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
(**

**I _{m} = 3.83
A** ----------- (**Answer**)

*Correct option is :*3.83 A

_{=====================================================}

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