An alternating emf $(ξ) = 220 \sin (50 π t)$ is applied to an inductive circuit having self inductance of $\sqrt{2} π^{-1} H$. What will be the reading of a.c, ammeter connected in the circuit?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2.2 A
Given:
Alternating emf: $\xi = 220 \sin(50\pi t)\ \text{V}$
Inductance: $L = \frac{\sqrt{2}}{\pi}\ \text{H}$
For a pure inductive circuit, rms current is:
$I_\text{rms} = \frac{V_\text{rms}}{X_L}$
where $X_L = \omega L$ is inductive reactance, $\omega = 50\pi\ \text{rad/s}$
Peak voltage: $V_0 = 220\ \text{V} \Rightarrow V_\text{rms} = \frac{V_0}{\sqrt{2}} = \frac{220}{\sqrt{2}} \approx 155.56\ \text{V}$
Inductive reactance: $X_L = \omega L = 50\pi \cdot \frac{\sqrt{2}}{\pi} = 50 \sqrt{2}\ \Omega \approx 70.71\ \Omega$
RMS current: $I_\text{rms} = \frac{155.56}{70.71} \approx 2.2\ \text{A}$
∴ Reading of the a.c. ammeter = $2.2\ \text{A}$