Q170 / 200 · MHT CET - Medical 2026 (26 April Shift 1) · Physics
In a series LCR circuit $R = X_L = 2X_C$. The impedance of the circuit and phase difference between 'V' and 'I' respectively will be
- A$\dfrac{R}{2}, \tan^{-1}(0.5)$
- Correct answerB$\dfrac{\sqrt{5}}{2}\ R, \tan^{-1}(0.5)$
- C$\dfrac{\sqrt{5}}{2}\ R, \tan^{-1}(0.2)$
- D$\dfrac{R}{2}, \tan^{-1}(0.2)$
Solution
Given $R = X_L = 2X_C$
$X_L = R$
$X_C = \dfrac{R}{2}$
The impedance of the series LCR circuit is given by:
$Z = \sqrt{R^2 + (X_L - X_C)^2}$
Substituting the values:
$Z = \sqrt{R^2 + \left(R - \dfrac{R}{2}\right)^2}$
$Z = \sqrt{R^2 + \left(\dfrac{R}{2}\right)^2} = \sqrt{R^2 + \dfrac{R^2}{4}} = \dfrac{\sqrt{5}}{2} R$
The phase difference $\phi$ between voltage and current is given by:
$\tan \phi = \dfrac{X_L - X_C}{R}$
$\tan \phi = \dfrac{R - \dfrac{R}{2}}{R} = \dfrac{\dfrac{R}{2}}{R} = 0.5$
$\phi = \tan^{-1}(0.5)$
Answer: $\dfrac{\sqrt{5}}{2}\ R, \tan^{-1}(0.5)$
$X_L = R$
$X_C = \dfrac{R}{2}$
The impedance of the series LCR circuit is given by:
$Z = \sqrt{R^2 + (X_L - X_C)^2}$
Substituting the values:
$Z = \sqrt{R^2 + \left(R - \dfrac{R}{2}\right)^2}$
$Z = \sqrt{R^2 + \left(\dfrac{R}{2}\right)^2} = \sqrt{R^2 + \dfrac{R^2}{4}} = \dfrac{\sqrt{5}}{2} R$
The phase difference $\phi$ between voltage and current is given by:
$\tan \phi = \dfrac{X_L - X_C}{R}$
$\tan \phi = \dfrac{R - \dfrac{R}{2}}{R} = \dfrac{\dfrac{R}{2}}{R} = 0.5$
$\phi = \tan^{-1}(0.5)$
Answer: $\dfrac{\sqrt{5}}{2}\ R, \tan^{-1}(0.5)$
Hence, the correct answer is option: 'B'.