A coil of inductance `L=50muH` and resistance = `0.5Omega` is connected to a battery of emf = 5V.
A resistance of `10Omega` is connected parallel to the coil. Now, at the same instant the connection of the battery is switched off. Then, the amount of heat generated in the coil after switching off the battery is

Options

A. 1.25 mJ
B. 2.5 mJ
C. 0.65 mJ
D. 0.12 mJ

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Calculate the current flowing through the inductor when the battery is connected.
The current \( I \) through the inductor can be calculated using Ohm's law. The total resistance in the circuit is the sum of the resistance of the coil \( R = 0.5 \, \Omega \) and the parallel resistance \( R_p = 10 \, \Omega \).

The equivalent resistance \( R_{eq} \) of the parallel combination of \( R_p \) and \( R \) is given by:

\[ \frac{1}{R_{eq}} = \frac{1}{R} + \frac{1}{R_p} \]

Substituting the values:

\[ \frac{1}{R_{eq}} = \frac{1}{0.5} + \frac{1}{10} \]

Calculating this gives:

\[ \frac{1}{R_{eq}} = 2 + 0.1 = 2.1 \implies R_{eq} = \frac{1}{2.1} \approx 0.476 \, \Omega \]

Now, using Ohm's law, the current \( I \) can be calculated as:

\[ I = \frac{E}{R_{eq}} = \frac{5 \, V}{0.476 \, \Omega} \approx 10.51 \, A \]

### Step 2: Calculate the energy stored in the inductor.
The energy \( U \) stored in the inductor is given by the formula:

\[ U = \frac{1}{2} L I^2 \]

Substituting the values:

\[ U = \frac{1}{2} \times 50 \times 10^{-6} \, H \times (10.51)^2 \]

Calculating this gives:

\[ U \approx \frac{1}{2} \times 50 \times 10^{-6} \times 110.53 \approx 0.00276 \, J = 2.76 \, mJ \]

### Step 3: Determine the heat generated in the coil after switching off the battery.
When the battery is switched off, the energy stored in the inductor will be dissipated as heat in the resistances. The heat \( Q \) developed in the coil can be calculated using the formula:

\[ Q = U \cdot \frac{R}{R + R_p} \]

Substituting the values:

\[ Q = 2.76 \, mJ \cdot \frac{0.5}{0.5 + 10} = 2.76 \, mJ \cdot \frac{0.5}{10.5} \]

Calculating this gives:

\[ Q \approx 2.76 \times \frac{0.5}{10.5} \approx 0.131 \, mJ \approx 0.12 \, mJ \]

### Final Answer:
The amount of heat generated in the coil after switching off the battery is approximately **0.12 mJ**.

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