Question

In an A.C. sub-circuit as shown in figure, the resistance R = 0.2 $\Omega$. At a certain instant $V_A - V_B = 0.5V, I = 0.5 A,$ and current is increasing at the rate of $\frac{\Delta I}{\Delta t} = 8 A/s.$ The inductance of the coil is:

• A. 0.01 H
• B. 0.02 H
• C. 0.05 H
• D. 0.5 H

Answer 1

Answer: (C)

0.05 H

Answer 2

The circuit shows an inductor (L) and a resistor (R) connected in series. The current (I) flows from point A to point B.

The voltage drop across the inductor ($V_L$) is given by: $V_L = L \frac{dI}{dt}$

Since the current is increasing, the induced EMF opposes the increase in current, thus acting as a voltage drop in the direction of current flow.

The voltage drop across the resistor ($V_R$) is given by Ohm's law: $V_R = IR$

Applying Kirchhoff's Voltage Law (KVL) from point A to point B: The total potential difference $V_A - V_B$ is the sum of the potential drops across the inductor and the resistor. $V_A - V_B = V_L + V_R$ $V_A - V_B = L \frac{dI}{dt} + IR$

We are given the following values: $R = 0.2 \, \Omega$ $V_A - V_B = 0.5 \, V$ $I = 0.5 \, A$ $\frac{dI}{dt} = 8 \, A/s$

Substitute these values into the KVL equation: $0.5 = L (8) + (0.5)(0.2)$ $0.5 = 8L + 0.1$

Now, solve for L: $8L = 0.5 - 0.1$ $8L = 0.4$ $L = \frac{0.4}{8}$ $L = 0.05 \, H$

The inductance of the coil is 0.05 H.