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Question #43488

Difficulty: 8/10 | 🟢 Free

The Question

A series circuit consisting of an inductor of inductance $L$ (in Henrys) and a resistor of resistance $R$ (in Ohms) is initially open. At time $t = 0\text{ s}$ , a switch is closed, connecting an alternating voltage source $V(t) = V_0 \cos(\omega t)$ (in Volts) across the circuit, where $\omega$ is the angular frequency (in rad/s). Assuming the initial current $I$ (in Amperes) in the inductor is zero, what is the value of the concavity of the current-time graph, defined as $\frac{d^2I}{dt^2}$ , exactly at $t = 0^+$ ?

Options

A.
$-\frac{V_0 R}{L^2}$
B.
$-\frac{V_0 \omega^2 R}{R^2 + \omega^2 L^2}$
C.
$-\frac{V_0 \omega}{L}$
D.
$0$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Graphical Concavity Significance|Transient Current in RL Circuit|Resistor Phase Invariance

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to distinguish between transient circuit dynamics and steady-state AC analysis.

A common lethal mistake is assuming the current follows the steady-state phasor solution $I(t) = (V_0/Z) \cos(\omega t - \phi)$ from $t=0$ , which incorrectly yields a concavity dependent on the angular frequency $\omega$ (Option B).

By asking for the instantaneous concavity ( $d^2I/dt^2$ ) at $t=0^+$ , the problem forces students to apply Kirchhoff's Voltage Law as a differential equation ( $L dI/dt + IR = V(t)$ ) and differentiate it.

The trap exploits the phase invariance assumption, revealing that the initial curvature of the current is entirely dictated by the resistive voltage drop of the growing current, independent of the source's frequency.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Motion in a Straight Line - Class 11 Part 1
Concept	Graphical Concavity Significance
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free