# Measurement of Energy and Power

 Question 1
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 1 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 2
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 2 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 3
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 3 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 4
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 4 Explanation:
$P_{loss}=\frac{20 \times 3600}{1200 \times 30}=2kW$
 Question 5
The coils of a wattmeter have resistances 0.01$\Omega$ and 1000$\Omega$; their inductances may be neglected. The wattmeter is connected as shown in the figure, to measure the power consumed by a load, which draws 25A at power factor 0.8. The voltage across the load terminals is 30V. The percentage error on the wattmeter reading is _________.
 A 0.85 B 0.15 C 0.25 D 0.55
GATE EE 2015-SET-2   Electrical and Electronic Measurements
Question 5 Explanation:
True value of power
$=VI \cos \phi$
$=30 \times 25 \times 0.8 =600W$
In the give diagram,
Measured power=Power loss in pressure coil +Power in loead
$=\frac{V^2}{R_P} +VI \cos \phi$
$=\frac{30^2}{1000}+600W=600.9W$
Therefore, percentage error in wattmeter reading [/latex]
$=\frac{\text{Measured value}-\text{True value}}{\text{True value}}$
$=\frac{600.9-600}{600}\times 100=0.15\%$
 Question 6
A 3-phase balanced load which has a power factor of 0.707 is connected to a balanced supply. The power consumed by the load is 5 kW. The power is measured by the two-wattmeter method. The readings of the two wattmeters are
 A 3.94 kW and 1.06 kW B 2.50 kW and 2.50 kW C 5.00 kW and 0.00 kW D 2.96 kW and 2.04 kW
GATE EE 2015-SET-2   Electrical and Electronic Measurements
Question 6 Explanation:
\begin{aligned} p.f. &= 0.707\\ \phi &= cos^{-1}(0.707)=45^{\circ}\\ P&= 5kW\\ W_1 +W_2&= 5\;\;...(i)\\ \tan \phi &=\frac{\sqrt{3(W_1-W_2)}}{(W-1+W_2)} \\ (W-1+W_2) \times 1 &= \sqrt{3}(W_1-W_2) \\ W_1-W_2 &= \frac{5}{\sqrt{3}} \;\;...(ii)\\ \text{From equation} & \text{ (i) and (ii),} \\ W-1&= 3.94kW\\ W_2&= 1.06kW \end{aligned}
 Question 7
An LPF wattmeter of power factor 0.2 is having three voltage settings 300 V, 150 V and 75 V, and two current settings 5 A and 10 A. The full scale reading is 150. If the wattmeter is used with 150 V voltage setting and 10 A current setting, the multiplying factor of the wattmeter is _______.
 A 1 B 2 C 3 D 2.5
GATE EE 2014-SET-3   Electrical and Electronic Measurements
Question 7 Explanation:
Given, $\cos \phi =0.2$ (LPF wattmeter)
Given ,, full scale reading of wattmeter,
$W_m=150$ Watt
Also for $V=150 \; \text{volt}$, $I=10 A$ settings,
$W=VI \cos \phi =150 \times 10 \times 0.2=300$
Therefore, multiplying factor of wattmeter,
$m=\frac{W}{W_m}=\frac{300}{150}=2$
 Question 8
While measuring power of a three-phase balanced load by the two-wattmeter method, the readings are 100 W and 250 W. The power factor of the load is ______.
 A 0.2 B 0.48 C 0.8 D 0.96
GATE EE 2014-SET-2   Electrical and Electronic Measurements
Question 8 Explanation:
Given, $W_1=250W$, $W_2=100W$
Power factor angle is given by
\begin{aligned} \phi &= \tan^{-1}\left [ \frac{\sqrt{3}(W_1-W_2)}{W_1+W_2} \right ]\\ &=\tan^{-1}\left [ \frac{\sqrt{3}(250-100)}{250+100} \right ] \\ &= \tan^{-1}\left [ \frac{\sqrt{3}\times 150}{350} \right ]\\ &= \tan^{-1}\left [ \frac{3\sqrt{3}}{7} \right ]=36.58^{\circ} \end{aligned}
Therefore, power factor of load $= \cos \phi =\cos 36.58^{\circ} =0.802=0.80$
 Question 9
Power consumed by a balanced 3-phase, 3-wire load is measured by the two wattmeter method. The first wattmeter reads twice that of the second. Then, the load impedance angle in radians is
 A $\pi /12$ B $\pi /8$ C $\pi /6$ D $\pi /3$
GATE EE 2014-SET-1   Electrical and Electronic Measurements
Question 9 Explanation:
For two wattmeter method, wattmeter readings are
\begin{aligned} W_1 &=V_LI_L \cos (30+\phi ) \\ W_2 &=V_LI_L \cos (30-\phi ) \\ For \; \phi =30^{\circ}, (or \;\pi/6) \\ W_1 &=V_LI_L \cos 60^{\circ} =\frac{V_LI_L}{2} \\ W_2 &=V_LI_L \cos 0^{\circ} =V_LI_L \\ \text {Therefore}\;\;W_2&=2W_1 \end{aligned}
 Question 10
For the circuit shown in the figure, the voltage and current expressions are
$v(t)=E_{1}sin(\omega t)+E_{3}sin(3\omega t)$ and $i(t)=I_{1}sin(\omega t- \phi _{1})+I_{3}sin(3\omega t-\phi _{3})+I_{5} sin(5\omega t)$
The average power measured by the wattmeter is
 A $\frac{1}{2}E_{1}I_{1}cos \phi _{1}$ B $\frac{1}{2}[E_{1}I_{1}cos \phi _{1}+E_{1}I_{3}cos \phi _{3}+E_{1}I_{5}]$ C $\frac{1}{2}[E_{1}I_{1}cos \phi _{1}+E_{3}I_{3}cos \phi _{3}]$ D $\frac{1}{2}[E_{1}I_{1}cos \phi _{1}+E_{3}I_{1}cos \phi _{3}]$
GATE EE 2012   Electrical and Electronic Measurements
Question 10 Explanation:
\begin{aligned} V(t) &= E_1 \sin \omega t +E_3 \sin 3\omega t\\ i(t)&= I_1 \sin (\omega t-\phi _1)+I_3 \sin (3\phi t-\phi _3)\\ & +I_5 \sin 5\omega t \end{aligned}
Average power,
$P_{avg}=\frac{1}{2 \pi}\int_{0}^{2 \pi}vi \; d(\omega t)$
The product of different frequency terms have zero average value.
$\therefore \;\; P_{avg}=\frac{1}{2}E_1I_1 \cos \phi _1 +\frac{1}{2}E_3I_3 \cos \phi _3$
There are 10 questions to complete.