The impedance of an LR circuit with $\mathrm{L}=\frac{60}{\pi} \mathrm{mH}$, $\mathrm{R}=8 \Omega$ and…
The impedance of an LR circuit with $\mathrm{L}=\frac{60}{\pi} \mathrm{mH}$, $\mathrm{R}=8 \Omega$ and frequency $50 \mathrm{~Hz}$ is
- $1.3 \Omega$
- $14.3 \Omega$
- $20 \Omega$
- $10 \Omega$
Solution
$\begin{aligned} & \mathrm{Z}=\sqrt{\mathrm{R}^2+\chi_{\mathrm{L}}^2}=\sqrt{\mathrm{R}^2+\omega^2 \mathrm{~L}^2} \\ & =\sqrt{8^2+\left(2 \pi \times 50 \times \frac{60}{\pi} \times 10^{-3}\right)} \\ & =\sqrt{8^2+6^2} \\ & =10 \Omega\end{aligned}$
Asked in: AP EAMCET 2022 (06 Jul Shift 1)
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