Question

In a series R–L circuit, the voltage of the battery is \(10\ \text{V}\). Resistance and inductance are \(10\ \Omega\) and \(10\ \text{mH}\) respectively. Find the energy stored in the inductor when the current reaches \(\dfrac{1}{e}\) times its maximum value.

• A. \(0.67\ \text{mJ}\)
• B. \(1.33\ \text{mJ}\)
• C. \(0.33\ \text{mJ}\)
• D. \(0.50\ \text{mJ}\)

Hint:

In R–L circuits, the maximum current is always \( \dfrac{V}{R} \). Energy stored in an inductor depends only on the \textbf{instantaneous current}, not directly on time.

Answer 1

The Correct Option is A

Concept: For a series R–L circuit connected to a DC source:
The steady-state (maximum) current is given by \( I_{\max} = \dfrac{V}{R} \)
The energy stored in an inductor is given by \( U = \dfrac{1}{2} L I^2 \)
Step 1: Calculate the maximum current in the circuit. \[ I_{\max} = \frac{V}{R} = \frac{10}{10} = 1\ \text{A} \]
Step 2: Determine the current at the specified instant. \[ I = \frac{1}{e} I_{\max} = \frac{1}{e}\ \text{A} \]
Step 3: Write the formula for the energy stored in the inductor. \[ U = \frac{1}{2} L I^2 \] Given: \[ L = 10\ \text{mH} = 0.01\ \text{H} \] Substituting values: \[ U = \frac{1}{2} \times 0.01 \times \left(\frac{1}{e}\right)^2 \]
Step 4: Evaluate the expression. \[ U = 0.005 \times \frac{1}{e^2} \approx 0.005 \times 0.135 = 6.75 \times 10^{-4}\ \text{J} \] \[ U \approx 0.675\ \text{mJ} \] \[ U \approx 0.67\ \text{mJ} \]