Question

In an uncompensated Electrodynamometer wattmeter, when a current
$\text{i(t) = [8 sin(}\omega \text{t - 30) + 12 sin(2}\omega \text{t - 10)] A}$ is applied to its current coil and a voltage
$\text{v(t) = [(250sin(}\omega \text{t + 10) + 300 sin(3}\omega \text{t - 40)] V}$ is applied to pressure coil, then the meter will read ______watt.

• A. 766.04

Answer 1

The correct option is A 766.04
Voltage applied across pressure coil,

$v\left(t\right)=\left[250sin\right(\omega t+10)+300sin(3\omega t-40\left)\right]V$

Current flowing through current coil,

$i\left(t\right)=\left[8sin\right(\omega t-30)+12sin(2\omega t-10\left)\right]A$

Power will only be due to fundamental component of voltage and current,

$P=V_{rms}\times I_{rms}\times cos\varphi$

$=\frac{250}{\sqrt{2}}\times \frac{8}{\sqrt{2}}\times cos(10°–(-30°\left)\right)$

$=\frac{250\times 8}{2}\times cos{40}^o=766.04watt$