Q174 / 180 · COMEDK 2022 · Physics
An alternating voltage $=200 \sin 100 t$ is applied to a series combination of $R=30 \Omega$ and an inductor of $400 \mathrm{mH}$. The power factor of the circuit is,
- A0.01
- Correct answerB0.6
- C0.05
- D0.042
Solution
The given alternating voltage is $V = 200 \sin(100t)$, which is of the form $V = V_m \sin(\omega t)$.
Comparing the two, the angular frequency is $\omega = 100 \text{ rad/s}$.
The resistance is $R = 30 \Omega$ and the inductance is $L = 400 \text{ mH} = 0.4 \text{ H}$.
The inductive reactance $X_L$ is given by $X_L = \omega L = 100 \times 0.4 = 40 \Omega$.
The impedance $Z$ of the series $RL$ circuit is $Z = \sqrt{R^2 + X_L^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \Omega$.
The power factor $\cos \phi$ is defined as $\cos \phi = \dfrac{R}{Z}$.
Substituting the values, $\cos \phi = \dfrac{30}{50} = 0.6$.
Answer: 0.6
Comparing the two, the angular frequency is $\omega = 100 \text{ rad/s}$.
The resistance is $R = 30 \Omega$ and the inductance is $L = 400 \text{ mH} = 0.4 \text{ H}$.
The inductive reactance $X_L$ is given by $X_L = \omega L = 100 \times 0.4 = 40 \Omega$.
The impedance $Z$ of the series $RL$ circuit is $Z = \sqrt{R^2 + X_L^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \Omega$.
The power factor $\cos \phi$ is defined as $\cos \phi = \dfrac{R}{Z}$.
Substituting the values, $\cos \phi = \dfrac{30}{50} = 0.6$.
Answer: 0.6
Hence, the correct answer is option: 'B'.