Question

An inductor has a 50 ohm reactance at 61 Hz. What will be the maximum current if this inductor is connected to a 54 Hz source that produces a 120-V rms voltage?

0.18 A

3.83 A

3.44 A

12.05 A

none of the above

Answer 1

Given that:

Inductive reactance, R = 50 Ω when f = 61 Hz

Therefore, Inductance (L) is given by:

L = (R) / (2πf)

L = (50) / (2 x 3.14 x 61)

L = 0.13045 H

Hence, Inductive reactance when f = 54 Hz is given by:

R = 2πf.L

R = (2 x 3.14 x 54) x 0.13045

R = 44.26 Ω

Therefore, rms current through the inductor is given by:

I_rms = (V_rms) / (R)

I_rms = (120 V) / (44.26 Ω)

I_rms = 2.71 A

Hence, Maximum current (I_m) is given by:

I_m = (I_rms) x sqrt (2)

I_m = (2.71 A) x (1.414)

I_m = 3.83 A ----------- (**Answer**)

• Correct option is : 3.83 A

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