Q113 / 120 · VITEEE 2017 · Physics

Resonance frequency of LCR series a.c. circuit is $\mathrm{f}_{0}$. Now the capacitance is made 4 times, then the new resonance frequency will become

  • A
    $\mathrm{f}_{0} / 4$
  • B
    $2 \mathrm{f}_{0}$
  • C
    $f_{0}$
  • Correct answer
    D
    $\mathrm{f}_{0} / 2$.

Solution

In LCR series circuit, resonance frequency $\mathrm{f}_{0}$ is given by $$ \begin{array}{c} L \omega=\frac{1}{\mathrm{C} \omega} \Rightarrow \omega^{2}=\frac{1}{\mathrm{LC}} \quad \therefore \quad \omega=\sqrt{\frac{1}{\mathrm{LC}}}=2 \pi \mathrm{f}_{0} \\ \therefore \quad \mathrm{f}_{0}=\frac{1}{2 \pi \sqrt{\mathrm{LC}}} \quad \text { or } \quad \mathrm{f}_{0} \alpha \frac{1}{\sqrt{\mathrm{C}}} \end{array} $$ When the capacitance of the circuit is made 4 times, its resonant frequency become $\mathrm{f}_{0}^{\prime}$ $$ \therefore \frac{\mathrm{f}_{0}^{\prime}}{\mathrm{f}_{0}}=\frac{\sqrt{\mathrm{C}}}{\sqrt{4 \mathrm{C}}} \quad \text { or } \quad \mathrm{f}_{0}^{\prime}=\frac{\mathrm{f}_{0}}{2} $$

Hence, the correct answer is option: 'D'.

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