Before You Begin

Question #42743

Difficulty: 8/10 | 🟢 Free

The Question

An ideal diode and a sinusoidal AC voltage source are connected in series to a junction $J$ . From junction $J$ , two parallel branches connect to the ground, containing ideal resistors $R_1 = 30 \ \Omega$ and $R_2 = 60 \ \Omega$ respectively. A moving-coil ammeter (which measures the average current) is connected in the main branch just before junction $J$ and reads a steady value of $2.0 \text{ A}$ . Assuming the diode operates as a perfect half-wave rectifier, what is the total average electrical power dissipated in the two resistors?

Options

A.
$20\pi^2 \text{ W}$
B.
$40\pi^2 \text{ W}$
C.
$80 \text{ W}$
D.
$10\pi^2 \text{ W}$

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

Kirchhoff’s Junction Rule|Rectifier Ripple Factor|Fourier Analysis (Basic)

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to distinguish between the DC component (average current) and the heating component (RMS current) in a pulsating, non-steady circuit.

Students often misapply Kirchhoff's Laws by treating the time-averaged current as a steady DC current, erroneously calculating power dissipation as $P = I_{avg}^2 R_{eq}$ .

By introducing a half-wave rectifier, the problem forces the realization that power depends on the waveform's form factor ( $I_{rms} = I_0/2$ vs $I_{avg} = I_0/\pi$ ), making the direct use of average current for Joule heating a lethal trap.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Current Electricity - Class 12
Concept	Kirchhoff’s Junction Rule
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free