Q149 / 150 · MHT CET 2026 (19 April Shift 1) · Physics
For a series LCR circuit inductive reactance $X_L$ is equal to resistance $R$ and also equal to twice the capacitive reactance $X_C$. The impedance of the circuit and the phase difference between voltage V and current i are respectively
- A$\sqrt{5}R,\ \tan^{-1}\left(\dfrac{1}{2}\right)$
- B$\sqrt{5}R,\ \tan^{-1}(2)$
- Correct answerC$\dfrac{\sqrt{5}R}{2},\ \tan^{-1}\left(\dfrac{1}{2}\right)$
- D$\dfrac{\sqrt{5}R}{2},\ \tan^{-1}(2)$
Solution
Given $X_L = R$ and $X_L = 2X_C$.
This gives $X_C = \dfrac{R}{2}$.
The impedance of the series LCR circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Substituting the values:
$Z = \sqrt{R^2 + \left(R - \dfrac{R}{2}\right)^2} = \sqrt{R^2 + \dfrac{R^2}{4}} = \dfrac{\sqrt{5}R}{2}$.
The phase difference $\phi$ between voltage and current is given by $\tan \phi = \dfrac{X_L - X_C}{R}$.
$\tan \phi = \dfrac{R - \dfrac{R}{2}}{R} = \dfrac{1}{2}$.
$\phi = \tan^{-1}\left(\dfrac{1}{2}\right)$.
Answer: $\dfrac{\sqrt{5}R}{2},\ \tan^{-1}\left(\dfrac{1}{2}\right)$
This gives $X_C = \dfrac{R}{2}$.
The impedance of the series LCR circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Substituting the values:
$Z = \sqrt{R^2 + \left(R - \dfrac{R}{2}\right)^2} = \sqrt{R^2 + \dfrac{R^2}{4}} = \dfrac{\sqrt{5}R}{2}$.
The phase difference $\phi$ between voltage and current is given by $\tan \phi = \dfrac{X_L - X_C}{R}$.
$\tan \phi = \dfrac{R - \dfrac{R}{2}}{R} = \dfrac{1}{2}$.
$\phi = \tan^{-1}\left(\dfrac{1}{2}\right)$.
Answer: $\dfrac{\sqrt{5}R}{2},\ \tan^{-1}\left(\dfrac{1}{2}\right)$
Hence, the correct answer is option: 'C'.