Q194 / 200 · MHT CET - Medical 2026 (22 April Shift 1) · Physics
In a series LCR circuit, the inductance L is 10 mH, capacitance C is 100$\mu$F and resistance R is 20 $\Omega$. The frequency at which impedance is 20 $\Omega$, is
- A$\dfrac{10^4}{2\pi}$
- B$\dfrac{10^4}{\pi}$
- Correct answerC$\dfrac{10^3}{2\pi}$
- D$\dfrac{10^3}{\pi}$
Solution
Given $L = 10 \text{ mH} = 10^{-2} \text{ H}$, $C = 100 \mu\text{F} = 10^{-4} \text{ F}$, and $R = 20 \Omega$.
The impedance of a series LCR circuit is given by $Z = \sqrt{R^2 + \left(X_L - X_C\right)^2}$.
Since $Z = 20 \Omega = R$, the circuit is in resonance, which implies $X_L = X_C$.
$\omega L = \dfrac{1}{\omega C} \Rightarrow \omega = \dfrac{1}{\sqrt{LC}}$
Substituting the values of $L$ and $C$:
$\omega = \dfrac{1}{\sqrt{10^{-2} \times 10^{-4}}} = \dfrac{1}{\sqrt{10^{-6}}} = 10^3 \text{ rad/s}$
The frequency $f$ is given by $f = \dfrac{\omega}{2\pi}$.
$f = \dfrac{10^3}{2\pi} \text{ Hz}$
Answer: $\dfrac{10^3}{2\pi}$
The impedance of a series LCR circuit is given by $Z = \sqrt{R^2 + \left(X_L - X_C\right)^2}$.
Since $Z = 20 \Omega = R$, the circuit is in resonance, which implies $X_L = X_C$.
$\omega L = \dfrac{1}{\omega C} \Rightarrow \omega = \dfrac{1}{\sqrt{LC}}$
Substituting the values of $L$ and $C$:
$\omega = \dfrac{1}{\sqrt{10^{-2} \times 10^{-4}}} = \dfrac{1}{\sqrt{10^{-6}}} = 10^3 \text{ rad/s}$
The frequency $f$ is given by $f = \dfrac{\omega}{2\pi}$.
$f = \dfrac{10^3}{2\pi} \text{ Hz}$
Answer: $\dfrac{10^3}{2\pi}$
Hence, the correct answer is option: 'C'.