Question:
In an LCR series circuit, the resonance frequency is \( f \). If the capacitance is made 4 times, what will be the new resonance frequency?
In an LCR series circuit, the resonance frequency is \( f \). If the capacitance is made 4 times, what will be the new resonance frequency?
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In an LCR circuit,
\[
f \propto \frac{1}{\sqrt{C}}
\]
So if capacitance increases \(k\) times, the frequency becomes \( \frac{1}{\sqrt{k}} \) times the original value.
Updated On: Apr 30, 2026
- \(4f\)
- \(2f\)
- \(f/2\)
- \(f/4\)
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The Correct Option is C
Solution and Explanation
Concept: For an LCR series circuit, the resonance frequency is given by \[ f = \frac{1}{2\pi\sqrt{LC}} \] where \(L\) is the inductance and \(C\) is the capacitance. Thus, resonance frequency is inversely proportional to the square root of capacitance.
Step 1: Write the formula for resonance frequency. \[ f = \frac{1}{2\pi\sqrt{LC}} \]
Step 2: Substitute the new capacitance. Given that the capacitance becomes \[ C' = 4C \] The new frequency becomes \[ f' = \frac{1}{2\pi\sqrt{L(4C)}} \]
Step 3: Simplify the expression. \[ f' = \frac{1}{2\pi\sqrt{4LC}} \] \[ f' = \frac{1}{2\pi \cdot 2\sqrt{LC}} \] \[ f' = \frac{1}{2}\left(\frac{1}{2\pi\sqrt{LC}}\right) \] \[ f' = \frac{f}{2} \] \[ \boxed{f' = \frac{f}{2}} \]
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