# Measurement of Energy and Power

 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

 A Unity B 0.5 leading C 0.866 leading D 0.707 lagging
GATE EE 2023   Electrical and Electronic Measurements
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
 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 _______.
 A 340.5 B 170.6 C 588.9 D 377.8
GATE EE 2019   Electrical and Electronic Measurements
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
 A 0.532 B 0.632 C 0.707 D 0.866
GATE EE 2018   Electrical and Electronic Measurements
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}
 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:
 A $W_1=1732 \; and \; W_2 =W_3=0$ B $W_1=0 , W_2 =1732 \; and \; W_3=0$ C $W_1=866, W_2 =0 , W_3=866$ D $W_1=W_2 =0 \; and \; W_3=1732$
GATE EE 2017-SET-1   Electrical and Electronic Measurements
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}
 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 _____________.
 A 1 B 2 C 3 D 4
GATE EE 2016-SET-2   Electrical and Electronic Measurements
Question 5 Explanation:
$P_{loss}=\frac{20 \times 3600}{1200 \times 30}=2kW$

There are 5 questions to complete.