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When 100 V DC is applied across a solenoid, a current of 1 A flows in it. When 100 V AC is applied across the same solenoid the current drops to 0.5 A. If the frequency of the AC source is 50 Hz, the impedance and inductance of the solenoid are:

Option: 1

200 \Omega \text { and } 0.55 \mathrm{H}


Option: 2

100 \Omega \text { and } 0.86 \mathrm{H}


Option: 3

200 \Omega \text { and } 1.0 \mathrm{H}


Option: 4

1100 \Omega \text { and } 0.93 \mathrm{H}


Answers (1)

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\begin{aligned} & \text { Impedance }=\frac{\mathrm{V}_{\mathrm{AC}}}{\mathrm{I}_{\mathrm{AC}}}=\frac{100}{0.5}=200 \Omega=\mathrm{Z} \\ & \mathrm{R}=\frac{\mathrm{V}_{\mathrm{DC}}}{\mathrm{I}_{\mathrm{DC}}}=\frac{100}{1}=100 \Omega \\ & \mathrm{X}_{\mathrm{L}}=\sqrt{\mathrm{Z}^2-\mathrm{R}^2}=100 \sqrt{3} \Omega=(2 \pi \mathrm{fL}) \\ & \therefore \quad \mathrm{L}=\frac{100 \sqrt{3}}{2 \pi \mathrm{f}}=\frac{100 \times 1.732}{2 \times 3.14 \times 50}=0.55 \mathrm{H} \end{aligned}

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shivangi.shekhar

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