Basic Electrical Engineering (3110005)

Question-3c-OR

BE | Semester-1 Winter-2019 | 11-01-2020

Q3) (c)

Reading of watt meter,

W_{1} = V_{RY} I_{R} \cos \emptyset_{1} = V_{L} I_{L} \cos (30 + \emptyset)

Reading of watt meter,

W_{2} = V_{B Y} I_{B} \cos \emptyset_{2} = V_{L} I_{L} \cos (30 - \emptyset)

Total Power,

P = W_{1} + W_{2}

\Rightarrow P = V_{L} I_{L} \cos (30 + \emptyset) + V_{L} I_{L} \cos (30 - \emptyset)

\Rightarrow P = V_{L} I_{L} [\cos (30 + \emptyset) + \cos (30 - \emptyset)]

\Rightarrow P = V_{L} I_{L} [\cos 30 \cos \emptyset + \sin 30 \sin \emptyset + \cos 30 \cos \emptyset - \sin 30 \sin \emptyset]

\Rightarrow P = V_{L} I_{L} [2 \cos 30 \cos \emptyset]

\Rightarrow P = V_{L} I_{L} [2 (\frac{\sqrt{3}}{2}) \cos \emptyset]

\Rightarrow P = \sqrt{3} V_{L} I_{L} \cos \emptyset

Now,

P = W_{1} - W_{2}

\Rightarrow P = V_{L} I_{L} \cos (30 + \emptyset) - V_{L} I_{L} \cos (30 - \emptyset)

\Rightarrow P = V_{L} I_{L} [\cos (30 + \emptyset) - \cos (30 - \emptyset)]

\Rightarrow P = V_{L} I_{L} [\cos 30 \cos \emptyset + \sin 30 \sin \emptyset - \cos 30 \cos \emptyset + \sin 30 \sin \emptyset]

\Rightarrow P = V_{L} I_{L} [2 \sin 30 \sin \emptyset]

\Rightarrow P = V_{L} I_{L} [2 (\frac{1}{2}) \sin \emptyset]

\Rightarrow P = V_{L} I_{L} \sin \emptyset

Now,

\frac{\sqrt{3} (W_{1} - W_{2})}{(W_{1} + W_{2})} = \frac{\sqrt{3} V_{L} I_{L} \sin \emptyset}{\sqrt{3} V_{L} I_{L} \cos \emptyset}

\Rightarrow \frac{\sqrt{3} (W_{1} - W_{2})}{(W_{1} + W_{2})} = \tan \emptyset

Effect of power factor on watt meter reading:

W_{1} = V_{L} I_{L} \cos (30 + \emptyset)

W_{2} = V_{L} I_{L} \cos (30 - \emptyset)

Thus, readings

W_{1}

and

W_{2}

will vary depending upon the power factor angle

\emptyset

Hence, prove that two-watt meters are sufficient to measure total power in 3-phase system.