Q152 / 180 · COMEDK 2025 (25 May Shift) · Physics
In a series LCR circuit, the inductive reactance $\mathrm{X}_{\mathrm{L}}$ is $24 \Omega$ and the capacitive reactance $\mathrm{X}_{\mathrm{C}}$ is $12 \Omega$. The resistance R in the circuit is $9 \Omega$. What is the power factor of the circuit?
- A$60^{\circ}$
- B$37^{\circ}$
- C$45^{\circ}$
- Correct answerD$53^{\circ}$
Solution
The power factor of a series LCR circuit is defined as $\cos \phi = \dfrac{R}{Z}$, where $Z$ is the impedance of the circuit.
The impedance $Z$ is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Given $R = 9 \Omega$, $X_L = 24 \Omega$, and $X_C = 12 \Omega$, we calculate $Z$:
$Z = \sqrt{9^2 + (24 - 12)^2} = \sqrt{81 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15 \Omega$.
The phase angle $\phi$ is given by $\tan \phi = \dfrac{X_L - X_C}{R}$.
Substituting the values, $\tan \phi = \dfrac{24 - 12}{9} = \dfrac{12}{9} = \dfrac{4}{3}$.
Since $\tan \phi = \dfrac{4}{3}$, the angle $\phi$ is $53^{\circ}$.
Answer: $53^{\circ}$
The impedance $Z$ is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Given $R = 9 \Omega$, $X_L = 24 \Omega$, and $X_C = 12 \Omega$, we calculate $Z$:
$Z = \sqrt{9^2 + (24 - 12)^2} = \sqrt{81 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15 \Omega$.
The phase angle $\phi$ is given by $\tan \phi = \dfrac{X_L - X_C}{R}$.
Substituting the values, $\tan \phi = \dfrac{24 - 12}{9} = \dfrac{12}{9} = \dfrac{4}{3}$.
Since $\tan \phi = \dfrac{4}{3}$, the angle $\phi$ is $53^{\circ}$.
Answer: $53^{\circ}$
Hence, the correct answer is option: 'D'.