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Question #42414

Difficulty: 8/10 | 🟢 Free

The Question

A time-varying periodic current $i(t)$ flows through a fixed resistor of resistance $R$ . The current waveform has a period of $2t_0$ and is defined as: $i(t) = \begin{cases} I_0 \left(\frac{t}{t_0}\right)^{3/2} & \text{for } 0 \le t < t_0 \\ 0 & \text{for } t_0 \le t < 2t_0 \end{cases}$ where $I_0$ is a positive constant in Amperes, $t_0$ is in seconds, and $R$ is in Ohms. What is the average thermal power dissipated in the resistor over a long time?

Options

A.
$\frac{I_0^2 R}{8}$
B.
$\frac{I_0^2 R}{25}$
C.
$\frac{2I_0^2 R}{25}$
D.
$\frac{I_0^2 R}{4}$

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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

Instantaneous vs Average|Mean vs. RMS Values|Calculus: Integration

Why This Question Is Tricky

Common Mistakes

This question targets a fundamental misconception in AC circuits and time-varying signals: conflating the average of a squared function (RMS basis) with the square of the average function.

Students falling into the primary trap will calculate the average current over the entire period ( $I_{avg} = I_0/5$ ) and incorrectly compute power as $P = I_{avg}^2 R = I_0^2 R / 25$ .

A secondary trap involves calculating the average current only during the active half-cycle ( $2I_0/5$ ), squaring it, and then halving it for the full period, yielding $2I_0^2 R / 25$ .

The problem forces students to return to first principles, integrating the instantaneous power $i^2(t)R$ over the full period $2t_0$ to find the true average thermal power.

Quick Facts About This Question

Question Information

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