Alternating Current

JEE Main level, single correct, medium

65 of 190

Question 65 of 190

A coil of negligible resistance is connected in series with 90Ω90 \Omega resistor across 120 V,60 Hz120 \mathrm{~V}, 60 \mathrm{~Hz} supply. A voltmeter reads 36 V36 \mathrm{~V} across resistance. Inductance of the coil is :

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Answer: (B) 0.76H0.76 \mathrm{H}


36=IrmsR36=120XL2+R2×RR=90Ω⇒36=120×90XL2+902XL2+902=300XL2=81900XL=286.18ωL=286.18 L=286.18376.8 L=0.76H\begin{aligned} & 36=I_{\mathrm{rms}} \mathrm{R} \\ & 36=\frac{120}{\sqrt{\mathrm{X}_{\mathrm{L}}^2+\mathrm{R}^2}} \times \mathrm{R} \\ & \mathrm{R}=90 \Omega \Rightarrow 36=\frac{120 \times 90}{\sqrt{\mathrm{X}_{\mathrm{L}}^2+90^2}} \\ & \sqrt{\mathrm{X}_{\mathrm{L}}^2+90^2}=300 \\ & \mathrm{X}_{\mathrm{L}}^2=81900 \\ & \mathrm{X}_{\mathrm{L}}=286.18 \\ & \omega \mathrm{L}=286.18 \\ & \mathrm{~L}=\frac{286.18}{376.8} \\ & \mathrm{~L}=0.76 \mathrm{H}\end{aligned}

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