Question

The current coil of a dynamometer type wattmeter is connected to a 24 V dc source in series with a 6 $\Omega$ resistor. The potential circuit is connected through an ideal half wave rectifier in series with a 50 Hz source of 100 V. The inductance of pressure circuit and current coil resistance are negligible. The reading of wattmeter is

• A. 80.5 W
• B. 180 W
• C. 362 W
• D. 148.7 W

Answer 1

The correct option is B 180 W

Current thorugh the current coil $=\frac{24}{6}=4A$
The pressure coil is energized by an ideal rectifier.

$\therefore$ The pressure coil carries during one half cycle and in other half cycle there is no current in it. This means that there is a deflecting torque on the meter during one half-cycle.

$\therefore$ Reading of wattmeter = average power over a cycle

$=\frac{1}{2\pi }\left[\underset{0}{\overset{2\pi }{\int }}vid\theta \right]$

$=\frac{1}{2\pi }\underset{0}{\overset{\pi }{\int }}\sqrt{2}\times 100\sin \theta \times 4d\theta =180.06W$