Q142 / 150 · MHT CET 2026 (20 April Shift 1) · Physics
When an inductance '$L$' and resistor '$R$' are connected in series to $25$V, $50$ Hz supply, a current of $0.5$ A flows in the circuit. The current lags behind in phase from applied voltage by $\left(\dfrac{\pi}{3}\right)$ radian. The value of R is $\left(\cos 60^\circ = \dfrac{1}{2}\right)$
- A$20\,\Omega$
- Correct answerB$25\,\Omega$
- C$40\,\Omega$
- D$50\,\Omega$
Solution
The impedance of the LR series circuit is given by:
$Z = \dfrac{V}{I}$
Substituting the given values:
$Z = \dfrac{25}{0.5} = 50 \, \Omega$
The power factor of the circuit is related to the resistance and impedance by:
$\cos \phi = \dfrac{R}{Z}$
Given that the current lags the voltage by $\phi = \dfrac{\pi}{3}$ radian ($60^\circ$):
$R = Z \cos \phi$
$R = 50 \times \cos 60^\circ$
$R = 50 \times \dfrac{1}{2} = 25 \, \Omega$
Answer: $25\,\Omega$
$Z = \dfrac{V}{I}$
Substituting the given values:
$Z = \dfrac{25}{0.5} = 50 \, \Omega$
The power factor of the circuit is related to the resistance and impedance by:
$\cos \phi = \dfrac{R}{Z}$
Given that the current lags the voltage by $\phi = \dfrac{\pi}{3}$ radian ($60^\circ$):
$R = Z \cos \phi$
$R = 50 \times \cos 60^\circ$
$R = 50 \times \dfrac{1}{2} = 25 \, \Omega$
Answer: $25\,\Omega$
Hence, the correct answer is option: 'B'.