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A virtual current of 4 A and 50 Hz flows in an AC circuit containing a coil. The power consumed in the coil is 240 W. If the virtual voltage across the coil is 100 V. Its inductance will be:

Option: 1

\frac{1}{3 \pi} \mathrm{H}


Option: 2

\frac{1}{5 \pi} \mathrm{H}


Option: 3

\frac{1}{7 \pi} \mathrm{H}


Option: 4

\frac{1}{9 \pi} \mathrm{H}


Answers (1)

best_answer

Power consumed in the coil
\mathrm{\begin{aligned} & \mathrm{P}=\mathrm{I}_{\mathrm{rms}}^2 \mathrm{R} \\ & \therefore \quad \mathrm{R}=\frac{\mathrm{P}}{\mathrm{I}_{\mathrm{rms}}^2}=\frac{240}{16}=15 \Omega \\ & \text { Impedance, } \mathrm{Z}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{100}{4}=25 \Omega \\ & \text { Now, } \mathrm{X}_{\mathrm{L}}=\sqrt{\mathrm{Z}^2-\mathrm{R}^2}=\sqrt{(25)^2-(15)^2}=20 \Omega \\ & \therefore \quad 2 \pi \mathrm{L}=20 \\ & \mathrm{~L}=\frac{20}{2 \pi \times 50}=\frac{1}{5 \pi} \mathrm{H} \end{aligned} }

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vishal kumar

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