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An inductor and a resistor are connected in series to an AC source. The current in circuit is $500 \mathrm{~mA}$, if the applied AC voltage is $8 \sqrt{2} \mathrm{~V}$ at a frequency of $\frac{175}{\pi} \mathrm{Hz}$ and the current in the circuit is $400 \mathrm{~mA}$, if the same AC voltage at a frequency of $\frac{225}{\pi} \mathrm{Hz}$ is applied. The values of the inductance and the resistance are respectively
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$60 \mathrm{mH}, \sqrt{71} \Omega$
$\begin{aligned} & \text { For an } L-R \text { circuit, } I=\frac{V}{Z}=\frac{V}{\sqrt{R^2+L^2 \omega^2}} \\ & \Rightarrow \quad R^2+L^2 \omega^2=\left(\frac{V}{I}\right)^2 \\ & \text { Here, } \quad I_1=500 \times 10^{-3} \mathrm{~A} \\ & \omega_1=\frac{175}{\pi} \times 2 \pi \frac{\mathrm{rad}}{\mathrm{s}}=350 \frac{\mathrm{rad}}{\mathrm{s}} \\ & V_1=8 \sqrt{2} \\ & \Rightarrow \quad R^2+L^2(350)^2=\left(\frac{8 \sqrt{2}}{500 \times 10^{-3}}\right)^2 \\ & \end{aligned}$
Solving, we get $R=\sqrt{71} \Omega$ and $L=60 \mathrm{mH}$
Solving, we get $R=\sqrt{71} \Omega$ and $L=60 \mathrm{mH}$
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