Question

An alternating $\operatorname{emf} E=400 \sin 100 \pi t$ is applied to a circuit containing an inductance of $\frac{\sqrt{2}}{\pi} H$. If an AC 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)

Here, $L=\frac{\sqrt{2}}{\pi} H ; \omega=100 \pi$;
$X_L=\omega L=100 \pi \times \frac{\sqrt{2}}{\pi}$
$X_L=100 \sqrt{2} \Omega$
$\because E_0=i_0 X_L$
or $i_0=\frac{440}{100 \sqrt{2}} A ; i_0=2.2 \sqrt{2} A$
Since ammeter is ac ammeter, it will read rms value of current,
$\therefore i_{r m s}=\frac{2.2 \sqrt{2}}{\sqrt{2}}=2.2 A$