Question

An alternating voltage $e=200\sin 100t$ is applied to a series combination and an inductor of $400mH$ and $R=30\Omega$. The power factor of the circuit is:

• A. $0.05$
• B. $0.2$
• C. $0.6$
• D. $0.042$

Answer 1

The correct option is C

$0.6$

Step 1: Given

Alternating voltage: $e=200\sin \omega t=200\sin 100t$

Inductance: $L=400mH=0.4H$

Resistance: $R=30\Omega$

Step 2: Formula Used

1. Inductive reactance is given by $X_L=\omega L$, where $\omega$ is the angular frequency and $L$ is inductance.
2. Impedance is given by $Z=\sqrt{R^2+{X_L}^2}$, where $R$ is resistance and $X_L$ is inductive reactance.
3. The phase difference is given by $\cos \varphi =\frac{R}{Z}$, where $R$ is resistance and $Z$ is impedance.

Step 3: Find the impedance

Calculate the inductive reactance using the formula

$$\begin{gathered} X_L=\omega L \\ =100\times 0.4 \\ =40\Omega \end{gathered}$$

Calculate the impedance using the formula and value of inductive reactance obtained

$$\begin{gathered} Z=\sqrt{{30}^2+{40}^2} \\ =\sqrt{900+1600} \\ =\sqrt{2500} \\ =50\Omega \end{gathered}$$

Step 4: Calculate the phase difference using the formula

$$\begin{gathered} \cos \varphi =\frac{30}{50} \\ =0.6 \end{gathered}$$

Hence, Option (C)is correct. The power factor of the circuit is $0.6$.