Question

A coil has an inductance of 2H and resistance of 4W, A 10V is applied across the coil. The energy stored in the magnetic field after the current has built up to its equilibrium value will be ____ × 10^{–2} J.

Answer 1

Correct Answer: 625

Given:

• Inductance (\( L \)) = 2 H
• Resistance (\( R \)) = 4 \( \Omega \)
• Applied voltage (\( V \)) = 10 V

Step 1: Analyze the Circuit at Steady State

At steady state, the inductor behaves as a short circuit, meaning it has zero resistance. The circuit becomes a simple resistor connected to the voltage source.

Step 2: Calculate the Current at Steady State

Using Ohm’s law, the steady-state current (\( I \)) through the resistor is given by:

\[ I = \frac{V}{R}. \]

Substitute \( V = 10 \, \text{V} \) and \( R = 4 \, \Omega \):

\[ I = \frac{10}{4} = \frac{5}{2} = 2.5 \, \text{A}. \]

Step 3: Calculate the Energy Stored in the Magnetic Field

The energy (\( E \)) stored in the magnetic field of the inductor is given by:

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

Substitute \( L = 2 \, \text{H} \) and \( I = 2.5 \, \text{A} \):

\[ E = \frac{1}{2} \cdot 2 \cdot \left(2.5\right)^2. \]

Calculate \( (2.5)^2 \):

\[ (2.5)^2 = 6.25. \]

Substitute back into the equation:

\[ E = \frac{1}{2} \cdot 2 \cdot 6.25 = 6.25 \, \text{J}. \]

Convert to scientific notation:

\[ E = 625 \times 10^{-2} \, \text{J}. \]

Final Answer:

The energy stored in the magnetic field is \( 625 \times 10^{-2} \, \text{J} \).