Exams › JEE Advanced › Physics

An induction coil has an impedance of 10 ohm. When an AC signal of frequency 1000 Hz is applied, the voltage leads the current by 45 deg. The inductance of the coil is

1. 1 / (2 * pi) H
2. 1 / (sqrt(2) * 200 * pi) H
3. 1 / (sqrt(2) * 20 * pi) H
4. 1 / (200 * pi) H

Correct answer: 1 / (sqrt(2) * 200 * pi) H

Solution

Phase angle phi = 45 deg: tan(45 deg) = X_L/R = 1 → X_L = R. Impedance Z = sqrt(R² + R²) = R*sqrt(2) = 10 Ω → R = 10/sqrt(2). X_L = R = 10/sqrt(2). L = X_L/omega = (10/sqrt(2))/(2*pi*1000) = 10/(2000*pi*sqrt(2)) = 1/(200*pi*sqrt(2)) H.

Related JEE Advanced Physics questions

• A resistor and an ideal inductor are connected in series in an AC circuit. Three ideal hot-wire voltmeters V1, V2, and V3 are connected across the resistor, the inductor, and the AC source respectively. Which relation correctly describes the readings?
• In an AC circuit, the power factor changes from 1/2 to 1/4 while the resistance remains unchanged. By what percentage should the impedance be increased to achieve this change in power factor?
• In an LC oscillating circuit, at a certain instant the charge on the capacitor is Q and the current through the inductor is I. At this instant the energy stored in the capacitor equals the energy stored in the inductor. What is the time period of the electrical oscillations?
• An AC source with emf V = V0*(sin(wt) + sin(3wt)) is connected in series with a capacitor C and a resistor R. The resulting current is expressed as i = i1*sin(wt + phi1) + i2*sin(3wt + phi2). Which of the following correctly compares i1 and i2?
• A series LCR circuit having a resistance of 100 ohm is driven by an AC source of 400 V (rms) at an angular frequency of 300 rad/s. When the capacitor is disconnected from the circuit, the current lags behind the applied voltage by 60 degrees. When the inductor is disconnected instead, the current leads the applied voltage by 60 degrees. Find the rms current and the power dissipated in the complete LCR circuit.
• A capacitor with reactance X_C = 10 ohm and an inductor with reactance X_L = 5 ohm are connected in parallel across an AC voltage source V = 100 V. Find the ratio of the current through the inductor to the total current drawn from the source.

⚔️ Practice JEE Advanced Physics free + battle 1v1 →