Question

An inductor coil with an internal resistance of 50 Ω stores magnetic field energy 180 mJ and dissipates energy as heat at the rate of 200 W when a constant current 9 passed through it. The inductance of the coil will be:

• A. 90 mH
• B. 120 mH
• C. 45 mH
• D. 30 mH
• E. 60 mH

Answer 1

The Correct Option is A

The correct answer is (A): 90 mH

Answer 2

To find the inductance of the coil, let's use the information provided:

1. Internal resistance (\( R \)): 50 Ω
2. Stored magnetic field energy (\( E \)): 180 mJ = 0.18 J
3. Power dissipation (\( P \)): 200 W
4. Current (\( I \)): 9 A

First, we need to confirm the power dissipation due to the resistance. The power dissipated by a resistor is given by:

\[ P = I^2 R \]

Substituting the given values:

\[ P = 9^2 \times 50 \]
\[ P = 81 \times 50 \]
\[ P = 4050 \text{ W} \]

Since the problem states that the power dissipation is 200 W, this discrepancy suggests that either the stated current or the stated power dissipation might be incorrect. Assuming the given power dissipation is correct and constant, we need to check the current using:

\[ I = \sqrt{\frac{P}{R}} \]

So,

\[ I = \sqrt{\frac{200}{50}} \]
\[ I = \sqrt{4} \]
\[ I = 2 \text{ A} \]

Now, we assume the current 2 A is correct and use it to find the inductance of the coil.

The energy stored in the inductor is given by:

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

Rearranging for \( L \):

\[ L = \frac{2E}{I^2} \]

Substitute the given values:

\[ L = \frac{2 \times 0.18}{2^2} \]
\[ L = \frac{0.36}{4} \]
\[ L = 0.09 \text{ H} \]

Therefore, the inductance of the coil is \( 0.09 \text{ H} \) or \( 90 \text{ mH} \).
So The correct Answer is Option  (A): 90 mH