When an inductor ' $L$ ' and a resistor ' $R$ ' in series are connected across a $15 \mathrm{~V}, 50…

When an inductor ' $L$ ' and a resistor ' $R$ ' in series are connected across a $15 \mathrm{~V}, 50 \mathrm{~Hz}$ a.c. supply, a current of $0.3 \mathrm{~A}$ flows in the circuit. The current differs in phase from applied voltage by $\left(\frac{\pi}{3}\right)^c$. The value of ' $R$ ' is $\left(\sin \frac{\pi}{6}=\cos \frac{\pi}{3}=\frac{1}{2}, \sin \frac{\pi}{3}=\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}\right)$
  1. $10 \Omega$
  2. $15 \Omega$
  3. $20 \Omega$
  4. $25 \Omega$

Solution

$\begin{array}{ll} & \text { Given: } E_v=15 \mathrm{~V}, \mathrm{f}=50 \mathrm{~Hz}, \mathrm{I}=0.3 \mathrm{~A}, \\ & \phi=\frac{\pi}{3} \mathrm{rad} \\ & \text { Impedance } Z=\frac{\mathrm{E}_{\mathrm{v}}}{\mathrm{I}}=\frac{15}{0.2}=50 \Omega \\ & \tan \phi=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} \\ & \tan \frac{\pi}{3}=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} \\ & \sqrt{3}=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} \\ & \mathrm{X}_{\mathrm{L}}=\sqrt{3} \mathrm{R} \\ & \mathrm{Impedance} Z=\sqrt{\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}^2} \\ & \mathrm{Z}=\sqrt{\mathrm{R}^2+(\sqrt{3} \mathrm{R})^2} \\ & \mathrm{Z}=\sqrt{4 \mathrm{R}^2} \\ 2 \mathrm{R}=\mathrm{Z} & \mathrm{R}=\frac{\mathrm{Z}}{2}=\frac{50}{2}=25 \Omega \\ \therefore \quad & \end{array}$

Asked in: MHT CET 2023 (12 May Shift 2)

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