
So impedence, $z=\sqrt{X_L^2+R^2}$
$=\sqrt{1^2+3^2}$
(Here given, $X_L=1 \Omega$ and $R=3 \Omega$ )
So,
$z=\sqrt{10}$
Given ; $V_{\mathrm{rms}}=10 \mathrm{~V}$ so,
$\begin{aligned} P_{\mathrm{avg}} & =\frac{V_{\mathrm{rms}}^2 R}{z^2}=\frac{100 \times 3}{10} \\ & =30 \mathrm{~W}\end{aligned}$