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Question #43478

Difficulty: 8/10 | 🟢 Free

The Question

An ideal inductor of inductance $L$ (in Henry) and a resistor of resistance $R$ (in Ohm) are connected in series to a DC voltage source of constant EMF $E$ (in Volt) and a switch. The switch is closed at $t = 0 \text{ s}$ . What is the average current $\langle i \rangle$ flowing through the circuit during the time interval from $t = 0$ to $t = \tau$ , where $\tau = \frac{L}{R}$ is the inductive time constant of the circuit?

Options

A.
$\frac{E}{eR} \text{ A}$
B.
$\frac{E}{2R}\left(1 - \frac{1}{e}\right) \text{ A}$
C.
$\frac{E}{R}\left(1 - \frac{1}{e}\right) \text{ A}$
D.
$\frac{E}{R}\left(1 - \frac{1}{2e}\right) \text{ A}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Average vs. Instantaneous Calculus|Transient Current in RL Circuit|Exponential Differentiation

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to distinguish between the arithmetic mean of boundary values and the true time-averaged value of a non-linear function.

A common trap is assuming that the average current during the transient phase is simply the midpoint of the initial current ( $0$ ) and the final current at $t=\tau$ ( $\frac{E}{R}(1 - 1/e)$ ), which yields $\frac{E}{2R}(1 - 1/e)$ .

However, because the current growth $i(t) = \frac{E}{R}(1 - e^{-t/\tau})$ is a concave exponential function, the true average requires integrating the instantaneous current to find the total charge flow.

Relying on linear approximations for exponential transient processes leads to significant conceptual and numerical errors.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Motion in a Straight Line - Class 11 Part 1
Concept	Average vs. Instantaneous Calculus
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free