Question 1 |
A 3-phase, star-connected, balanced load is supplied from a 3-phase, 400V (rms), balanced voltage source with phase sequence R-Y-B, as shown in the figure. If the wattmeter reading is -400 \mathrm{~W} and the line current is \mathrm{I}_{\mathrm{R}}=2 \mathrm{~A}(\mathrm{rms}), then the power factor of the load per phase is
Unity | |
0.5 leading | |
0.866 leading | |
0.707 lagging |
Question 1 Explanation:
By observation of wattmeter connection, the wattmeter read reactive power
\because Wattmeter reading,
W=Y_{Y B} I_{R} (Angle between V_{Y B} and I_{R}.
Phasor diagram :
\therefore \quad W=V_{\mathrm{L}} \mathrm{I}_{\mathrm{L}} \cos \left(90^{\circ}-\beta\right)
=V_{L} I_{L} \sin \phi
Given: \quad W=-400 \mathrm{~W}
\therefore \quad-400=400 \times 2 \times \sin \phi
\Rightarrow \quad \phi=-30^{\circ}
Thus, P.F. =\cos \phi=0.866 leading
\because Wattmeter reading,
W=Y_{Y B} I_{R} (Angle between V_{Y B} and I_{R}.
Phasor diagram :
\therefore \quad W=V_{\mathrm{L}} \mathrm{I}_{\mathrm{L}} \cos \left(90^{\circ}-\beta\right)
=V_{L} I_{L} \sin \phi
Given: \quad W=-400 \mathrm{~W}
\therefore \quad-400=400 \times 2 \times \sin \phi
\Rightarrow \quad \phi=-30^{\circ}
Thus, P.F. =\cos \phi=0.866 leading
Question 2 |
The voltage across and the current through a load are expressed as follows
v(t)=-170sin(377t-\frac{\pi}{6})V
i(t)=8 cos(377t+\frac{\pi}{6})A
The average power in watts (round off to one decimal place) consumed by the load is _______.
v(t)=-170sin(377t-\frac{\pi}{6})V
i(t)=8 cos(377t+\frac{\pi}{6})A
The average power in watts (round off to one decimal place) consumed by the load is _______.
340.5 | |
170.6 | |
588.9 | |
377.8 |
Question 2 Explanation:
\begin{aligned} v(t) &=-170 \sin \left ( 377t-\frac{\pi}{6} \right )V=V_{pc}\\ i(t) &= 8 \cos \left ( 377t+\frac{\pi}{6} \right )A=I_{cc}\\ i(t) &= 8 \sin \left ( 377t+\frac{2\pi}{3} \right )A=I_{cc} \\ P_{avg}&=\frac{1}{2} \times (-170)(8) \times \cos\left ( -\frac{\pi}{6}-\frac{2 \pi}{3} \right ) \\ &= \frac{1}{2} \times (-170)(8) \times \cos(150^{\circ}) \\ &=588.9W \end{aligned}
Question 3 |
Two wattmeter method is used for measurement of power in a balanced three-phase load
supplied from a balanced three-phase system. If one of the wattmeters reads half of the
other (both positive), then the power factor of the load is
0.532 | |
0.632 | |
0.707 | |
0.866 |
Question 3 Explanation:
In two wattmeter method,
\tan \phi =\frac{\sqrt{3}(W_1-W_2)}{W_1+W_2}
\begin{aligned} \text{Given}, \; W_2 &=\frac{W_1}{2} \\ \tan \phi &=\frac{\sqrt{3}\left (W_1-\frac{W_1}{2} \right ) }{\left (W_1+\frac{W_1}{2} \right )} \\ \phi &= 30^{\circ}\\ \cos \phi &=\cos 30^{\circ}=0.866 \end{aligned}
\tan \phi =\frac{\sqrt{3}(W_1-W_2)}{W_1+W_2}
\begin{aligned} \text{Given}, \; W_2 &=\frac{W_1}{2} \\ \tan \phi &=\frac{\sqrt{3}\left (W_1-\frac{W_1}{2} \right ) }{\left (W_1+\frac{W_1}{2} \right )} \\ \phi &= 30^{\circ}\\ \cos \phi &=\cos 30^{\circ}=0.866 \end{aligned}
Question 4 |
The load shown in the figure is supplied by a 400 V (line to line) 3-phase source (RYB
sequence). The load is balanced and inductive, drawing 3464 VA. When the switch S is in
position N, the three watt-meters W_1,W_2 \; and \; W_3 read 577.35W each. If the switch is moved to
position Y, the readings of the watt-meters in watts will be:
W_1=1732 \; and \; W_2 =W_3=0 | |
W_1=0 , W_2 =1732 \; and \; W_3=0 | |
W_1=866, W_2 =0 , W_3=866 | |
W_1=W_2 =0 \; and \; W_3=1732 |
Question 4 Explanation:
When switch is on Neutral side
\begin{aligned} W &= W_1+W_2+W_3\\ &=577.35+577.35+577.35 \\ &=1732 \; Watts \\ \text{VA load} &= 3464 =3 V_{ph}I_{ph}\\ \Rightarrow \; V_{ph}I_{ph} &=1154.66 \end{aligned}
Each \Omega meter reading
\begin{aligned} &= V_{RN}I_{R} \cos (\angle V_{RN} \; \text{and}\; I_R)\\ 577.35&= V_{ph}I_{ph} \cos (\phi )\\ \cos (\phi ) &=\frac{577.35}{1154.66}=05 \\ \phi &= \cos ^{-1} (0.5)=60^{\circ} \end{aligned}
When switch is on Y-phase side,
\begin{aligned} W_1&= V_{RY}I_R \cos (\angle V_{RY} \; \text{and}\; I_R)\\ &= \sqrt{3} V_{ph}I_{ph} \cos (30^{\circ}+\phi ) \\ &= \sqrt{3} \times 1154.66 \times \cos (30^{\circ}+ 60^{\circ})\\ &= 0 \text{Watt}\\ W_3 &= V_{BY}I_R \cos (\angle V_{BY} \; \text{and}\; I_R)\\ &=\sqrt{3} V_{ph}I_{ph} \cos (30^{\circ}-\phi )\\ &=\sqrt{3} \times 1154.66 \times \cos (30^{\circ}- 60^{\circ}) \\ &= 1732 \text{Watt}\\ W_1&=0\\ W_2&=0\\ W_3&=1732 \text{Watt} \end{aligned}
\begin{aligned} W &= W_1+W_2+W_3\\ &=577.35+577.35+577.35 \\ &=1732 \; Watts \\ \text{VA load} &= 3464 =3 V_{ph}I_{ph}\\ \Rightarrow \; V_{ph}I_{ph} &=1154.66 \end{aligned}
Each \Omega meter reading
\begin{aligned} &= V_{RN}I_{R} \cos (\angle V_{RN} \; \text{and}\; I_R)\\ 577.35&= V_{ph}I_{ph} \cos (\phi )\\ \cos (\phi ) &=\frac{577.35}{1154.66}=05 \\ \phi &= \cos ^{-1} (0.5)=60^{\circ} \end{aligned}
When switch is on Y-phase side,
\begin{aligned} W_1&= V_{RY}I_R \cos (\angle V_{RY} \; \text{and}\; I_R)\\ &= \sqrt{3} V_{ph}I_{ph} \cos (30^{\circ}+\phi ) \\ &= \sqrt{3} \times 1154.66 \times \cos (30^{\circ}+ 60^{\circ})\\ &= 0 \text{Watt}\\ W_3 &= V_{BY}I_R \cos (\angle V_{BY} \; \text{and}\; I_R)\\ &=\sqrt{3} V_{ph}I_{ph} \cos (30^{\circ}-\phi )\\ &=\sqrt{3} \times 1154.66 \times \cos (30^{\circ}- 60^{\circ}) \\ &= 1732 \text{Watt}\\ W_1&=0\\ W_2&=0\\ W_3&=1732 \text{Watt} \end{aligned}
Question 5 |
An energy meter, having meter constant of 1200 revolutions/kWh, makes 20 revolutions in 30 seconds for a constant load. The load, in kW, is _____________.
1 | |
2 | |
3 | |
4 |
Question 5 Explanation:
P_{loss}=\frac{20 \times 3600}{1200 \times 30}=2kW
There are 5 questions to complete.