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Question #43452

Difficulty: 8/10 | 🟢 Free

The Question

A full-wave rectifier is connected to an AC source of frequency $f = 50\text{ Hz}$ and peak voltage $V_m = 150\pi\text{ V}$ . The output of the rectifier is fed into an L-C filter consisting of a series inductor $L = \frac{1.25}{\pi}\text{ H}$ and a parallel capacitor $C = \frac{100}{\pi}\text{ }\mu\text{F}$ connected across a load resistor $R = 1000\text{ }\Omega$ . Assuming $R \gg X_C$ at the fundamental ripple frequency, calculate the amplitude of the fundamental wattless (reactive) current drawn from the rectifier by the filter circuit.

Options

A.
$1.00\text{ A}$
B.
$0.67\text{ A}$
C.
$1.50\text{ A}$
D.
$8.00\text{ A}$

Check Your Answer

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

Rectifier Ripple Factor|Wattless Current|Vector Cross-Products

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to apply Fourier series concepts to rectifier circuits and evaluate harmonic-dependent AC impedance.

A common trap is assuming the fundamental ripple frequency is $\omega$ instead of $2\omega$ , which drastically alters the reactances (leading to the $8.00\text{ A}$ trap).

Another trap is treating the L-C filter as a scalar sum of impedances ( $X_L + X_C$ ) rather than a vector sum ( $X_L - X_C$ ), ignoring the $180^\circ$ phase shift between the inductor and capacitor, which leads to the $0.67\text{ A}$ trap.

Finally, students must recognize that in the approximation $R \gg X_C$ , the ripple current is almost entirely wattless (reactive) and oscillates between the reactive components.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Semiconductor Electronics: Materials, Devices, and Simple Circuits - Class 12
Concept	Rectifier Ripple Factor
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free