Question:medium

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

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