An inductor of self inductance 1 H is connected in series with a resistor of $100 \pi$ ohm and an ac supply of $100 \pi$ volt, 50 Hz . Maximum current flowing in the circuit is _________ A.
Answer (integer)
1
Solution
<p>Impedance of circuit</p>
<p>$$\begin{aligned}
& \mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}\right)^2}=\sqrt{\mathrm{R}^2+\left(\omega_{\mathrm{L}}\right)^2} \\
& =\sqrt{(100 \pi)^2+(2 \pi \times 50 \times 1)^2} \\
& =\sqrt{(100 \pi)^2+(100 \pi)^2} \\
& =\sqrt{2} \times 100 \pi \\
& \mathrm{I}_{\mathrm{rms}}=\frac{\mathrm{V}}{2}=\frac{100 \pi}{\sqrt{2} \times 100 \pi}=\frac{1}{\sqrt{2}} \\
& \mathrm{I}_{\max }=\sqrt{2} \mathrm{I}_{\mathrm{rms}}=\sqrt{2} \times \frac{1}{\sqrt{2}}=1 \text { Ampere }
\end{aligned}$$</p>
<p>Correct Answer : 1</p>
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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