Question

An alternating emf $E = 440 \sin(100\pi t)$ is applied to a circuit containing an inductance of $\frac{\sqrt{2}}{\pi}$ H. If an a.c. ammeter is connected in the circuit, its reading will be:

• A. 4.4 A
• B. 1.55 A
• C. 2.2 A
• D. 3.11 A

Answer 1

Answer: (C)

2.2 A

Answer 2

Current $I = \frac{V}{\omega L}$
$I = \frac{440}{100\pi \times \frac{\sqrt{2}}{\pi}}$
$I = \frac{44}{10\sqrt{2}}$
$⇒$ $I_{\text{rms}} = \frac{I}{\sqrt{2}}$
$=\frac{44}{20}$
$=2.2 A$
So, the correct option is (C): 2.2 A