Question

When a choke coil of inductance L and small resistance R, is connected to a DC voltage supply of 100 V and negligible resistance, the current in the coil is 50 A. When the same coil is connected to an AC voltage supply of 100 V, $\frac{50}{\pi}$ H: frequency, in series with a capacitor of capacitance 50 µF, the current in the coil is still 50

Answer 1

2

Answer 2

1. DC Circuit Analysis: For a DC supply, an inductor offers no opposition to current flow (its inductive reactance $X_L = 0$). The coil's resistance $R$ is the only impedance. Using Ohm's Law for DC: $V_{DC} = I_{DC} \times R$ $100 \text{ V} = 50 \text{ A} \times R$ $R = \frac{100}{50} = 2 \Omega$

2. AC Circuit Analysis: For the AC supply, the total impedance $Z_{total}$ of the series circuit (coil + capacitor) is given by Ohm's Law: $Z_{total} = \frac{V_{AC}}{I_{AC}}$ $Z_{total} = \frac{100 \text{ V}}{50 \text{ A}} = 2 \Omega$

3. Reactance Calculations: The angular frequency $\omega$ is calculated from the given frequency $f = \frac{50}{\pi}$ Hz: $\omega = 2\pi f = 2\pi \left(\frac{50}{\pi}\right) = 100$ rad/s

   The capacitive reactance $X_C$ is: $X_C = \frac{1}{\omega C} = \frac{1}{100 \text{ rad/s} \times 50 \times 10^{-6} \text{ F}} = \frac{1}{0.005} = 200 \Omega$

   The inductive reactance $X_L$ is given by $X_L = \omega L = 100L$.

4. Impedance Equation: The total impedance of a series RLC circuit is $Z_{total} = \sqrt{R^2 + (X_L - X_C)^2}$. Substituting the known values: $2 \Omega = \sqrt{(2 \Omega)^2 + (100L - 200 \Omega)^2}$ Squaring both sides: $4 = 4 + (100L - 200)^2$ $(100L - 200)^2 = 0$ $100L - 200 = 0$ $100L = 200$ $L = \frac{200}{100} = 2$ H

The inductance of the coil $L$ is 2 H.