In series LCR circuit, the capacitance is changed from C to 4 C. To keep the resonance frequency unchanged, the new inductance should be:

Solution

Physics sir · 4 steps

Resonance depends only on the product LC: if C goes up four times, L must come down four times.

1. Resonance condition. A series LCR circuit resonates when X_L = X_C, which gives a frequency that depends only on L and C.

   f₀ = 1/2π√LC

2. Keep LC fixed. For f₀ to stay the same, the product LC must stay the same. So L' × 4C = L × C, which gives L' = L/4.
3. Find the change. The inductance falls from L to L/4. The amount it is reduced by is L - L/4 = 3/4L.
4. So. The inductance must be reduced by 3/4L, which is option (c).

The idea to keep. For resonance problems, hold LC constant. Then read the options carefully: "reduced to" gives the new value, "reduced by" gives the change.

Answer from the NTA final answer key. Solution by PrepWiser.