An alternating emf $\mathrm{E}=440 \sin 100 \pi \mathrm{t}$ is applied to a circuit containing an inductance of $\frac{\sqrt{2}}{\pi} \mathrm{H}$. If an a.c. ammeter is connected in the circuit, its reading will be :
Solution
<p>$I = {V \over {\omega L}}$</p>
<p>$$ = {{440} \over {100\pi \times {{\sqrt 2 } \over \pi }}} = {{44} \over {10\sqrt 2 }}$$</p>
<p>$\Rightarrow {I_{rms}} = {I \over {\sqrt 2 }} = {{44} \over {20}} = 2.2\,A$</p>
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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