In 'two-wattmeter method' of power calculation of a 3-phase balanced star connected system, what is the power factor of the system if both the wattmeter’s show a positive reading?

Greater than 0 but less than 0.5

Greater than 0.5 but less than 1

In two wattmeter method of power calculation of a 3-phase balanced star connected system, what is the power factor of the system, if both the wattmeters show the same positive reading?

Greater than 0 but less than 0.5

Greater than 0.5 but less than 1

In a crossed coil power factor meter the moving coils should be mounted on which type of shaft and at what angles to each other?

Different shafts and 45° angles

Different shafts and right angles

Different shafts and 45° angles

In case of the two-wattmeter method for measuring the power of a balanced 3ϕ load, the whole power is measured by one of the two wattmeters when:

The power factor is less than 0.5 but more than 0

The power factor is more than 0.5 but less than 1

The power factor is less than 0.5 but more than 0

In the two-wattmeter method of 3ϕ power measurement, if the phase sequence of the supply is reversed:

one of the meters will show a negative reading

there won't be a change in meter readings

the reading of wattmeters will be interchanged

In ‘Two-wattmeter method’ of power calculation of a 3-phase balanced star connected system, what is the power factor of the system, if one of the wattmeter’s shows negative reading and the other shows a positive reading?

Greater than or equal 0 but less than 0.5

Greater than 0.5 but less than equal to 1

In a low power factor wattmeter, sometimes compensating coil is connected in order to

reduce the error caused by eddy current

neutralize the capacitive effect of pressure coil

compensate for inductance of pressure coil

compensate for power loss in the pressure coil

reduce the error caused by eddy current

For balanced lagging power factor, according to two wattmeter method,W1 = VL IL cos (30 - ϕ)W2 = VL IL cos (30 + ϕ)The total reactive volt-amperes Q is given by

Q = \(\sqrt 3 {V_L}{I_L}\sin \phi\)

For a delta-connected load being measured for power by the two-wattmeter method, if W1 = VLIL cos (30 - ϕ) and W2 = VLIL cos (30 + ϕ), Then total 3-phase power is:

\(\sqrt 3 {V_{ph}}{I_{ph}}\cos \left( {30 + \phi } \right)\)

\(\sqrt 3 {V_L}{I_L}\cos \phi \)

\(\sqrt 3 {V_{ph}}{I_{ph}}\cos \left( {30 + \phi } \right)\)

\(3{V_{ph}}{I_{ph}}\sin \left( {30 + \phi } \right)\)

\(\sqrt 3 {V_L}{I_L}\sin \left( {30 + \phi } \right)\)

According to Blondel’s theorem, if a network is supplied using ‘n’ conductor’s the total power is ________ of the reading of n wattmeter’s so arranged that a current coil of each wattmeter is in each line and the corresponding pressure coil is connected between that line and a common point

Sum of products of two wattmeter’s reading taken together

In 3ϕ power measurement for a balanced load using the two-wattmeter method, the reactive power is given by:

the sum of both the wattmeter readings

√2 times the difference of the readings of the two wattmeters

√3 times the sum of the readings of the two wattmeters

√3 times the difference of the readings of the two wattmeters

For a star-connected load being measured for power by the two-wattmeter method, if W1 = VLIL cos (30 - ϕ) and W2 = VLIL cos (30 + ϕ), then the total 3-phase power will be:

\(\sqrt 3 {V_{ph}}{I_{ph}}\sin \left( {30 + \phi } \right)\)

\(3{V_{ph}}{I_{ph}}\sin \left( \phi \right)\)

\(\sqrt 3 {V_L}{I_L}\cos \left( \phi \right)\)

\(\sqrt 3 {V_{ph}}{I_{ph}}\sin \left( {30 + \phi } \right)\)

\(\sqrt 3 {V_L}{I_L}\cos \left( {30 + \phi } \right)\)

37 + Mcqs in Measurement Of Power in Electrical Engineering Page 1 McqOptions

1.	In 'two-wattmeter method' of power calculation of a 3-phase balanced star connected system, what is the power factor of the system if both the wattmeter’s show a positive reading?
A.	0
B.	Greater than 0 but less than 0.5
C.	3.5
D.	Greater than 0.5 but less than 1
Answer» E.

Discussion

2.	A 3-phase 10 kVA load has a power factor of 0.342. The power is measured by the two-wattmeter method.Find the reading of each wattmeter when the power factor is leading.
A.	W1 = -1 kW and W2 = 4.4 kW
B.	W1 = -4.4 kW and W2 = 1 kW
C.	W1 = 1 kW and W2 = 4.4 kW
D.	W1 = 4.4 kW and W2 = 1 kW
Answer» B. W1 = -4.4 kW and W2 = 1 kW

Discussion

3.	In two wattmeter method of power calculation of a 3-phase balanced star connected system, what is the power factor of the system, if both the wattmeters show the same positive reading?
A.	0
B.	Greater than 0 but less than 0.5
C.	Greater than 0.5 but less than 1
D.	1
Answer» E.

Discussion

4.	A device capable of detecting voltage, current and the angle between the voltage and the current to provide power readings directly in watts is known as a/an:
A.	ammeter
B.	wattmeter
C.	power factor meter
D.	voltmeter
Answer» C. power factor meter

Discussion

5.	A balanced 3-phase star-connected load draws power from a 440 V supply. The two watt-meters connected indicate W1 = 4.2 kW and W2 = 0.8 kW. Calculate the current in the circuit in Amperes
A.	10.731
B.	10.137
C.	10.271
D.	10.371
Answer» C. 10.271

Discussion

6.	For a balanced load, the reactive power is measured by two wattmeter method with readings W1 and W2, the total reactive power is given by:
A.	(W1 – W2)
B.	(W1 + W2)
C.	√3 (W1 + W2)
D.	√3 (W1 – W2)
Answer» E.

Discussion

7.	In a crossed coil power factor meter the moving coils should be mounted on which type of shaft and at what angles to each other?
A.	Different shafts and right angles
B.	Same shaft and right angles
C.	Different shafts and 45° angles
D.	same shaft and 45° angels
Answer» C. Different shafts and 45° angles

Discussion

8.	A three-phase 10 kVA load has a power factor of 0.342. The power is measured by the two-wattmeter method.Find the reading of each wattmeter when the power factor is lagging.
A.	W1 = 1 kW and W2 = 4.4 kW
B.	W1 = -1 kW and W2 = -4.4 kW
C.	W1 = 4.4 kW and W2 = -1 kW
D.	W1 = -4.4 kW and W2 = 1 kW
Answer» D. W1 = -4.4 kW and W2 = 1 kW

Discussion

9.	For a delta-connected load being measured for power by the two-wattmeter method, if Iph will lead Vph by angle ϕ then it is the case of __________.
A.	Short circuit
B.	Leading power factor
C.	Lagging power factor
D.	Open circuit
Answer» C. Lagging power factor

Discussion

10.	A dynamometer instrument is mainly used as
A.	dc ammeter
B.	wattmeter
C.	dc voltmeter
D.	ohmmeter
Answer» C. dc voltmeter

Discussion

11.	In the two-wattmeter method, the readings of the two wattmeters are 500 W, 500 W respectively. The load power factor in a balanced 3-phase 3-wire circuit is:
A.	0.8
B.	0
C.	1
D.	0.3
Answer» D. 0.3

Discussion

12.	In case of the two-wattmeter method for measuring the power of a balanced 3ϕ load, the whole power is measured by one of the two wattmeters when:
A.	The power factor is unity
B.	The power factor is more than 0.5 but less than 1
C.	The power factor is 0.5
D.	The power factor is less than 0.5 but more than 0
Answer» D. The power factor is less than 0.5 but more than 0

Discussion

13.	A delta-connected balanced load is supplied from a three-phase balanced 400 V, 50 Hz AC supply. The reading on the two wattmeters is 970 W and 480 W respectively. Each phase of the load consists of resistance and inductance connected in series. Calculate the total active power consumed.
A.	14.5 kW
B.	1,450 kW
C.	1.45 kW
D.	145 kW
Answer» D. 145 kW

Discussion

14.	In a two-wattmeter measurement of power in a three-phase circuit, both the wattmeters show equal readings. In this case, the load power factor is:
A.	Zero
B.	Unity
C.	0.8 Lagging
D.	0.8 Leadings
Answer» C. 0.8 Lagging

Discussion

15.	In the two-wattmeter method of 3ϕ power measurement, if the phase sequence of the supply is reversed:
A.	one of the meters will show a negative reading
B.	the meters will not read
C.	there won't be a change in meter readings
D.	the reading of wattmeters will be interchanged
Answer» E.

Discussion

16.	If two watt-meters used to measure three phase power read equal reading, then the power factor is:
A.	0.866
B.	0.5
C.	0
D.	1
Answer» E.

Discussion

17.	________ is a measuring device which can evaluate and record the electrical power passing through a circuit in a certain time.By implementing it, we can know how much amount of electrical energy is used by a consumer or a residence or an electrically powered device or a business.
A.	DC meter
B.	Air flow meter
C.	Watt hour meter
D.	AC meter
Answer» D. AC meter

Discussion

18.	In ‘Two-wattmeter method’ of power calculation of a 3-phase balanced star connected system, what is the power factor of the system, if one of the wattmeter’s shows negative reading and the other shows a positive reading?
A.	Greater than or equal 0 but less than 0.5
B.	0.5
C.	Greater than 0.5 but less than equal to 1
D.	Greater than 1
Answer» B. 0.5

Discussion

19.	In a low power factor wattmeter, sometimes compensating coil is connected in order to
A.	neutralize the capacitive effect of pressure coil
B.	compensate for inductance of pressure coil
C.	compensate for power loss in the pressure coil
D.	reduce the error caused by eddy current
Answer» D. reduce the error caused by eddy current

Discussion

20.	For balanced lagging power factor, according to two wattmeter method,W1 = VL IL cos (30 - ϕ)W2 = VL IL cos (30 + ϕ)The total reactive volt-amperes Q is given by
A.	Q = 0
B.	Q = \(\sqrt 3 {V_L}{I_L}\sin \phi\)
C.	Q = 1
D.	Q = VLILsin ϕ
Answer» C. Q = 1

Discussion

21.	A 3 phase – 110 V motor has a power factor of 0.5. The two watt-meters connected measure the total input of 50 kW. Calculate the reading (in kW) of each wattmeter.
A.	0, 150
B.	100, 50
C.	0, 50
D.	50, 50
Answer» D. 50, 50

Discussion

22.	For a delta-connected load being measured for power by the two-wattmeter method, if W1 = VLIL cos (30 - ϕ) and W2 = VLIL cos (30 + ϕ), Then total 3-phase power is:
A.	\(\sqrt 3 {V_L}{I_L}\cos \phi \)
B.	\(\sqrt 3 {V_{ph}}{I_{ph}}\cos \left( {30 + \phi } \right)\)
C.	\(3{V_{ph}}{I_{ph}}\sin \left( {30 + \phi } \right)\)
D.	\(\sqrt 3 {V_L}{I_L}\sin \left( {30 + \phi } \right)\)
Answer» B. \(\sqrt 3 {V_{ph}}{I_{ph}}\cos \left( {30 + \phi } \right)\)

Discussion

23.	Electric heater utilizes __________ effect of current.
A.	Chemical
B.	Heating
C.	Mechanical
D.	Filed
Answer» C. Mechanical

Discussion

24.	According to Blondel’s theorem, if a network is supplied using ‘n’ conductor’s the total power is ________ of the reading of n wattmeter’s so arranged that a current coil of each wattmeter is in each line and the corresponding pressure coil is connected between that line and a common point
A.	Sum
B.	Product
C.	Sum of products of two wattmeter’s reading taken together
D.	Difference
Answer» B. Product

Discussion

25.	In 3ϕ power measurement for a balanced load using the two-wattmeter method, the reactive power is given by:
A.	the sum of both the wattmeter readings
B.	√2 times the difference of the readings of the two wattmeters
C.	√3 times the sum of the readings of the two wattmeters
D.	√3 times the difference of the readings of the two wattmeters
Answer» E.

Discussion

26.	A balanced three-phase star-connected load draws power from a 440V supply. The two connected wattmeters’, W1 and W2, indicate 5 kW and 1,200 W respectively. Calculate the current in the circuit.
A.	11.86 A
B.	15.56 A
C.	1.11 A
D.	10.11 A
Answer» B. 15.56 A

Discussion

27.	Phase voltage and current of a 3ϕ, three-wire star-connected system, with an inductive load of power factor 0.707 (lag), is 150 V and \(30\sqrt 3 {\rm{A}}\). If the power in the system is being measured using two wattmeter’s, the difference in meter readings is:
A.	10.46 KW
B.	2.26 KW
C.	3.95 KW
D.	9.54 KW
Answer» E.

Discussion

28.	In the two-wattmeter method of three phase power measurement of a balanced load, if the reading of one metre is – 200 W, then then power factor of the load is ______.
A.	greater than 0.5
B.	less than 0.5
C.	equal to 0.5
D.	equal to zero
Answer» C. equal to 0.5

Discussion

29.	A 3-phase balanced supply system is connected to a 3-phase unbalanced load. Power supplied to this load can be measured using:1. two wattmeters2. one wattmeter3. three wattmetersWhich of these statement/s is/are correct?
A.	1 and 2
B.	1 and 3
C.	2 and 3
D.	3 alone
Answer» C. 2 and 3

Discussion

30.	Baretters and Bolometers are used for measurement of
A.	VSWR
B.	Transmission losses
C.	Microwave power
D.	None of the above
Answer» D. None of the above

Discussion

31.	A balanced three-phase star-connected load draws power from a 440 V supply. The two connected wattmeters W1 and W2, indicate 5 kW and 1200 W. Calculate the total power.
A.	5 kW
B.	6,200 kW
C.	62 kW
D.	6,200 W
Answer» E.

Discussion

32.	For a star-connected load being measured for power by the two-wattmeter method, if W1 = VLIL cos (30 - ϕ) and W2 = VLIL cos (30 + ϕ), then the total 3-phase power will be:
A.	\(3{V_{ph}}{I_{ph}}\sin \left( \phi \right)\)
B.	\(\sqrt 3 {V_L}{I_L}\cos \left( \phi \right)\)
C.	\(\sqrt 3 {V_{ph}}{I_{ph}}\sin \left( {30 + \phi } \right)\)
D.	\(\sqrt 3 {V_L}{I_L}\cos \left( {30 + \phi } \right)\)
Answer» C. \(\sqrt 3 {V_{ph}}{I_{ph}}\sin \left( {30 + \phi } \right)\)

Discussion

33.	In the circuit shown, the pressure coils of two wattmeters are connected to:
A.	neutral wire
B.	B
C.	Y
D.	R
Answer» C. Y

Discussion

34.	A three-phase star-connected balanced load of (4 + j3) Ω per phase is connected across three-phase, 50 Hz, 400 V AC supply. If the two-wattmeter method is used to determine input power, find each wattmeter reading.
A.	18.34 kW and 7.26 kW
B.	18.34 kW and 46 kW
C.	23 kW and 23 kW
D.	25.60 kW and 23.23 kW
Answer» B. 18.34 kW and 46 kW

Discussion

35.	One B.O.T. is:
A.	1 kWh
B.	0.1 kWh
C.	0.01 kWh
D.	0.001 kWh
Answer» B. 0.1 kWh

Discussion

36.	A wattmeter having a range of 4000 W has an error of ±2% of full-scale deflection. If the true power is 2000 W, what would be the range of the reading?
A.	1920-2000 W
B.	2000-2080 W
C.	1920-2040 W
D.	1920-2080 W
Answer» E.

Discussion

37.	A balanced three-phase star-connected load draws power from a 440 V supply. The two connected wattmeters, W1 and W2, Indicate 5 kW and 1,200 W respectively. Calculate the power factor of the system.
A.	0.54
B.	0.98
C.	0.69
D.	0.75
Answer» D. 0.75

Discussion

Explore topic-wise MCQs in Electrical Engineering.

In 'two-wattmeter method' of power calculation of a 3-phase balanced star connected system, what is the power factor of the system if both the wattmeter’s show a positive reading?

A 3-phase 10 kVA load has a power factor of 0.342. The power is measured by the two-wattmeter method.Find the reading of each wattmeter when the power factor is leading.

In two wattmeter method of power calculation of a 3-phase balanced star connected system, what is the power factor of the system, if both the wattmeters show the same positive reading?

A device capable of detecting voltage, current and the angle between the voltage and the current to provide power readings directly in watts is known as a/an:

A balanced 3-phase star-connected load draws power from a 440 V supply. The two watt-meters connected indicate W1 = 4.2 kW and W2 = 0.8 kW. Calculate the current in the circuit in Amperes

For a balanced load, the reactive power is measured by two wattmeter method with readings W1 and W2, the total reactive power is given by:

In a crossed coil power factor meter the moving coils should be mounted on which type of shaft and at what angles to each other?

A three-phase 10 kVA load has a power factor of 0.342. The power is measured by the two-wattmeter method.Find the reading of each wattmeter when the power factor is lagging.

For a delta-connected load being measured for power by the two-wattmeter method, if Iph will lead Vph by angle ϕ then it is the case of __________.

A dynamometer instrument is mainly used as

In the two-wattmeter method, the readings of the two wattmeters are 500 W, 500 W respectively. The load power factor in a balanced 3-phase 3-wire circuit is:

In case of the two-wattmeter method for measuring the power of a balanced 3ϕ load, the whole power is measured by one of the two wattmeters when:

A delta-connected balanced load is supplied from a three-phase balanced 400 V, 50 Hz AC supply. The reading on the two wattmeters is 970 W and 480 W respectively. Each phase of the load consists of resistance and inductance connected in series. Calculate the total active power consumed.

In a two-wattmeter measurement of power in a three-phase circuit, both the wattmeters show equal readings. In this case, the load power factor is:

In the two-wattmeter method of 3ϕ power measurement, if the phase sequence of the supply is reversed:

If two watt-meters used to measure three phase power read equal reading, then the power factor is:

________ is a measuring device which can evaluate and record the electrical power passing through a circuit in a certain time.By implementing it, we can know how much amount of electrical energy is used by a consumer or a residence or an electrically powered device or a business.

In ‘Two-wattmeter method’ of power calculation of a 3-phase balanced star connected system, what is the power factor of the system, if one of the wattmeter’s shows negative reading and the other shows a positive reading?

In a low power factor wattmeter, sometimes compensating coil is connected in order to

For balanced lagging power factor, according to two wattmeter method,W1 = VL IL cos (30 - ϕ)W2 = VL IL cos (30 + ϕ)The total reactive volt-amperes Q is given by

A 3 phase – 110 V motor has a power factor of 0.5. The two watt-meters connected measure the total input of 50 kW. Calculate the reading (in kW) of each wattmeter.

For a delta-connected load being measured for power by the two-wattmeter method, if W1 = VLIL cos (30 - ϕ) and W2 = VLIL cos (30 + ϕ), Then total 3-phase power is:

Electric heater utilizes __________ effect of current.

In 3ϕ power measurement for a balanced load using the two-wattmeter method, the reactive power is given by:

A balanced three-phase star-connected load draws power from a 440V supply. The two connected wattmeters’, W1 and W2, indicate 5 kW and 1,200 W respectively. Calculate the current in the circuit.

Phase voltage and current of a 3ϕ, three-wire star-connected system, with an inductive load of power factor 0.707 (lag), is 150 V and \(30\sqrt 3 {\rm{A}}\). If the power in the system is being measured using two wattmeter’s, the difference in meter readings is:

In the two-wattmeter method of three phase power measurement of a balanced load, if the reading of one metre is – 200 W, then then power factor of the load is ______.

A 3-phase balanced supply system is connected to a 3-phase unbalanced load. Power supplied to this load can be measured using:1. two wattmeters2. one wattmeter3. three wattmetersWhich of these statement/s is/are correct?

Baretters and Bolometers are used for measurement of

A balanced three-phase star-connected load draws power from a 440 V supply. The two connected wattmeters W1 and W2, indicate 5 kW and 1200 W. Calculate the total power.

For a star-connected load being measured for power by the two-wattmeter method, if W1 = VLIL cos (30 - ϕ) and W2 = VLIL cos (30 + ϕ), then the total 3-phase power will be:

In the circuit shown, the pressure coils of two wattmeters are connected to:

A three-phase star-connected balanced load of (4 + j3) Ω per phase is connected across three-phase, 50 Hz, 400 V AC supply. If the two-wattmeter method is used to determine input power, find each wattmeter reading.

One B.O.T. is:

A wattmeter having a range of 4000 W has an error of ±2% of full-scale deflection. If the true power is 2000 W, what would be the range of the reading?

A balanced three-phase star-connected load draws power from a 440 V supply. The two connected wattmeters, W1 and W2, Indicate 5 kW and 1,200 W respectively. Calculate the power factor of the system.