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Alternating Current question

2024 · 1 Feb · Shift 1 · Q77

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Alternating Current
2024 · 1 Feb · Shift 1 · Q77

Alternating Current question

2024 · 1 Feb · Shift 1 · Q77

JEE MainPhysicsAlternating CurrentMCQ+4 / −1

In series LCR circuit, the capacitance is changed from

C

to

4 C

. To keep the resonance frequency unchanged, the new inductance should be:

A
increased by $2 \mathrm{~L}$
B
reduced by $\frac{1}{4} \mathrm{~L}$
C
reduced by $\frac{3}{4} \mathrm{~L}$
D
increased to $4 \mathrm{~L}$

View written solutionFree

Correct answer: C

For a series LCR circuit, the resonance frequency is

f_0 = \frac{1}{2\pi\sqrt{LC}}.

To keep the resonance frequency unchanged, the quantity $LC$ must remain constant.

So initially,

LC = \text{constant}.

The capacitance is changed from $C$ to $4C$ . Let the new inductance be $L'$ . Then for the same resonance frequency,

L'\cdot 4C = LC.

Solving for $L'$ :

L' = \frac{LC}{4C} = \frac{L}{4}.

So the new inductance must become $\frac{L}{4}$ .

Change in inductance:

\Delta L = L - \frac{L}{4} = \frac{3L}{4}.

Thus the inductance is reduced by

\frac{3}{4}L.

Checking options:

A: increased by $2L$ — incorrect
B: reduced by $\frac{1}{4}L$ — incorrect
C: reduced by $\frac{3}{4}L$ — correct
D: increased to $4L$ — incorrect

Therefore, the correct option is C.

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