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: Understand the question.
A coil has resistance and also some inductance. When an AC signal of frequency $1000\ \text{Hz}$ flows, the voltage leads the current by $45^\circ$. We must find the inductance $L$ of the coil.

Step 2: Why voltage leads current.
A coil behaves like a resistor and an inductor together. The inductor makes the voltage lead the current. The amount of this lead is set by the phase angle. The bigger the inductive reactance compared to the resistance, the bigger the angle.

Step 3: Write the phase angle rule.
For a coil (resistance $R$ and inductive reactance $X_L$):
\[ \tan\phi = \frac{X_L}{R} \]

Step 4: Find the reactance.
Here $\phi = 45^\circ$ and $R = 100\ \Omega$. Since $\tan 45^\circ = 1$:
\[ 1 = \frac{X_L}{100} \quad\Rightarrow\quad X_L = 100\ \Omega \]

Step 5: Connect reactance to inductance.
The inductive reactance depends on frequency and inductance:
\[ X_L = 2\pi f L \]
Put $X_L = 100$ and $f = 1000$:
\[ 100 = 2\pi (1000) L = 2000\pi L \]

Step 6: Solve for $L$.
\[ L = \frac{100}{2000\pi} = \frac{1}{20\pi}\ \text{H} = \frac{0.05}{\pi}\ \text{H} \]
This matches option (2).
\[ \boxed{L = \frac{0.05}{\pi}\ \text{H}} \]