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:
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:
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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.
Updated On: Mar 19, 2026
- \( \dfrac{12}{t}e^{-3t}\,\text{V} \)
- \( 6(1-e^{-0.2t})\,\text{V} \)
- \( 12e^{-5t}\,\text{V} \)
- 6e⁻5tV
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The Correct Option is C
Solution and Explanation
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
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