AIIMS2018Physics-Current Electricity

AIIMS 2018 Physics RL Circuits MCQ Question

Type: MCQ-numerical-Medium-Class 12

The expression of current is, I = \frac{ε\varepsilon}{R} \left[¹⁻ e^{-tR/L}$$\right] Substitute values, I = \frac{2}{3} \left[¹⁻ e^{-(2)(1.5)/(3)}$$\right] = \frac{2}{3} \left[¹⁻ \frac{1}{e}\right] = 0.4 $$\textA{ A}

Correct Answer

Option A

Detailed Explanation

Option A is correct because the calculated current II is derived from the formula I=εR[1etR/L]I = \frac{\varepsilon}{R} \left[1 - e^{-tR/L}\right], where substituting the values ε=2V\varepsilon = 2 \, \text{V}, R=3ΩR = 3 \, \Omega, t=2st = 2 \, \text{s}, and L=1.5HL = 1.5 \, \text{H} yields I=0.4AI = 0.4 \, \text{A}. The other options are not applicable as they do not provide any alternative answers or values relevant to the problem, reinforcing that option A is the only valid solution based on the calculations performed. Understanding this equation is crucial as it describes the transient behavior of current in an RL circuit, illustrating how current approaches its maximum value over time.

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