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

Given:

• Internal resistance, \( R = 50 \, \Omega \)
• Magnetic field energy stored, \( E = 180 \, \text{mJ} = 0.180 \, \text{J} \)
• Power dissipated as heat, \( P = 200 \, \text{W} \)

Step 1: Find the Current (\( I \))

The power dissipated in the resistor is given by:

\[ P = I^2 R \]

Solving for \( I \):

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

Step 2: Relate Energy to Inductance (\( L \))

The energy stored in the inductor's magnetic field is:

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

Solving for \( L \):

\[ L = \frac{2E}{I^2} = \frac{2 \times 0.180}{2^2} = \frac{0.360}{4} = 0.090 \, \text{H} = 90 \, \text{mH} \]

Answer 2

Step 1: Recall the formulas for magnetic field energy and power dissipation.

The magnetic field energy stored in an inductor is given by:

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

where \( U \) is the energy stored, \( L \) is the inductance, and \( I \) is the current through the inductor.

The power dissipated as heat in the resistor is given by:

\[ P = I^2 R, \]

where \( P \) is the power dissipated, \( R \) is the resistance, and \( I \) is the current.

Step 2: Solve for the current \( I \).

We are given that the power dissipated as heat is \( P = 200 \, \text{W} \) and the resistance is \( R = 50 \, \Omega \). Substituting into the formula for power:

\[ 200 = I^2 \cdot 50. \]

Solve for \( I^2 \):

\[ I^2 = \frac{200}{50} = 4. \]

Taking the square root:

\[ I = \sqrt{4} = 2 \, \text{A}. \]

Step 3: Solve for the inductance \( L \).

We are given that the magnetic field energy stored is \( U = 180 \, \text{mJ} = 0.180 \, \text{J} \). Using the formula for magnetic field energy:

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

Substitute \( U = 0.180 \, \text{J} \) and \( I^2 = 4 \):

\[ 0.180 = \frac{1}{2} L \cdot 4. \]

Simplify:

\[ 0.180 = 2L. \]

Solve for \( L \):

\[ L = \frac{0.180}{2} = 0.090 \, \text{H} = 90 \, \text{mH}. \]

Final Answer: The inductance of the coil is \( \mathbf{90 \, \text{mH}} \), which corresponds to option \( \mathbf{(A)} \).