Question

An inductor of inductance L = 400mH and resistors of resistance R₁ = 2Ω and R₂ = 2Ω are connected to a battery of emf 12V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at t = 0. The potential drop across L as a function of time is: 

• A. \( \dfrac{12}{t} e^{-3t}\,\text{V} \)
• B. \( 6(1-e^{-t/0.2})\,\text{V} \)
• C. \( 12e^{-5t}\,\text{V} \)
• D. 6e⁻5tV

Hint:

In RL circuits, voltage across the inductor decays exponentially after switching.

Answer 1

The Correct Option is D

Step 1: Equivalent resistance in series with inductor: R = R₁ + R₂ = 4Ω
Step 2: Time constant: τ = (L)/(R) = (0.4)/(4) = 0.1s
Step 3: Voltage across inductor: VL = V₀ e⁻t/τ = 6e⁻5tV