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NEET PHYSICSELECTROMAGNETIC INDUCTIONMedium

Question

An ideal inductor-resistor-battery circuit is switched on at t=0t=0 s. At time tt, the current is i=i0(1et/τ)i=i_0(1-e^{-t/\tau}), where i0i_0 is the steady-state value. The time at which the current becomes 0.5i00.5i_0 is: [Given ln(2)=0.693\ln(2)=0.693]

A

6.93×1036.93 \times 10^3 s

B

6.936.93 ms

C

69.369.3 s

D

6.936.93 s

Step-by-Step Solution

The growth of current in an LR circuit is governed by the equation: i=i0(1et/τ)i = i_0 (1 - e^{-t/\tau})

1. Set the Condition: We need to find the time tt when the current ii reaches 50% of its steady-state value i0i_0. 0.5i0=i0(1et/τ)0.5 i_0 = i_0 (1 - e^{-t/\tau})

2. Simplify: 0.5=1et/τ0.5 = 1 - e^{-t/\tau} et/τ=10.5=0.5e^{-t/\tau} = 1 - 0.5 = 0.5

3. Solve for t: Taking the natural logarithm (ln) on both sides: t/τ=ln(0.5)-t/\tau = \ln(0.5) t/τ=ln(2)-t/\tau = -\ln(2) t=τln(2)t = \tau \ln(2)

4. Calculation: Given ln(2)=0.693\ln(2) = 0.693, we have: t=0.693τt = 0.693 \tau

(Note: The provided question text is missing the specific value of the time constant τ\tau. However, for the answer to be 6.93 ms, the time constant τ\tau must be 10 ms. Assuming τ=10 ms\tau = 10 \text{ ms}, then t=0.693×10 ms=6.93 mst = 0.693 \times 10 \text{ ms} = 6.93 \text{ ms}.)

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ELECTROMAGNETIC INDUCTION. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSELECTROMAGNETIC INDUCTIONinductorresistorbatterycircuitswitchedcurrentiiettau

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