Discussion
A pure inductor of self inductance 500 mH is connected to an alternating voltage supply of \(100\sqrt{2}\pi \sin 100\pi t\) volt. The current in the circuit in ampere is
Answer: E
The key gives the current as 2 A, its rms value.
The inductive reactance is
\[ X_{L} = \omega L = 100\pi \times 0.5 = 50\pi\ \Omega \]
The peak voltage is \(100\sqrt{2}\pi\) V, so
\[ \begin{aligned} I_{0} &= \frac{100\sqrt{2}\pi}{50\pi} = 2\sqrt{2}\ \text{A} \\ I_{rms} &= \frac{I_{0}}{\sqrt{2}} = 2\ \text{A} \end{aligned} \]
An alternating current is normally quoted by its rms value, which is what an ammeter reads; \(2\sqrt{2}\) A is the peak value.
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