Q144 / 150 · MHT CET 2026 (20 April Shift 1) · Physics
In a series LCR circuit, the voltage across R is $100$ V, $R = 1\,\text{K}\,\Omega$ and $C = 2\,\mu\text{F}$. The angular frequency $\omega$ is $200$ rad s$^{-1}$. At resonance, the voltage across '$L$' is
- A$150$ V
- B$200$ V
- Correct answerC$250$ V
- D$300$ V
Solution
The current in the series LCR circuit is given by $I = \dfrac{V_R}{R}$.
Substituting the given values:
$I = \dfrac{100}{1000} = 0.1$ A
The capacitive reactance is $X_C = \dfrac{1}{\omega C}$.
$X_C = \dfrac{1}{200 \times 2 \times 10^{-6}} = \dfrac{10^6}{400} = 2500 \, \Omega$
The voltage across the capacitor is $V_C = I X_C$.
$V_C = 0.1 \times 2500 = 250$ V
At resonance, the inductive reactance equals the capacitive reactance ($X_L = X_C$), so the voltage across the inductor equals the voltage across the capacitor.
$V_L = V_C = 250$ V
Answer: $250$ V
Substituting the given values:
$I = \dfrac{100}{1000} = 0.1$ A
The capacitive reactance is $X_C = \dfrac{1}{\omega C}$.
$X_C = \dfrac{1}{200 \times 2 \times 10^{-6}} = \dfrac{10^6}{400} = 2500 \, \Omega$
The voltage across the capacitor is $V_C = I X_C$.
$V_C = 0.1 \times 2500 = 250$ V
At resonance, the inductive reactance equals the capacitive reactance ($X_L = X_C$), so the voltage across the inductor equals the voltage across the capacitor.
$V_L = V_C = 250$ V
Answer: $250$ V
Hence, the correct answer is option: 'C'.