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A pure inductor of 25 \mathrm{mH} is connected to an ac source of \mathrm{220 V}. Given the frequency of the source as \mathrm{50 Hz}, the rms current in the circuit is:

Option: 1

7 A


Option: 2

14 A


Option: 3

28 A


Option: 4

42 A


Answers (1)

\mathrm{Here, \mathrm{L}=25 \mathrm{mH}=25 \times 10^{-3} \mathrm{H} \mathrm{v}=50 \mathrm{~Hz}, \mathrm{~V}_{\mathrm{ms}}=220 \mathrm{~V}}
The inductive reactance is \mathrm{X_L=2 \pi v l=2 \times \frac{22}{7} \times 50 \times 25 \times 10^{-3} \Omega}
The rms current in the circuit is \mathrm{I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{X_{\mathrm{L}}}=\frac{220}{2 \times \frac{22}{7} \times 50 \times 25 \times 10^{-3}}=\frac{7 \times 1000}{2 \times 5 \times 25} \mathrm{~A}=28 \mathrm{~A}}

Posted by

Ramraj Saini

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