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Question #43366

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The Question

A parallel plate capacitor has plate area $A$ and separation $d$ . The space between the plates is filled with a non-uniform, lossy dielectric. The real permittivity $\epsilon(x)$ and electrical conductivity $\sigma(x)$ of the dielectric vary with the distance $x$ from one plate ( $0 \le x \le d$ ) as: $\epsilon(x) = \epsilon_0 \left(1 + \frac{x}{d}\right)^2, \quad \sigma(x) = \sigma_0 \left(1 + \frac{x}{d}\right)^2$ The capacitor is connected to an AC voltage source of angular frequency $\omega$ . The source has an internal resistance $R_0$ and an internal inductance $L_0$ . To ensure that the maximum possible average power is transferred from the source to the lossy capacitor, what must be the values of $R_0$ and $L_0$ ? (Assume edge effects are negligible).

Options

A.
$R_0 = \frac{\sigma_0 d}{2A(\sigma_0^2 + \omega^2 \epsilon_0^2)}$ , $L_0 = \frac{\epsilon_0 d}{2A(\sigma_0^2 + \omega^2 \epsilon_0^2)}$
B.
$R_0 = \frac{d}{2\sigma_0 A}$ , $L_0 = \frac{d}{2\omega^2 \epsilon_0 A}$
C.
$R_0 = \frac{3\sigma_0 d}{7A(\sigma_0^2 + \omega^2 \epsilon_0^2)}$ , $L_0 = \frac{3\epsilon_0 d}{7A(\sigma_0^2 + \omega^2 \epsilon_0^2)}$
D.
$R_0 = \frac{\sigma_0 d}{A(\sigma_0^2 + \omega^2 \epsilon_0^2)}$ , $L_0 = \frac{\epsilon_0 d}{A(\sigma_0^2 + \omega^2 \epsilon_0^2)}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Maximum Power Transfer|Non-Uniform Dielectric Filling|Capacitive Reactance Dependency

Why This Question Is Tricky

Common Mistakes

This question tests the application of the Maximum Power Transfer theorem to complex, distributed impedances.

The primary trap lies in incorrectly modeling the equivalent circuit of the non-uniform lossy dielectric.

Students often calculate the total equivalent series resistance $R_{eq}$ and total equivalent series capacitance $C_{eq}$ independently by integrating $dR$ and $dC^{-1}$ , and then erroneously combine them in series as $Z_L = R_{eq} - j/(\omega C_{eq})$ (Option B).

However, physically, the conduction and displacement currents flow in parallel through each infinitesimal slice $dx$ , meaning the local resistance and capacitance are in parallel, while the slices themselves are in series.

Furthermore, the problem requires applying the complex conjugate matching condition ( $Z_S = Z_L^*$ ) rather than merely equating magnitudes, testing a deeper understanding of AC power transfer.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Current Electricity - Class 12
Concept	Maximum Power Transfer
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free