Q143 / 150 · MHT CET 2026 (17 April Shift 1) · Physics
The LC series resonant circuit produces a resonant frequency '$f$'. If L is tripled and 'C' is increased by 3C, the resonant frequency will be
- A$\dfrac{f}{3}$
- Correct answerB$\dfrac{f}{2\sqrt{3}}$
- C$6f$
- D$\dfrac{f}{3\sqrt{2}}$
Solution
The resonant frequency of an LC series circuit is given by $f = \dfrac{1}{2\pi\sqrt{LC}}$.
Given that the inductance $L$ is tripled, the new inductance is $L' = 3L$.
The capacitance $C$ is increased by $3C$, so the new capacitance is $C' = C + 3C = 4C$.
The new resonant frequency $f'$ is given by:
$f' = \dfrac{1}{2\pi\sqrt{L'C'}}$
Substituting the values of $L'$ and $C'$:
$f' = \dfrac{1}{2\pi\sqrt{(3L)(4C)}} = \dfrac{1}{2\pi\sqrt{12LC}}$
$f' = \dfrac{1}{2\sqrt{3}} \left( \dfrac{1}{2\pi\sqrt{LC}} \right)$
$f' = \dfrac{f}{2\sqrt{3}}$
Answer: $\dfrac{f}{2\sqrt{3}}$
Given that the inductance $L$ is tripled, the new inductance is $L' = 3L$.
The capacitance $C$ is increased by $3C$, so the new capacitance is $C' = C + 3C = 4C$.
The new resonant frequency $f'$ is given by:
$f' = \dfrac{1}{2\pi\sqrt{L'C'}}$
Substituting the values of $L'$ and $C'$:
$f' = \dfrac{1}{2\pi\sqrt{(3L)(4C)}} = \dfrac{1}{2\pi\sqrt{12LC}}$
$f' = \dfrac{1}{2\sqrt{3}} \left( \dfrac{1}{2\pi\sqrt{LC}} \right)$
$f' = \dfrac{f}{2\sqrt{3}}$
Answer: $\dfrac{f}{2\sqrt{3}}$
Hence, the correct answer is option: 'B'.