Question

An inductive coil has a resistance of $100\ \Omega$. When an a.c. signal of frequency $1000\ \text{Hz}$ is applied to the coil the voltage leads the current by $45^\circ$. The inductance of the coil is ($\tan 45^\circ = 1$)

• A. $\frac{0.25}{\pi}\ \text{H}$
• B. $\frac{0.05}{\pi}\ \text{H}$
• C. $0.25\pi\ \text{H}$
• D. $0.5\pi\ \text{H}$

Hint:

Whenever the voltage leads the current by exactly $45^\circ$ in an RL circuit (or lags by $45^\circ$ in an RC circuit), the reactance is strictly equal to the resistance ($X_L = R$ or $X_C = R$).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given the resistance, frequency, and the phase angle by which voltage leads current in an RL circuit. We need to find the inductance of the coil.

Step 2: Key Formula or Approach:
For a series RL circuit, the tangent of the phase angle ($\phi$) is the ratio of inductive reactance ($X_L$) to resistance ($R$):
$$\tan \phi = \frac{X_L}{R}$$
The inductive reactance is related to inductance ($L$) by $X_L = \omega L = 2\pi f L$.

Step 3: Detailed Explanation:
Given parameters:
Resistance $R = 100\ \Omega$
Frequency $f = 1000\ \text{Hz}$
Phase angle $\phi = 45^\circ$
First, find the inductive reactance $X_L$:
$$\tan 45^\circ = \frac{X_L}{100}$$
Since $\tan 45^\circ = 1$:
$$1 = \frac{X_L}{100} \implies X_L = 100\ \Omega$$
Now, use the formula for inductive reactance to find $L$:
$$X_L = 2\pi f L$$
$$100 = 2\pi (1000) L$$
$$100 = 2000\pi L$$
$$L = \frac{100}{2000\pi} = \frac{1}{20\pi}\ \text{H}$$
Convert the fraction to a decimal to match the options:
$$L = \frac{0.05}{\pi}\ \text{H}$$

Step 4: Final Answer:
The inductance of the coil is $\frac{0.05}{\pi}\ \text{H}$, matching option (B).