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  • A coil has resistance 30  and inductive reactance 20 <img alt="\Omega" src="https://entrancecorner.oncod

A coil has resistance 30 \Omega and inductive reactance 20 \Omega at 50 Hz frequency. If an ac source of 200 V, 100 Hz is connected across the coil, the current in the coil will be:

Option: 1

4.0 A


Option: 2

8.0 A


Option: 3

\frac{20}{\sqrt{13}} \mathrm{~A}


Option: 4

2.0 A


Answers (1)

best_answer

Here,\mathrm{ \mathrm{R}=30 \Omega, \mathrm{X}_{\mathrm{L}}=20 \Omega, \mathrm{v}=50 \mathrm{~Hz} }
\mathrm{\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=2 \pi v \mathrm{~L} \text { or } 20=2 \pi \times 50 \times \mathrm{L} }
When \mathrm{v^{\prime}=100 \mathrm{~Hz} }, then
\mathrm{\begin{aligned} & \mathrm{X}_{\mathrm{L}}^{\prime}=2 \pi \times 100 \times \mathrm{L}=2(2 \pi \times 50 \mathrm{~L}) \\ & =2 \times 20=40 \Omega \quad \text { (Using (i)) } \\ & \text { Current in the coil is } \mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{V}_{\mathrm{rms}}}{\sqrt{\mathrm{R}^2+\mathrm{X}_{\mathrm{L}}^{\prime 2}}}=\frac{200}{\sqrt{30^2+40^2}}=4 \mathrm{~A} \end{aligned} }

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Ritika Jonwal

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