Question:
Resonance frequency of an LCR series AC circuit is \(f_0\). Now inductance is reduced to \(\dfrac{1}{4}\) times and capacitance is increased to \(16\) times, then the resonance frequency becomes:
Resonance frequency of an LCR series AC circuit is \(f_0\). Now inductance is reduced to \(\dfrac{1}{4}\) times and capacitance is increased to \(16\) times, then the resonance frequency becomes:
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The resonance frequency of an LCR circuit varies inversely with
\[
\sqrt{LC}
\]
If either inductance or capacitance increases, the resonance frequency decreases.
Updated On: Jun 26, 2026
- \(\dfrac{f_0}{4}\)
- \(\dfrac{f_0}{2}\)
- \(2f_0\)
- \(4f_0\)
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The Correct Option is B
Solution and Explanation
Step 1: Write the formula for resonance frequency.
The resonance frequency of an LCR circuit is \[ f_0=\frac{1}{2\pi\sqrt{LC}} \] where \[ L=\text{inductance} \] and \[ C=\text{capacitance} \]
Step 2: Apply the changed values of inductance and capacitance.
The new inductance is \[ L'=\frac{L}{4} \] The new capacitance is \[ C'=16C \] Therefore, the new resonance frequency is \[ f'=\frac{1}{2\pi\sqrt{L'C'}} \] Substituting the values, \[ f'=\frac{1}{2\pi\sqrt{\left(\frac{L}{4}\right)(16C)}} \]
Step 3: Simplify the expression.
Inside the square root, \[ \left(\frac{L}{4}\right)(16C)=4LC \] Thus, \[ f'=\frac{1}{2\pi\sqrt{4LC}} \] Since \[ \sqrt{4LC}=2\sqrt{LC} \] we get \[ f'=\frac{1}{2\pi\cdot2\sqrt{LC}} \] \[ f'=\frac{1}{2}\left(\frac{1}{2\pi\sqrt{LC}}\right) \] But \[ \frac{1}{2\pi\sqrt{LC}}=f_0 \] Hence, \[ f'=\frac{f_0}{2} \]
Step 4: Final conclusion.
Therefore, the new resonance frequency becomes \[ \boxed{\frac{f_0}{2}} \]
The resonance frequency of an LCR circuit is \[ f_0=\frac{1}{2\pi\sqrt{LC}} \] where \[ L=\text{inductance} \] and \[ C=\text{capacitance} \]
Step 2: Apply the changed values of inductance and capacitance.
The new inductance is \[ L'=\frac{L}{4} \] The new capacitance is \[ C'=16C \] Therefore, the new resonance frequency is \[ f'=\frac{1}{2\pi\sqrt{L'C'}} \] Substituting the values, \[ f'=\frac{1}{2\pi\sqrt{\left(\frac{L}{4}\right)(16C)}} \]
Step 3: Simplify the expression.
Inside the square root, \[ \left(\frac{L}{4}\right)(16C)=4LC \] Thus, \[ f'=\frac{1}{2\pi\sqrt{4LC}} \] Since \[ \sqrt{4LC}=2\sqrt{LC} \] we get \[ f'=\frac{1}{2\pi\cdot2\sqrt{LC}} \] \[ f'=\frac{1}{2}\left(\frac{1}{2\pi\sqrt{LC}}\right) \] But \[ \frac{1}{2\pi\sqrt{LC}}=f_0 \] Hence, \[ f'=\frac{f_0}{2} \]
Step 4: Final conclusion.
Therefore, the new resonance frequency becomes \[ \boxed{\frac{f_0}{2}} \]
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