Question: Q. 1. An inductor of reactance $1 \Omega$ and a resistor of $2 \Omega$ are connected in series to the terminals of a $6 V(r m s)$ AC source. The power dissipated in the circuit is (a) $8 \mathrm{~W}$. (b) $12 \mathrm{~W}$. (c) $14.4 \mathrm{~W}$. (d) $18 \mathrm{~W}$.
[NCERT Exemplar]
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Solution:
Ans. Correct option : (c)
Explanation :
Here, $E_{\mathrm{rms}}=6 \mathrm{~V}, X_{L}=1 \Omega, R=2 \Omega, p$
We know that, average power dissipated in the circuit is given by,
$$ \begin{aligned} P_{a v} & =E_{r m s} I_{r m s} \cos \phi \ I_{r m s} & =\frac{I_{0}}{\sqrt{2}}=\frac{E_{\text {sms }}}{Z} \ Z & =\sqrt{R^{2}+X_{L}^{2}}=\sqrt{4+1}=\sqrt{5} \Omega \ \therefore \quad I_{r m s} & =\frac{6}{\sqrt{5}} \mathrm{~A} \end{aligned} $$
$$ \begin{aligned} & \cos \phi=\frac{R}{Z}=\frac{2}{\sqrt{5}} \ & P_{a v}=6 \times \frac{6}{\sqrt{5}} \times \frac{2}{\sqrt{5}}=14.4 \mathrm{~W} \end{aligned} $$
