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Previous Year Questions- Measurement of Energy and Power - 1 | Electrical and Electronic Measurements - Electrical Engineering (EE) PDF Download

Q1: In the circuit shown,  Z1 = 50∠ −90° and Z2 = 200∠ −30°Ω. It is supplied by a three phase 400 V source with the phase sequence being R-Y-B. Assume the watt meters W1 and  W2 to be ideal. The magnitude of the difference between the readings of  W1 and W2 in watts is ______ (rounded off to 2 decimal places). (2024)
Previous Year Questions- Measurement of Energy and Power - 1 | Electrical and Electronic Measurements - Electrical Engineering (EE)(a) 253.65
(b) 742.25
(c) 692.8
(d) 562.25
Ans:
(c)
Sol: Previous Year Questions- Measurement of Energy and Power - 1 | Electrical and Electronic Measurements - Electrical Engineering (EE)Previous Year Questions- Measurement of Energy and Power - 1 | Electrical and Electronic Measurements - Electrical Engineering (EE)VRY = 400∠0°
VYB = 400∠ − 120°
VBR = 400∠ − 240°
VPC = VRB = 400∠ − 60°
Previous Year Questions- Measurement of Energy and Power - 1 | Electrical and Electronic Measurements - Electrical Engineering (EE)ILI = ICC(1) = I1 + I2 = 8∠30° + 2∠30°
= 10∠30° ⇒ ICC(1)
W = VRB x Icos (∠VRB and ILO)
W= 400 x 10 x cos(-60° - 30°) = 0 Watt
W= VYB x I2cos(∠VYB and I2)
= 400 x 2 x cos(-120° - 30°) = -692.8 Watt
W2 = 0 - (-692.8) = 692.8 Watt

Q2: 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 W and the line current is IR = 2 A(rms), then the power factor of the load per phase is (2023)
Previous Year Questions- Measurement of Energy and Power - 1 | Electrical and Electronic Measurements - Electrical Engineering (EE)(a) Unity
(b) 0.5 leading
(c) 0.866 leading
(d) 0.707 lagging
Ans:
(c)
Sol: By observation of wattmeter connection, the wattmeter read reactive power
∵ Wattmeter reading,
W = YYBIR (Angle between VYB and IR.
Phasor diagram :
Previous Year Questions- Measurement of Energy and Power - 1 | Electrical and Electronic Measurements - Electrical Engineering (EE)∴ W = VLILcos(90° − β)
= VLILsinϕ
Given: W = −400 W
∴ − 400 = 400 × 2 × sinϕ
⇒ ϕ = −30°
Thus, P.F. = cosϕ = 0.866 leading

Q3: The voltage across and the current through a load are expressed as follows
v(t) = −170sin(377t − (π/6))V
i(t) = 8cos(377t + (π/6))A
The average power in watts (round off to one decimal place) consumed by the load is _______.   (2019)
(a) 340.5
(b) 170.6
(c) 588.9
(d) 377.8
Ans:
(c)
Sol: v(t)= -170sin (377t - (π/6))V = Vpc
i(t) = 8cos (377t + (π/6))A = ICC
i(t) = 8sin (377t + (2π/3))A = ICC
Pavg = (1/2) x (-170)(8) x cos(- (π/6) - (2π/3))
= (1/2) x (-170)(8) x cos (150°)
= 588.9 W

Q4: 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  (2018)
(a) 0.532
(b) 0.632
(c) 0.707
(d) 0.866
Ans: 
(d)
Sol: In two wattmeter method,
Previous Year Questions- Measurement of Energy and Power - 1 | Electrical and Electronic Measurements - Electrical Engineering (EE)Given, W2 = W1/2
Previous Year Questions- Measurement of Energy and Power - 1 | Electrical and Electronic Measurements - Electrical Engineering (EE)ϕ = 30°
cosϕ = cos30° = 0.866

Q5: 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 W1, W2 and W3 read 577.35W each. If the switch is moved to position Y, the readings of the watt-meters in watts will be:  (SET-1(2017))
Previous Year Questions- Measurement of Energy and Power - 1 | Electrical and Electronic Measurements - Electrical Engineering (EE)(a) W= 1732 and W= W= 0
(b) W= 0, W= 1732 and W= 0
(c) W= 866, W= 0, W= 866
(d) W= W= 0 and W= 1732
Ans:
(d)
Sol: When switch is on Neutral side
W = W1 + W2 + W3
= 577.35 + 577.35 + 577.35
= 1732 Watts
VA load = 3464 = 3VphIph
⇒ VphIph = 1154.66
Each Ω meter reading
= VRNIRcos(∠VRN and IR)
577.35  = VphIph cos(ϕ)
Previous Year Questions- Measurement of Energy and Power - 1 | Electrical and Electronic Measurements - Electrical Engineering (EE)ϕ = cos-1(0.5) = 60°
When switch is on Y-phase side,
W1 = VRYIRcos(∠VRY and IR)
Previous Year Questions- Measurement of Energy and Power - 1 | Electrical and Electronic Measurements - Electrical Engineering (EE)= 0 Watt
W3 = VBYIRcos(∠VBY and IR)
Previous Year Questions- Measurement of Energy and Power - 1 | Electrical and Electronic Measurements - Electrical Engineering (EE)= 1732 Watt
W1 = 0
W2 = 0
W3 = 1732 Watt

Q6: 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 _____________.  (SET-2 (2016))
(a) 1
(b) 2
(c) 3
(d) 4
Ans:
(b)
Sol: Ploss = (20 x 3600)/(1200 x 30)  = 2 kW

Q7: The coils of a wattmeter have resistances 0.01Ω and 1000Ω; 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 _________.  (SET-2(2015))
Previous Year Questions- Measurement of Energy and Power - 1 | Electrical and Electronic Measurements - Electrical Engineering (EE)(a) 0.85
(b) 0.15
(c) 0.25
(d) 0.55
Ans:
(b)
Sol: True value of power
= VI cos ϕ
= 30 × 25 × 0.8 = 600W
In the give diagram,
Measured power = Power loss in pressure coil + Power in loead
= V2/RP + VI Cos ϕ
= (302/1000) + 600 W = 600.9 W
Therefore, percentage error in wattmeter reading [/latex]
Previous Year Questions- Measurement of Energy and Power - 1 | Electrical and Electronic Measurements - Electrical Engineering (EE)
Q8: 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 (SET-2 (2015))
(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
Ans:
(a)
Sol: p.f. = 0.707
ϕ = cos−1(0.707) = 45°
P = 5kW
W+ W2 = 5 ...(i)
Previous Year Questions- Measurement of Energy and Power - 1 | Electrical and Electronic Measurements - Electrical Engineering (EE)Previous Year Questions- Measurement of Energy and Power - 1 | Electrical and Electronic Measurements - Electrical Engineering (EE)From equation (i) and (ii),
W - 1 = 3.94 kW
W2 = 1.06 kW

Q9: 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 _______. (SET-3 (2014))
(a) 1
(b) 2
(c) 3
(d) 2.5
Ans:
(b)
Sol: Given, cosϕ = 0.2 (LPF wattmeter)
Given ,, full scale reading of wattmeter,
Wm = 150 Watt
Also for V = 150 volt, I = 10A settings,
wattmeter reading,
 W = V I cosϕ = 150 × 10 × 0.2 = 300
Therefore, multiplying factor of wattmeter,
Previous Year Questions- Measurement of Energy and Power - 1 | Electrical and Electronic Measurements - Electrical Engineering (EE)
Q10: 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 ______.  (SET-2(2014))
(a) 0.20
(b) 0.48
(c) 0.80
(d) 0.96
Ans:
(c)
Sol: Given, W1 = 250W, W2 = 100W
Power factor angle is given by
Previous Year Questions- Measurement of Energy and Power - 1 | Electrical and Electronic Measurements - Electrical Engineering (EE)Therefore, power factor of load = cosϕ = cos36.58° = 0.802 = 0.80

Q11: 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 (SET-1 (2014))
(a) π/12
(b) π/8
(c) π/6
(d) π/3
Ans:
(c)
Sol: For two wattmeter method, wattmeter readings are
W1 = VLILcos(30 + ϕ)
W2 = VLILcos(30 + ϕ)
Forϕ = 30°, (or π/6)
W1 = VLILcos60° = Previous Year Questions- Measurement of Energy and Power - 1 | Electrical and Electronic Measurements - Electrical Engineering (EE)
W2 = VLILcos0° = VLIL
Therefore W2 = 2W1

Q12: For the circuit shown in the figure, the voltage and current expressions are
v(t) = E1sin(ωt) + E3sin(3ωt) and i(t) = I1sin(ωt − ϕ1) + I3sin(3ωt − ϕ3) + I5sin(5ωt)
The average power measured by the wattmeter is (2012)
Previous Year Questions- Measurement of Energy and Power - 1 | Electrical and Electronic Measurements - Electrical Engineering (EE)(a) (1/2) E1I1cosϕ1
(b) (1/2)
[E1I1cosϕ1 + E1I3cosϕ3 + E1I5]
(c) (1/2)[E1I1cosϕ+ E3I3cosϕ3]
(d) (1/2)[E1I1cosϕ+ E3I1cosϕ3]  
Ans:
(c)
Sol: V(t) = E1sinωt + E3sin3ωt
i(t) = I1sin(ωt − ϕ1) + I3sin(3ϕt − ϕ3) + Isin 5ωt
Average power,
Pavg = 1/2π Previous Year Questions- Measurement of Energy and Power - 1 | Electrical and Electronic Measurements - Electrical Engineering (EE)
The product of different frequency terms have zero average value.
∴ Pavg = (1/2) E1Icos ϕ1 + (1/2)E3Icos ϕ3

Q13: 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. (2011)
(a) (1) is true but (2) is false
(b) (1) is false but (2) is true
(c) both (1) and (2) are true
(d) both (1) and (2) are false
Ans:
(b)
Sol: The current coil carries a current of I + IP 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 IP 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.
Previous Year Questions- Measurement of Energy and Power - 1 | Electrical and Electronic Measurements - Electrical Engineering (EE)
Q14: A wattmeter is connected as shown in figure. The wattmeter reads. (2010)
Previous Year Questions- Measurement of Energy and Power - 1 | Electrical and Electronic Measurements - Electrical Engineering (EE)(a) Zero always
(b) Total power consumed by Zand Z2
(c) Power consumed by Z1
(d) Power consumed by Z2
Ans:
(d)
Sol: Previous Year Questions- Measurement of Energy and Power - 1 | Electrical and Electronic Measurements - Electrical Engineering (EE)Potential coil draws negligible current. So current through Z1 and  Z2 is same.
current through current coil = Icc = I
Voltage across potential coil = Vpc
Voltage across Z1 = Vpc = V
Wattmeter reads power consumed by Z2, as voltage across potential coil = Voltage across Z2
Current through current coil = Current through Z2.

Q15: 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 (2009)
Previous Year Questions- Measurement of Energy and Power - 1 | Electrical and Electronic Measurements - Electrical Engineering (EE)(a)  0
(b) 1600 Watt
(c) 800 Watt
(d) 400 Watt
Ans:
(c)
Sol: Assuming phase angle sequence abc
Line to line voltage Vl−l = 400V
Taking Vab as the reference
Vab = Vl−l∠0° = 400∠0°V
Vbc  = 400∠ − 120°V
Vca = 400∠ − 240°V
Current through current coil
Previous Year Questions- Measurement of Energy and Power - 1 | Electrical and Electronic Measurements - Electrical Engineering (EE)Voltage across pressure ciol
Bpc = Vbc = 400∠ - 120°V
ϕ = angle between Icc and Vpc
= -120 - (-240°) = 120°
Wattmeter reading,
|Vpc||Icc| cosϕ = 400 x 4 x cos(120°)
= -800 W

The document Previous Year Questions- Measurement of Energy and Power - 1 | Electrical and Electronic Measurements - Electrical Engineering (EE) is a part of the Electrical Engineering (EE) Course Electrical and Electronic Measurements.
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FAQs on Previous Year Questions- Measurement of Energy and Power - 1 - Electrical and Electronic Measurements - Electrical Engineering (EE)

1. What are the different units used for measuring energy in electrical engineering?
Ans. In electrical engineering, energy is commonly measured in joules (J), kilowatt-hours (kWh), and watt-hours (Wh). The joule is the SI unit of energy, while kilowatt-hours and watt-hours are often used in practical applications to measure energy consumption over time.
2. How is power calculated in electrical circuits?
Ans. Power in electrical circuits can be calculated using the formula P = VI, where P represents power in watts (W), V is voltage in volts (V), and I is current in amperes (A). For alternating current (AC) circuits, power can also be calculated using P = VI cos(φ), where φ is the phase angle between the current and voltage.
3. What is the difference between active power and reactive power?
Ans. Active power, measured in watts (W), is the actual power consumed by electrical devices to perform useful work. Reactive power, measured in reactive volt-amperes (VAR), is the power that oscillates between the source and the load, which does not perform any useful work but is necessary for maintaining voltage levels in the system.
4. How do you convert between different units of energy, such as joules and kilowatt-hours?
Ans. To convert joules to kilowatt-hours, you can use the conversion factor: 1 kWh = 3.6 million joules (3.6 x 10^6 J). Therefore, to convert joules to kilowatt-hours, divide the energy in joules by 3.6 million. For example, 7.2 million joules is equivalent to 2 kWh (7.2 x 10^6 J ÷ 3.6 x 10^6 J/kWh).
5. What role does the power factor play in electrical systems?
Ans. The power factor is a measure of how effectively electrical power is being converted into useful work output. It is defined as the ratio of active power to apparent power (P/S) and ranges from 0 to 1. A power factor closer to 1 indicates efficient utilization of electrical power, while a lower power factor indicates that more reactive power is present, which can lead to higher energy costs and reduced system efficiency.
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