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Question #43064

Difficulty: 8/10 | 🟢 Free

The Question

An AC voltage source $v(t) = 200 \sin(\omega t)\text{ V}$ is connected in series with a resistor $R = 100\ \Omega$ and a symmetric Zener clipper. The clipper consists of two identical ideal Zener diodes connected back-to-back. Each Zener diode has a breakdown voltage of $100\text{ V}$ and a forward voltage drop of $0\text{ V}$ . What is the average power supplied by the AC source to the circuit?

Options

A.
$\left( \frac{400}{3} - \frac{100\sqrt{3}}{\pi} \right) \text{ W}$
B.
$\left( 200 - \frac{400}{\pi} \right) \text{ W}$
C.
$\left( \frac{400}{3} + \frac{100\sqrt{3}}{\pi} \right) \text{ W}$
D.
$\left( 300 - \frac{800}{\pi} \right) \text{ W}$

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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

Wattless Current|Ohm's Law Validity|Zener vs Avalanche Breakdown|Binomial Approximation

Why This Question Is Tricky

Common Mistakes

This question tests the fundamental definition of average AC power.

A common trap is blindly applying the formula $P = V_{rms} I_{rms} \cos\phi$ , which is only valid for purely sinusoidal voltages and currents of the same frequency.

When a non-linear element like a Zener diode is introduced, the current waveform becomes clipped and non-sinusoidal, rendering the standard power factor concept invalid.

To find the true average power, students must revert to first principles and integrate the instantaneous power $P = \frac{1}{T} \int v(t)i(t)dt$ , carefully accounting for the piecewise continuous nature of the current due to the Zener breakdown dead-zone.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Alternating Current - Class 12
Concept	Wattless Current
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free