JEE Main 2021 · 22 July · evening · Single correct · +4 / −1
Match List - I with List - II
Choose the correct answer from the options given below
| List - I | List - II | ||
|---|---|---|---|
| (a) | ω L > 1/(ω C) | (i) | Current is in phase with emf |
| (b) | ω L = 1/(ω C) | (ii) | Current lags behind the applied emf |
| (c) | ω L < 1/(ω C) | (iii) | Maximum current occurs |
| (d) | Resonant frequency | (iv) | Current leads the emf |
- (a) a(ii), b(i), c(iv), d(iii)
- (b) a(ii), b(i), c(iii), d(iv)
- (c) a(iii), b(i), c(iv), d(ii)
- (d) a(iv), b(iii), c(ii), d(i)
Answer and solution
Answer (a) a(ii), b(i), c(iv), d(iii)
Solution
Physics sir · 3 steps
Compare the two reactances: whichever one is bigger decides whether the circuit acts inductive or capacitive.
- Phase angle. In a series LCR circuit the phase angle between emf and current is φ, where tan φ = (X_L - X_C)/R. Its sign tells you who is ahead.
tan φ = (ω L - 1/(ω C))/R, Z = √(R² + (ω L - 1/(ω C))²)
Condition Circuit behaves as Matches (a) ω L > 1/(ω C) inductive, φ > 0 (ii) current lags emf (b) ω L = 1/(ω C) purely resistive, φ = 0 (i) current in phase (c) ω L < 1/(ω C) capacitive, φ < 0 (iv) current leads emf (d) resonant frequency Z = R, the least possible (iii) maximum current - Resonance. At resonance the reactances cancel, so Z drops to its minimum value R. The current I = V/Z is then the largest it can be.
- So. The matching is a(ii), b(i), c(iv), d(iii), which is option (a).
The idea to keep. Inductor wins means current lags, capacitor wins means current leads, and a tie means resonance: in phase with maximum current. Remember ELI the ICE man.
Answer from the NTA final answer key. Solution by PrepWiser.