Question

An electric lamp is designed to operate at $110 \, \mathrm{V} $ DC and $11 \, \mathrm{A}$ current. If the lamp is operated on $220 \, \mathrm{V}$, $50 \, \mathrm{Hz}$ AC source with a coil in series, then find the inductance of the coil.Correct Answer: 
The inductance of the coil is $L = 0.064 \, \mathrm{H}$.

Hint:

To calculate the inductance of a coil in an AC circuit, determine the total impedance, separate the resistive and reactive components, and use the relationship $X_L = \omega L$.

Answer 1

The lamp is designed for DC operation with the following specifications: \[ V_{\mathrm{DC}} = 110 \, \mathrm{V}, \quad I = 11 \, \mathrm{A}. \] The power consumed by the lamp is: \[ P = V_{\mathrm{DC}} \cdot I = 110 \cdot 11 = 1210 \, \mathrm{W}. \] When the lamp is connected to an AC source of $220 \, \mathrm{V}$ and $50 \, \mathrm{Hz}$ with a coil in series, the total impedance $Z$ of the circuit is given by: \[ Z = \frac{V_{\mathrm{AC}}}{I} = \frac{220}{11} = 20 \, \Omega. \] The impedance of the circuit is the combination of the resistance of the lamp and the inductive reactance of the coil: \[ Z = \sqrt{R^2 + X_L^2}. \] The resistance of the lamp is: \[ R = \frac{V_{\mathrm{DC}}}{I} = \frac{110}{11} = 10 \, \Omega. \] The inductive reactance $X_L$ is given by: \[ X_L = \sqrt{Z^2 - R^2}. \] Substituting the values: \[ X_L = \sqrt{20^2 - 10^2} = \sqrt{400 - 100} = \sqrt{300} = 10\sqrt{3} \, \Omega. \] The inductive reactance is related to the inductance $L$ by: \[ X_L = \omega L, \] where $\omega = 2\pi f$ is the angular frequency of the AC source. For $f = 50 \, \mathrm{Hz}$: \[ \omega = 2\pi \cdot 50 = 100\pi \, \mathrm{rad/s}. \] Substituting for $L$: \[ L = \frac{X_L}{\omega} = \frac{10\sqrt{3}}{100\pi}. \] Simplifying: \[ L = \frac{\sqrt{3}}{10\pi} \, \mathrm{H}. \] Approximating: \[ L \approx 0.064 \, \mathrm{H}. \]