Q148 / 150 · MHT CET 2026 (15 April Shift 1) · Physics
A series LCR circuit is connected to an a.c. source of $220$ V, $50$ Hz. The circuit contains a resistance $R = 80\,\Omega$, an inductor of inductive reactance $70\,\Omega$ and capacitor of capacitive reactance $130\,\Omega$. The power factor of the circuit is $\dfrac{x}{10}$. The value of x is
- Correct answerA$8$
- B$6$
- C$4$
- D$7$
Solution
The impedance of the series LCR circuit is given by
$Z = \sqrt{R^2 + (X_C - X_L)^2}$
Substituting the given values:
$Z = \sqrt{80^2 + (130 - 70)^2}$
$Z = \sqrt{80^2 + 60^2}$
$Z = \sqrt{6400 + 3600} = \sqrt{10000} = 100\,\Omega$
The power factor of the circuit is
$\cos \phi = \dfrac{R}{Z}$
$\cos \phi = \dfrac{80}{100} = \dfrac{8}{10}$
Comparing this with $\dfrac{x}{10}$, we get $x = 8$.
Answer: $8$
$Z = \sqrt{R^2 + (X_C - X_L)^2}$
Substituting the given values:
$Z = \sqrt{80^2 + (130 - 70)^2}$
$Z = \sqrt{80^2 + 60^2}$
$Z = \sqrt{6400 + 3600} = \sqrt{10000} = 100\,\Omega$
The power factor of the circuit is
$\cos \phi = \dfrac{R}{Z}$
$\cos \phi = \dfrac{80}{100} = \dfrac{8}{10}$
Comparing this with $\dfrac{x}{10}$, we get $x = 8$.
Answer: $8$
Hence, the correct answer is option: 'A'.