Question

Find the maximum value of current when an inductance of one henry is connected to an a.c. source of 200 volts, 50 Hz. ​

Answer 1

We have to find the maximum value of current when an inductance of one Henry is connected to an A.C source of 200 volts, 50 Hz.

solution : RMS voltage of source, V_rms = 200 volts.

so peak voltage, V_peak = √2 V_rms

= 1.41 × 200 = 282 volts.

frequency of oscillation of AC circuit, f = 50 Hz

so angular frequency, ω = 2πf

= 2π × 50 = 100π = 100 × 3.14 = 314 rad/s

so X_L = ωL

= 314 rad/s × 1 H = 314 ohm

now I_peak = V_peak/X_L

= 282/314 ≈ 0.9 A

Therefore the maximum value of current is 0.9A.

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