Problems

1. The input power to a 3-phase a.c. motor is measured as 5kW. If the

voltage and current to the motor are 400V and 8.6A respectively,
determine the power factor of the system?

Power P=5000W,

line voltage VL = 400 V,

line current, IL = 8.6A and

power, P =√3 VLIL cos φ

Hence

power factor = cos φ = P √3 VLIL

= 5000 √3 (400) (8.6)

= 0.839

2. Two wattmeters are connected to measure the input power to a

balanced 3-phase load by the two-wattmeter method. If the instrument
readings are 8kW and 4kW, determine (a) the total power input and (b)
the load power factor.
(a)Total input power,

P=P1 +P2 =8+4=12kW

(b) tan φ =√3(P1 − P2)/(P1 + P2)

=√3 (8 – 4) / (8 + 4)

=√3 (4/12)

=√3(1/3)

= 1/ √3

Hence φ= tan−1 1 √3 =30◦

Power factor= cos φ= cos 30◦ =0.866

3. Two wattmeters connected to a 3-phase motor indicate the total power

input to be 12kW. The power factor is 0.6. Determine the readings of
each wattmeter.

If the two wattmeters indicate P1 and P2 respectively

Then P1 + P2 = 12kW ---(1)

tan φ =√3(P1 − P2)/(P1 + P2)

And power factor=0.6= cos φ.

 Angle φ= cos−10.6=53.13◦ and

tan 53.13◦ =1.3333.

Hence

1.3333 =√3(P1 − P2)/12

From which,

P1 − P2 = 12(1.3333) /√3

i.e. P1 −P2 =9.237kW ----(2)

Adding Equations (1) and (2) gives:

2P1 = 21.237

i.e P1 = 21.237/2

= 10.62kW Hence wattmeter 1 reads 10.62kW From

Equation (1), wattmeter 2 reads

(12−10.62)=1.38kW

4. Three loads, each of resistance 30, are connected in star to a 415 V, 3-

phase supply. Determine

(a) the system phase voltage, (b) the phase current and (c) the line
current.
A ‘415 V, 3-phase supply’ means that 415 V is the line voltage, VL

(a) For a star connection, VL =√3Vp Hence phase voltage, Vp = VL/√3

= 415 /√3

= 239.6 V or 240 V

correct to 3 significant figures

(b) Phase current, Ip = Vp/Rp

= 240/30

=8 A

(c) For a star connection, Ip = IL Hence the line current, IL = 8 A

5. Three identical coils, each of resistance 10ohm and inductance 42mH

are connected (a) in star and (b) in delta to a 415V, 50 Hz, 3-phase
supply. Determine the total power dissipated in each case.

(a) Star connection

Inductive reactance,

XL =2πf L =2π (50) (42×10−3) =13.19

Phase impedance,

Zp =√(R2 +XL2)

=√(102 +13.192) =16.55

Line voltage, VL =415 V

And phase voltage,

VP =VL/√3=415/√3=240 V.

Phase current,

Ip =Vp/Zp =240/16.55=14.50 A. Line current,

IL =Ip =14.50 A.

Power factor= cos φ=Rp/Zp =10/16.55 =0.6042 lagging.

Power dissipated,

P =√3 VLIL cos φ =√3 (415) (14.50)(0.6042) = 6.3kW (Alternatively,

P =3I2R =3(14.50)2(10)=6.3kW)

(b) Delta connection

VL = Vp = 415 V,

Zp = 16.55_, cos φ = 0.6042 lagging (from above). Phase current,

Ip =Vp/Zp =415/16.55=25.08A. Line current,

IL =√3Ip =√3(25.08)=43.44A.

Power dissipated,

P =√3 VLIL cos φ

=√3 (415)(43.44)(0.6042) = 18.87kW

(Alternatively,

P =3I2R

=3(25.08)2(10) =18.87 kW)

6. A 415V, 3-phase a.c. motor has a power output of 12.75kW and

operates at a power factor of 0.77 lagging and with an efficiency of 85
per cent. If the motor is delta-connected, determine (a) the power input,
(b) the line current and (c) the phase current.

(a) Efficiency=power output/power input.

Hence
 (85/100)=12.750 power input from which, Power input = 12.
750 × 10085

= 15 000W or 15Kw

(b) Power, P=√3 VLIL cos φ, hence

(c) line current,

IL = P/ √3 (415) (0.77)

= 15 000/ √3 (415) (0.77)

= 27.10A

(d) For a delta connection, IL =√3 Ip,

Hence

Phase current, Ip = IL/√3

= 27.10 /√3

= 15.65A
 7. A 400V, 3-phase star connected alternator supplies a delta-connected
load, each phase of which has a resistance of 30_ and inductive
reactance 40_. Calculate (a) the current supplied by the alternator and
(b) the output power and the kVA of the alternator, neglecting losses in
the line between the alternator and load.

A circuit diagram of the alternator and load is shown in Fig.

(a) Considering the load:

Phase current, Ip =Vp/Zp

Vp =VL for a delta connection,

Hence Vp =400V.

Phase impedance,

Zp =√ (R2+XL2)

=√ (302 +402) =50

Figure

Hence Ip =Vp/Zp =400/50=8A.

For a delta-connection,

Line current, IL =√3 Ip =√3 (8) =13.86 A.

Hence 13.86A is the current supplied by the alternator.

(b) Alternator output power is equal to the power Dissipated by the load

I.e. P =√3 VLIL cos φ, Where cos φ = Rp/Zp = 30/50 = 0.6.

Hence P =√3 (400) (13.86) (0.6) = 5.76kW.

Alternator output kVA,

S =√3 VLIL =√3 (400) (13.86)

9.60 kVA.

Two Wattmeter Method of Power

Measurement
Two Wattmeter Method can be employed to measure the power in a 3 phase, three-wire star or
delta connected the balanced or unbalanced load.

In two wattmeter method, the current coils of the wattmeter are connected with any two lines,
say R and Y and the potential coil of each wattmeter is joined on the same line, the third line i.e.
B as shown below in figure (A):

The total instantaneous power absorbed by the three loads Z1, Z2 and Z3, is equal to the sum of
the powers measured by the two wattmeters, W 1 and W2.

Contents:


o Measurement of Power by Two Wattmeter Method in Star Connection
 o Measurement of Power by Two Wattmeter Method in Delta Connection

Measurement of Power by Two Wattmeter Method in Star

Connection
Considering the above figure (A) in which Two Wattmeter W 1 and W2 are connected, the
instantaneous current through the current coil of Wattmeter, W 1 is given by the equation shown
below:

The instantaneous potential difference across the potential coil of Wattmeter, W 1 is given as:

Instantaneous power measured by the Wattmeter, W1 is

The instantaneous current through the current coil of Wattmeter, W 2 is given by the equation:

The instantaneous potential difference across the potential coil of Wattmeter, W 2 is given as:

Instantaneous power measured by the Wattmeter, W2 is:

Therefore, the total power measured by the two wattmeters W 1 and W2 will be obtained by
adding the equation (1) and (2).
Where, P – the total power absorbed in the three loads at any instant.

Measurement of Power by Two Wattmeter Method in Delta

Connection
Considering the delta connected circuit shown in the figure below:

The instantaneous
current through the coil of the wattmeter, W1 is given by the equation:

Instantaneous power measured by the Wattmeter, W1 will be:

Therefore, the instantaneous power measured by the wattmeter, W 1 will be given as:

The instantaneous current through the current coil of the Wattmeter, W 2 is given as:

The instantaneous potential difference across the potential coil of wattmeter, W 2 :

Therefore, the instantaneous power measured by Wattmeter, W 2 will be:

Hence, to obtain the total power measured by the two wattmeter the two equations, i.e. equation
(3) and (4) has to be added.

Where P is the total power absorbed in the three loads at any instant.

The power measured by the Two Wattmeter at any instant is the instantaneous power absorbed
by the three loads connected in three phases. In fact, this power is the average power drawn by
the load since the Wattmeter reads the average power because of the inertia of their moving
system.
Two Wattmeter Method:
In two wattmeter method, a three phase balanced voltage is to a balanced three phase
load where the current in each phase is assumed lagging by an angle of Ø behind the
corresponding phase voltage.

The schematic diagram for the measurement of three phase power using two wattmeter
method is shown below.

From the figure, it is obvious that current through the Current Coil (CC) of Wattmeter
W = I , current though Current Coil of wattmeter W = I whereas the potential difference
1 R 2 B

seen by the Pressure Coil (PC) of wattmeter W = V (Line Voltage) and potential
1 RB

difference seen by Pressure Coil of wattmeter W = V . The phasor diagram of the

2 BY

above circuit is drawn by taking VR as reference phasor as shown below.

From the above phasor diagram,

Angle between the current I and voltage V = (30° – Ø)
R RB
Two Wattmeter Method:
In two wattmeter method, a three phase balanced voltage is to a balanced three phase load where the
current in each phase is assumed lagging by an angle of Ø behind the corresponding phase voltage.

The schematic diagram for the measurement of three phase power using two wattmeter method is
shown below.

Two Wattmeter Method- three-phase-power-measurement

From the figure, it is obvious that current through the Current Coil (CC) of Wattmeter W1 = IR , current
though Current Coil of wattmeter W2 = IB whereas the potential difference seen by the Pressure Coil
(PC) of wattmeter W1= VRB (Line Voltage) and potential difference seen by Pressure Coil of wattmeter
W2 = VBY. The phasor diagram of the above circuit is drawn by taking VR as reference phasor as shown
below.

Two Wattmeter Method- power-measurement

From the above phasor diagram,

Angle between the current IRand voltage VRB = (30° – Ø)

Angle between current IYand voltage VYB = (30° + Ø)

Therefore, Active power measured by wattmeter W1 = VRBIR Cos (30° – Ø)

Similarly, Active power measured by wattmeter W2 = VYBIYCos(30° + Ø)

As the load is balanced, therefore magnitude of line voltage will be same irrespective of phase taken i.e.
VRY, VYB and VRB all will have same magnitude. Also for Star / Y connection line current and phase
current are equal, say IR = IY = IB = I

Let VRY = VYB = VRB = VL

Therefore,

W1 = VRBIRCos (30° – Ø)

= VLICos(30° – Ø)

In the same manner,

W2 = VLICos(30° + Ø)

Hence, total power measured by wattmeters for the balanced three phase load is given as,

W = W1 + W2

= VLI×Cos(30° – Ø) + VLI×Cos(30° + Ø)

= VLI [Cos(30° – Ø) + Cos(30° + Ø)]

= 2VLI×Cos30°CosØ ……………….[ CosC + CosD = 2Cos(C+D)/2×Cos(C-D)/2 ]

 =√3VLICosØ

Therefore, total power measured by wattmeters W = √3VLICosØ

Now, suppose you are asked to find the power factor of the load when individual power measured by
the wattmeters are given, then we should proceed as

W1 + W2 = √3VLICosØ ……………………………..(1)

Similarly,

W1 – W2= VLI×Cos(30° – Ø) + VLI×Cos(30° + Ø)

= VLI [Cos(30° – Ø) + Cos(30° + Ø)]

= 2VLI×Sin30°SinØ ………[ CosC – CosD = 2Sin(C+D)/2×Sin(D-C)/2 ]

= VLISinØ

Hence,

W1 – W2 = VLISinØ ………………………………(2)

Dividing equation (2) by equation (1),

(W1 – W2) / (W1 + W2) = VLISinØ / √3VLICosØ

(W1 – W2) / (W1 + W2) = (tanØ) /√3

Hence,

tanØ = √3[(W1 – W2) / (W1 + W2)]

From the above equation, we can find the value of Ø and hence the power factor Cos Ø of the load.

Hope you understand the method of measurement of three phase power using two wattmeter method.
Now we will consider three cases and will observe the how the individual wattmeter measures the
power in each case.

Case1: When the power factor of load is unity.

As the power factor of load is unity, hence Ø = 0

Therefore,

Power measured by first wattmeter W1= VLI Cos(30° – 0)

= VLI Cos30°

= 0.866 VLI

Power measured by second wattmeter W2= VLI Cos(30° + 0)

 = VLI Cos30°

= 0.866 VLI

Thus we see that, when the power factor of load is unity then both the wattmeter reads the same value.

Case2: When power factor of load is 0.5 lagging.

As power factor is 0.5 hence CosØ = 0.5 i.e. Ø = 60°

Therefore,

Power measured by first wattmeter W1= VLI Cos(30° – 60°)

= VLI Cos30° ……[Cos (-Ɵ) = CosƟ ]

= 0.866 VLI

Power measured by second wattmeter W2= VLI Cos(30° + 60°)

= VLI Cos90°

=0
Thus we see that, when power factor of load is 0.5 lagging then power is only measured by first
wattmeter and reading of second wattmeter is ZERO.

Case3: When power factor of load is zero.

As power factor of load is zero, hence CosØ = 0 i.e. Ø = 90°

Therefore,

Power measured by first wattmeter W1= VLI Cos(30° – 90°)

= VLI Cos60°

= 0.5 VLI

Power measured by second wattmeter W2= VLI Cos(30° + 90°)

= –VLI Cos60°

= -0.5 VLI

Thus we see that, when power factor of load is zero then one wattmeter reads +ive while second
wattmeter reads –ive. As second wattmeter reads –ive hence wattmeter won’t read anything practically,
therefore for second wattmeter we need to interchange the leads of either Pressure Coil or Current Coil
so that second wattmeter may read value. As we have interchanged the connection of leads of either PC
or CC, hence second wattmeter will read +ive but while calculating the total power measured we must
take the reading of second wattmeter as –ive.

It shall be noted that when 60° <Ø < 90°, reading of one wattmeter will be positive while the reading of
second wattmeter will be negative.