JEE Main 2023 · 12 April · morning · Single correct · +4 / −1
Given below are two statements:
Statement I : When the frequency of an a.c source in a series LCR circuit increases, the current in the circuit first increases, attains a maximum value and then decreases.
Statement II : In a series LCR circuit, the value of power factor at resonance is one.
In the light of given statements, choose the most appropriate answer from the options given below.
- (a) Both Statement I and Statement II are False.
- (b) Statement I is incorrect but Statement II is true.
- (c) Both Statement I and Statement II are true.
- (d) Statement I is correct but Statement II is false.
Answer and solution
Answer (c) Both Statement I and Statement II are true.
Solution
Physics sir · 4 steps
Watch the impedance as frequency rises: it dips to R at resonance and climbs again on both sides.
- Current and impedance. The current is I = V/Z, with Z = √(R² + (X_L - X_C)²). Here X_L = ω L grows with frequency and X_C = 1/(ω C) shrinks with it.
- Statement I. At low frequency X_C is large, so Z is large and the current is small. As the frequency rises, the gap X_L - X_C closes until X_L = X_C at resonance, where Z = R is smallest and the current is largest. Beyond that X_L keeps growing, Z grows again and the current falls, so Statement I is true.
cos φ = R/Z = R/√(R² + (X_L - X_C)²)
- Statement II. At resonance X_L = X_C, so Z = R and cos φ = R/R = 1. Current and voltage are in phase, so Statement II is also true.
- So. Both statements are true, which is option (c).
The idea to keep. In series LCR, resonance is where X_L = X_C: impedance is least (Z = R), current is greatest, and the power factor is exactly 1. Move away in either direction and the current drops.
Answer from the NTA final answer key. Solution by PrepWiser.