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  4. A pure inductor of 25 mH is connected to an ac source of 220 V Given the frequency of the sour ce as 50 Hz, the rms current in the circuit is

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Q. A pure inductor of 25 mH is connected to an ac source of 220 V Given the frequency of the sour ce as 50 Hz, the rms current in the circuit is

Alternating Current

Solution:

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