(a) What is the self-inductance of the inductor?

Question

An inductor of inductance 'L' is connected to an AC source, V = 100 sin ωt. The graph below represents the variation of inductive reactance (XL) of the inductor with the frequency of an alternating source.

(a) What is the self-inductance of the inductor? 

(b) If the ac source is replaced by a battery such that V = 100 V, then what is the inductive reactance of the inductor? Give reason. 

(c) When the frequency is 50 Hz, what is the average power dissipated by the inductor over a complete cycle in the circuit? Justify your answer. 

(d) This inductor is connected in series with a resistance of 15 Ω and a capacitor of 5 μF. The frequency of the alternating source is varied such that the power dissipated in the circuit becomes maximum. Calculate the frequency and the phase difference between alternating voltage and current when the power dissipated is the maximum.

Answer 1

(a) X_L = 2πf L

L = X_L /2πf 

L = 20/(2 × 3.14 × 100) = 0.032 H

(b) A battery is a source of direct current and thus f = 0 Hz. 

As X_L = 2πfL, the inductive reactance of the inductor becomes zero.

(c) P_avg = V_rms I_rms cosφ 

where φ is the phase difference between current and voltage in the circuit. 

Phase difference is 90° for pure inductive circuit.

∴ P_avg = 0

(d) Power dissipated in an LCR circuit is maximum when X_L= X_C

f = 1/2π√(LC) 

f = 0.398 × 10³ Hz

f = 398 Hz

Under this condition of resonance, the circuit behaves as a pure resistive circuit. Hence phase difference between current and voltage is 0°.