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 R₁ as a function of time is: 

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

Hint:

In RL circuits, current (and voltage across resistors) varies exponentially: I(t) = I₀(1-e⁻t/τ), τ=(L)/(R) Always calculate the correct time constant.

Answer 1

The Correct Option is C

Step 1: When the switch is closed, current in the inductor branch increases gradually due to self-induction.
Step 2: The time constant of the RL circuit is: τ = (L)/(R₂) = (0.4)/(2) = 0.2s
Step 3: The angular decay constant is: (R₂)/(L) = (2)/(0.4) = 5s⁻1
Step 4: The potential drop across R₁ decreases exponentially as current is diverted into the inductor branch: VR₁ = 12e⁻5tV