Question

A coil of negligible resistance is connected in series with $90 \, \Omega$ resistor across $120 \, \text{V}, \, 60 \, \text{Hz}$ supply. A voltmeter reads $36 \, \text{V}$ across resistance. Inductance of the coil is:

• A. $0.76 \, \text{H}$
• B. $2.86 \, \text{H}$
• C. $0.286 \, \text{H}$
• D. $0.91 \, \text{H}$

Answer 1

Answer: (A)

$0.76 \, \text{H}$

Answer 2

The circuit contains a resistor ($R$) and an inductor ($L$) in series. The total voltage ($V$) is the RMS voltage of the supply. The current in the circuit is:
$I_{\text{rms}} = \frac{V_R}{R},$
where $V_R = 36 \, \text{V}$ and $R = 90 \, \Omega$. Substituting:
$I_{\text{rms}} = \frac{36}{90} = 0.4 \, \text{A}.$
The total impedance of the circuit is:
$Z = \frac{V}{I_{\text{rms}}} = \frac{120}{0.4} = 300 \, \Omega.$
The impedance is related to the resistance and inductive reactance by:
$Z = \sqrt{R^2 + X_L^2}.$
Substituting $Z = 300 \, \Omega$ and $R = 90 \, \Omega$:
$300 = \sqrt{90^2 + X_L^2}.$
Squaring both sides:
$300^2 = 90^2 + X_L^2 \implies X_L^2 = 300^2 - 90^2 = 81900.$
Thus:
$X_L = \sqrt{81900} \approx 286.18 \, \Omega.$
The inductive reactance is given by:
$X_L = \omega L \quad \text{where} \quad \omega = 2\pi f.$
For $f = 60 \, \text{Hz}$:
$\omega = 2\pi \cdot 60 \approx 376.8 \, \text{rad/s}.$
Solving for $L$:
$L = \frac{X_L}{\omega} = \frac{286.18}{376.8} \approx 0.76 \, \text{H}.$
Thus, the inductance of the coil is:
$L = 0.76 \, \text{H}.$