Q153 / 160 · TS EAMCET 2019 (06 May Shift 1) · Physics
A coil is connected to an AC source with peak emf, $8 \mathrm{~V}$ and frequency $\frac{30}{\pi} \mathrm{Hz}$. The coil has resistance of $8 \Omega$. If the average power dissipated by the coil is $0.4 \mathrm{~W}$, then the inductance of the coil is
- A$0.8 \mathrm{H}$
- B$2.0 \mathrm{H}$
- C$1.4 \mathrm{H}$
- Correct answerD$0.4 \mathrm{H}$
Solution
Given, peak emf, $V_0=8 \mathrm{~V}$
Frequency of AC source, $f=\frac{30}{\pi} \mathrm{Hz}$ and resistance, $R=8 \Omega$
$V_{m s}=\frac{8}{\sqrt{2}} \mathrm{~V} \quad\left(\because V_{m s}=\frac{V_0}{\sqrt{2}}\right)$
$\begin{aligned} & \therefore \quad V_{\text {rms }} \times I_{\text {rms }} \times \cos \phi=0.4 \\ & V_{\text {rms }} \times \frac{V_{\text {rms }}}{Z} \times \frac{R}{Z}=0.4 \\ &\left(\because \text { Power factor, } \cos \phi=\frac{R}{Z}\right)\end{aligned}$
$\begin{aligned} & \frac{V_{\text {mms }}^2}{Z^2} \times R=0.4 \\ & \frac{\left(\frac{8}{\sqrt{2}}\right)^2 \times 8}{X_L^2+R^2}=0.4 \\ & X_L^2+R^2=640 \\ & X_L^2+64=640 \\ & \omega^2 L^2=640-64 \quad\left(\because X_L=\omega L\right)\end{aligned}$
$\begin{aligned} L^2 & =\frac{576}{\omega^2} \\ \Rightarrow L=\frac{24}{\omega}=\frac{24}{2 \pi \times \frac{30}{\pi}} & =\frac{24}{60}=0.4 \mathrm{H}\end{aligned}$
The inductance of the coil is $0.4 \mathrm{H}$.
Hence, the correct answer is option: 'D'.