Q.23. When an inductor ‘L’ and a resistor ‘R’ in series are connected across a 15 V,50 Hz a.c. supply, a current of 0.3 A flows in the circuit. The current differs in phase from applied voltage by ( \frac{\pi}{3})^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 )
A. 10Ω
B 15Ω
C. 20Ω
D. 25Ω
Answer :- D. 25Ω
Explanation :-
\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}{0.3}=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} \text{Impedance } \mathrm{Z}=\sqrt{\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}^2} Z=\sqrt{R^2+(\sqrt{3} 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