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

The Correct Option is A

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:
\[L = 0.76 \, \text{H}.\]