GATE EE · Electrical And Electronic Measurements

Measurement Of Energy And Power – GATE EE Electrical And Electronic MeasurementsPractice Questions & PYQs

Generate GATE-level questions on Power and Energy measurement. Focus on: 1. Electrodynamometer Wattmeters: LPf and HPF wattmeters, Errors in wattmeters. 2. Three-phase power measurement using Two-wattmeter method. 3. Induction type Energy meters: Construction, Principle, and Compensation for errors.

27 questions · 20 PYQs · 0 AI practice · GATE EE 2027

🎯 These are sample questions

Get the full Measurement Of Energy And Power experience — 100% free

✓ Attempt tracking & analytics
✓ Timer & exam-mode simulation
✓ Adaptive difficulty (Easy → Hard)
✓ Bookmarks, notes & mistake review

Just sign in to unlock everything. Free for all students.

🚀 Start Practice Mode Sign in free →

Q1GATE 2025NAT

In an experiment to measure the active power drawn by a single-phase RL Load connected to an AC source through a $2 \Omega$ resistor, three voltmeters are connected as shown in the figure below. The voltmeter readings are as follows: $\mathrm{V}_{\text {source }}=200 \mathrm{~V}, \mathrm{~V}_{\mathrm{R}}=9 \mathrm{~V}, \mathrm{~V}_{\text {Load }}=199 \mathrm{~V}$ . Assuming perfect resistors and ideal voltmeters, the Loadactive power measured in this experiment, in W, is ___________ (round off to one decimal place).

🚀 Solve in Practice Mode

📖 Explanation

Here $|\mathrm{I}|=\frac{\left|\mathrm{V}_{\mathrm{R}}\right|}{2}=\frac{9}{2}=4.5 \mathrm{~A}[\mathrm{RMS}]$ From resultant of vectors's formulae

\begin{aligned} \mathrm{V}_{\text {source }} & =\sqrt{\mathrm{V}_{\mathrm{R}}^{2}+\mathrm{V}_{\mathrm{Load}}^{2}+2 \mathrm{~V}_{\mathrm{R}} \cdot \mathrm{~V}_{\mathrm{Load}} \cos \phi_{\mathrm{L}}} \\ 200 & =\sqrt{9^{2}+199^{2}+2(9)(199) \cos \phi_{\mathrm{L}}} \\ \cos \phi_{\mathrm{L}} & =\frac{(200)^{2}-9^{2}-199^{2}}{2(9)(199)}=\frac{318}{3582}=0.08877(\mathrm{lag}) \\ \mathrm{P} & =\left|\mathrm{V}_{\mathrm{Load}}\right| \mathrm{I} \mid \cos \phi_{\mathrm{L}} \\ & =|199|\left|\frac{9}{2}\right|(0.08877) \\ \mathrm{P} & =79.5 \mathrm{~W} \end{aligned}

Explanation Figure 2

Q2GATE 2024NAT

In the circuit shown, $Z_{1}=50 \angle-90^{\circ}$ and $Z_{2}=200 \angle-30^{\circ} \Omega$ . It is supplied by a three phase $400 \mathrm{~V}$ source with the phase sequence being R-Y-B. Assume the watt meters $W_{1}$ and $W_{2}$ to be ideal. The magnitude of the difference between the readings of $W_{1}$ and $W_{2}$ in watts is ______ (rounded off to 2 decimal places).

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} V_{R Y} & =400 \angle 0^{\circ} \\ V_{Y B} & =400 \angle-120^{\circ} \\ V_{B R} & =400 \angle-240^{\circ} \\ V_{P C} & =V_{R B}=400 \angle-60^{\circ} \\ I_{1} & =\frac{V_{R B}}{Z_{1}}=\frac{400 \angle-60^{\circ}}{50 \angle-90^{\circ}}=8 \angle 30^{\circ} \\ I_{2} & =\frac{V_{R Y}}{Z_{2}}=\frac{400 \angle 0^{\circ}}{200 \angle-30^{\circ}}=2 \angle 30^{\circ} \Rightarrow I_{C C(2)} \\ I_{L 1} & =I_{C C(1)}=I_{1}+I_{2}=8 \angle 30^{\circ}+2 \angle 30^{\circ} \\ & =10 \angle 30^{\circ} \Rightarrow I_{C C(1)} \\ W & =V_{R B} \times I_{2 \theta} \cos \left(\angle V_{R B} \text { and } I_{L O}\right) \\ W_{1} & =400 \times 10 \times \cos \left(-60^{\circ}-30^{\circ}\right)=0 \text { Watt } \\ W_{2} & =V_{Y B} \times I_{2} \cos \left(\angle V_{Y B} \text { and } I_{2}\right) \\ & =400 \times 2 \times \cos \left(-120^{\circ}-30^{\circ}\right)=-692.8 \text { Watt } \\ W_{2} & =0-(-692.8)=692.8 \text { Watt } \end{aligned}

Explanation Figure 2

Q3GATE 2023MCQ

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

🚀 Solve in Practice Mode

📖 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

Explanation Figure 2

Q4GATE 2019NAT

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 _______.

🚀 Solve in Practice Mode

📖 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}

Q5GATE 2018MCQ

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

🚀 Solve in Practice Mode

📖 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}

Q6GATE 2017MCQ

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:

🚀 Solve in Practice Mode

📖 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}

Q7GATE 2016NAT

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 _____________.

🚀 Solve in Practice Mode

📖 Explanation

$P_{loss}=\frac{20 \times 3600}{1200 \times 30}=2kW$

Q8GATE 2015MCQ

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

🚀 Solve in Practice Mode

📖 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}

Q9GATE 2015MCQ

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 _________.

🚀 Solve in Practice Mode

📖 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] $=\frac{\text{Measured value}-\text{True value}}{\text{True value}}$ $=\frac{600.9-600}{600}\times 100=0.15\%$

Q10GATE 2014NAT

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 ______.

🚀 Solve in Practice Mode

📖 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 $= \cos \phi =\cos 36.58^{\circ} =0.802=0.80$

Q11GATE 2014MCQ

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

🚀 Solve in Practice Mode

📖 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}

Q12GATE 2014NAT

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 _______.

🚀 Solve in Practice Mode

📖 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, wattmeter reading, $W=VI \cos \phi =150 \times 10 \times 0.2=300$ Therefore, multiplying factor of wattmeter, $m=\frac{W}{W_m}=\frac{300}{150}=2$

Q13GATE 2012MCQ

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

🚀 Solve in Practice Mode

📖 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. $\therefore \;\; P_{avg}=\frac{1}{2}E_1I_1 \cos \phi _1 +\frac{1}{2}E_3I_3 \cos \phi _3$

Q14GATE 2011MCQ

Consider the following statement (1) The compensating coil of a low power factor wattmeter compensates the effect of the impedance of the current coil. (2) The compensating coil of a low power factor wattmeter compensates the effect of the impedance of the voltage coil circuit.

🚀 Solve in Practice Mode

📖 Explanation

The current coil carries a current of $I +I_P$ and produces a filed corresponding to this current. The compensating coil is connected in series with the pressure coil circuit and is made as nearly as possible identical and coincident with the current coil. It is so connected that it opposes the field of the current coil. The compensating coil carries a current $I_P$ and produces a field corresponding to this current. This field acts as in opposition to the current coil field. Thus the resultant field is due to current $I$ only. Hence the error caused by the pressure coil current flowing in the current coil is neutralized.

Explanation Figure 1

Q15GATE 2010MCQ

A wattmeter is connected as shown in figure. The wattmeter reads.

🚀 Solve in Practice Mode

📖 Explanation

Potential coil draws negligible current. So current through $Z_1$ and $Z_2$ is same. current through current coil $=I_{cc}=I$ Voltage across potential coil $=V_{pc}$ Voltage across $Z_1=V_{pc}=V$ Wattmeter reads power consumed by $Z_2$ , as voltage across potential coil = Voltage across $Z_2$ Current through current coil = Current through $Z_2$ .

Explanation Figure 2

Q16GATE 2009MCQ

The figure shows a three-phase delta connected load supplied from a 400V, 50 Hz, 3-phase balanced source. The pressure coil (PC) and current coil (CC) of a wattmeter are connected to the load as shown, with the coil polarities suitably selected to ensure a positive deflection. The wattmeter reading will be

🚀 Solve in Practice Mode

📖 Explanation

Assuming phase angle sequence abc Line to line voltage $V_{l-l}=400V$ Taking $V_{ab}$ as the reference

\begin{aligned} V_{ab} &=V_{l-l}\angle 0^{\circ}=400 \angle 0^{\circ}V\\ V_{bc}&=400\angle -120^{\circ}V \\ V_{ca} &= 400\angle -240^{\circ}V\\ \text{Current}&\text{ through current coil} \\ I{cc}&=\frac{V_{ca}}{Z_2}=\frac{400\angle -240^{\circ}}{100+j0}\\ &=4\angle -240^{\circ}A\\ \text{Voltage }&\text{ across pressure ciol}\\ V_{pc}&=V_{bc}=400\angle -120^{\circ}V \\ \phi &=\text{angle between} I_{cc} \text{and} V_{pc}\\ &=-120-(-240^{\circ})=120^{\circ}\\ \text{Wattmeter }&\text{reading,}\\ |V_{pc}||I_{cc}| \cos \phi &=400 \times 4 \times \cos (120^{\circ})\\ &=-800W \end{aligned}

Q17GATE 2009MCQ

The pressure coil of a dynamometer type wattmeter is

🚀 Solve in Practice Mode

Q18GATE 2006MCQ

A sampling wattmeter (that computes power from simultaneously sampled values of voltage and current) is used to measure the average power of a load. The peak to peak voltage of the square wave is 10 V and the current is a triangular wave of 5 A p-p as shown in the figure. The period is 20 ms. The reading in W will be

🚀 Solve in Practice Mode

Q19GATE 2005MCQ

Two wattmeters, which are connected to measure the total power on a three-phase system supplying a balanced load, read 10.5 kW and -2.5 kW, respectively. The total power and the power factor, respectively, are

🚀 Solve in Practice Mode

📖 Explanation

$P_1=10.5 kW$ and $P_2=-2.5kW$ Total power $=P_1+P_2$ $\;\;\;\;\; =10.5-2.5=8kW$ Power factor,

\begin{aligned} \cos \phi &= \cos \left ( \tan ^{-1}\sqrt{3}\frac{P_1-P_2}{P_1+P_2} \right )\\ &= \cos \left ( \tan ^{-1}\sqrt{3}\frac{10.5-(-2.5)}{10.5+(-2.5)} \right ) \\ &= 0.334 \end{aligned}

Q20GATE 2004MCQ

A single-phase load is connected between R and Y terminals of a 415 V, symmetrical, 3-phase, 4-wire system with phase sequence RYB. A wattmeter is connected in the system as shown in figure. The power factor of the load is 0.8 lagging. The wattmeter will read

🚀 Solve in Practice Mode

📖 Explanation

Line of line voltage $=450V$ Phase to neutral voltage $=\frac{415}{\sqrt{3}}V$ Taking $V_R$ as the reference

\begin{aligned} V_{RN} &=\frac{415}{\sqrt{3}} \angle 0^{\circ}V\\ V_{yN} &=\frac{415}{\sqrt{3}} \angle -120^{\circ}V\\ \text{and } V_{BN} &=\frac{415}{\sqrt{3}} \angle -240^{\circ}\\ V_{Ry} &=V_{RN}-V_{yN} \\ &=\frac{415}{\sqrt{3}} \angle 0^{\circ}- \frac{415}{\sqrt{3}} \angle -120^{\circ} \\ &=415 \angle 30^{\circ}V \\ & \text{Load current,}\\ I_L&= \frac{V_{Ry}}{Z}=\frac{415 \angle 30^{\circ}}{100 \angle 36.87^{\circ}}\\ &= 4.15 \angle -6.87^{\circ} A \end{aligned}

Current through current coil, $I_{cc}=I_L=4.15\angle -6.87^{\circ}A$ Voltage across pressure coil $V_{BN}=\frac{415}{\sqrt{3}}\angle -240^{\circ}V$ Phase angle between $V_{BN} \text{and} I_{cc}$ $\phi =-240^{\circ}-(-6.87^{\circ})=233.13^{\circ}$ Wattmeter reading $=|V_{BN}||I_L| \cos \phi$ $\;\;\; =\frac{415}{\sqrt{3}} \times 4.15 \times \cos 233.13$ $\;\;\; \approx -597W$

Want unlimited AI-generated Measurement Of Energy And Power questions?

Sign up free and practice with adaptive difficulty — Easy, Medium, Hard. New questions every session.

Start practising for free →