1

CHAPTER 2 (Part 2)

Single-Phase Circuits

BEKG 2433 1

BEKG 2433 2

7. Effect of frequency variations

8. Power in AC circuits P, Q & S

9. Power to resistive, inductive, capacitive loads

12. Apparent power (S)

13. Relation between P,Q & S

14. Power to resistance and reactance loads

15. Power Factor

16. Measurement of Power

This part will cover the subtopic as below:

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BEKG 2433 3

AC Circuit with Series Loads

VS

Z3 Z2 Zn Z1

Total Impedance : ZT = Z1 + Z2 + Z3 +…….+ Zn

Is

Source Current : IS = VS / ZT

Voltage Across each Impedance : Vn = IS Zn

1

321

1.....

111

n

TZZZZ

Z

BEKG 2433 4

AC Circuit with Parallel Loads

VS

Total Impedance :

Is

Source Current : IS = I1 + I2 + I3 +……+In

Current Across each Impedance : In = VS / Zn

Z1 Z3 Z2 Zn

I1 I3 I2 In

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BEKG 2433 5

7. Effect of frequency

Variation of Inductive Reactance with frequency

XL= ω L = 2 π f L

Inductive reactance is directly proportional to frequency & inductance

“If frequency is doubled, reactance doubles”

“If frequency is halved, reactance halves and so on”

“If inductance is doubled, reactance doubles”

“If inductance is halved, reactance halves and so on”

BEKG 2433 6

Example:

A circuit has 50 ohms inductive reactance. If

both the inductance and the frequency are

doubled, what is the new XL?

Answer: 200Ω

4

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Variation of Capacitive Reactance with frequency

Capacitive reactance is inversely proportional with frequency & Capacitance

• The higher the frequency, the lower the reactance, and vice versa

• At 0Hz, capacitive reactance is infinite

• If the capacitance is doubled, reactance will be halves

CfCXC

2

11

BEKG 2433 8

8. Power in AC Circuits

• What is power?

• Electrical appliances rated in Watt, HP (horsepower), VA (volt-

ampere) or VAR.

• In DC circuits, the equation for power, P = VI Watts.

• In AC circuits, we have real power or active power (P), reactive power in

VAR (Q) and apparent power in VA (S).

• Active power is the power that does useful work- light a lamp, heat, turning

of an electric motor, etc

• Reactive power is the power that does not do useful work or “phantom

power”. Normally exist when the load contains inductor or capacitor

• Power is defined as the rate of flow of energy past a

given point

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Complex Power

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Another way to easily calculate power

Real Power, P (a scalar with unit: Watt)

RMS based – thermally equivalent to DC power

Reactive Power, Q (a vector with unit: Var)

Oscillating power into & out of the load because of its reactive elements

(L &C)

Positive value for inductive (lagging) & negative fro capacitive(leading)

Complex Power, S (a vector with unit: VA) - overall power delivered

into/out of the system

Apparent power – is the magnitude of the complex power.

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Voltage phasor:

Current phasor:

Complex Power Formulas

vV rmsV

iI rmsI

S Q

P

sinIV

cosIV

sinIVcosIV

IVIV

rmsrms

rmsrms

rmsrmsrmsrms

rmsrmsrmsrms

**

Q

P

jQPS

jS

IVVIS iviv

iInote rmsI*:

. ivfplet

Thus, the complex power is the product of the voltage & the conjugate

of the current:

22powerapparent of magnitude QPSS

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Power Factor & Power Triangle

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• A ratio between real power & apparent power :

• Power factor can only have value from 0 to 1.

• Low power factor : value close to 0 (e.g 0.2, 0.3)

• indicates poor utilization of electric

• High power factor : value close to 1 (0.9, 0.8, 1, )

• electrical capacity is being utilized effectively

• Unity power factor : value = 1

• electrical capacity is being utilized at the MOST effective

•A measurement of how efficient a facility uses electrical energy.

)cos( ..factor power ivfp

)( anglefactor power ivpf

Sfp

P

powerapparent

power real.

The Power Factor & Power Triangle..cont’d

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VIcosP

tanP

sin VIQ

VIS

P

Qtan

P.

cospowerapparent

power real.

1

22

iv

QP

P

Sfp

fp

This triangle is to summarize & give relations of the real, reactive, apparent and power factor concept.

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The Power Factor & Power Triangle..cont’d

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( )=δ2

( )=δ1

Where δ = power factor phase Angle For δ1 : θv > θi For δ2 : θv < θi

Power Consumption in Devices

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2Resistor Resistor

2Inductor Inductor L

2

Capacitor Capacitor C

Capaci

Resistors only consume real power:

,

Inductors only "consume" reactive power:

,

Capacitors only "generate" reactive power:

1.C

P I R

Q I X

Q I X XC

Q

2

Capacitortor

C

(Note-some define negative.). CXV

X

8

Applying Ohms Law in Complex power formulas

*

2

*

2

*

**

22**

S

VZ

Z

V

Z

VVVIS

or

IjXIRZIIVIS

ZIV

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22 and

: note

XRZ

ZjXRZ

R

X

Z

Impedance triangle

Conservation of Power (balance power)

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At every node (bus) in the system:

Sum of real power into node must equal zero,

Sum of reactive power into node must equal zero.

This is a direct consequence of Kirchhoff’s current

law, which states that the total current into each node

must equal zero.

Conservation of real power and conservation of reactive

power also follows since S = VI*.

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The Complex Power Balance

The total complex/apparent power delivered to the loads in parallel

Is the sum of the complex powers delivered to each branch:

V Z3 Z2 Z1

I3 I2 I1 I

*

3

*

2

*

1

*

321

* ][

VIVIVI

IIIVVIS

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Example 1

V Z3 Z2 Z1

I3 I2 I1 I V = 120000 V

Z1 = 60 + j0

Z2 = 6 + j12

Z3 = 30 – j30

Find the power absorbed by each load and the total

complex power

Answers:

varW

varW

varW

varW

000,72000,96

000,24000,24

000,96000,48

0000,24

3

2

1

jS

jS

jS

jS

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BEKG 2433 19

9. Power to Pure Resistive Loads

RIR

VIV

VVIVttIV

dtt

IVdttIVP

tItVtitvtp

rmsrms

rmsrmsmmmm

mm

mmmm

mm

22

2

0

2

0

2

0

2

2224

2sin

22

)2

2cos1(

2sin

2

)sin)(sin()()()(

General equation for the power

BEKG 2433 20

10. Power to Pure Inductive Loads

Energy stored - power flows from source to load

Energy released - power goes back to the source

04

2cos

2

)2

2sin(

2cossin

2

)cos)(sin()()()(

2

0

2

0

2

0

tIV

dtt

IVdtttIVP

tItVtitvtp

mm

mmmm

mm

Power merely absorbed and returned in load due to its reactive properties

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BEKG 2433 21

where V = IXL or I = V/XL

L

LLX

VXIVIQ

22

Unit for reactive power is VAR

(By convention, the reactive power for inductance is positive or leading)

We can see that inductors dissipates zero power, yet the fact that it drops voltage and draws current gives the deceptive impression that it actually does dissipate power.

This “phantom power” is called reactive power, and it is measured in a unit called Volt-Amps-Reactive (VAR), rather than watts. The mathematical symbol for reactive power is (unfortunately) the capital letter Q

BEKG 2433 22

11. Power to Capacitive Loads

Energy stored - power flows from source to load

Energy released - power goes back to the source

04

2cos

2

)2

2sin(

2cossin

2

)cos)(sin()()()(

2

0

2

0

2

0

tIV

dtt

IVdtttIVP

tItVtitvtp

mm

mmmm

mm

Again, power is merely absorbed and returned in load due to its reactive properties

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BEKG 2433 23

C

CCX

VXIVIQ

22 Unit for reactive power is VAR

(By convention, the reactive power for capacitance is negative or lagging)

Again we can see that reactive load such as capacitor dissipate zero power

and the fact that it drops voltage and draws current gives the same

deceptive impression that it actually do dissipate power.

BEKG 2433 24

12. Apparent Power (S)

• We have seen that in pure resistive circuit – Active power, real power, average power (P) appear

• We also have seen that in pure inductive or pure capacitive circuit – Reactive power, (QL) or (QC) appear

• How do both active & reactive power interact when we have R & L , R & C, or R, L & C in a circuit?

• The combination of reactive power and active power is called Apparent power, and it is the product of a circuit's voltage and current, power being supplied to the load without reference to phase angle

S = V I Unit for apparent power is VA

13

13. Relation between P,Q & S

BEKG 2433 25

22CQPS

22LQPS

Purely Resistive (Unity)

Resistive + Inductive (Lagging)

Resistive + Capacitive (Leading)

jXL

“Power Triangle”

Z = R

X = 0

V = IR S = P

Q = 0

jIXL

θ P = V2/R

jQL =V2/XL θ

R

Z = R+jXL

jIXC

θ

IR

V = IZ IZ

θ

P

θ R

jXC

Z = R-jXL

θ IR

IZ

V = IZ

jQC =V2/XC

BEKG 2433 26

14. Power in a circuit with Resistance and Reactance loads

)2cos(2

1cos

2

1

)2cos(cos2

1

)sin()(sin(

)()()(

tIVIV

tIV

tItV

titvtp

mmmm

mm

mm

)sin(

sin

tII

tVV

m

m

CosIVIV

IVP rmsrmsmm

mm cos22

cos2

1

0

cosVIP

R

X1tan

14

Summary

BEKG 2433 27

22 QPIVS

P – Active power, real power or average power. (Watt)

P = V I cos

Q – Reactive power. QL or QC. (Unit: VAR)

Q = V I sin = P tan

S – Apparent power. (Unit: VA)

Where, V & I is the rms value & θ is the phase angle between V and I

**Apparent power – is the magnitude of the complex power. Complex Power:

sinIV

cosIV

sinIVcosIV

IVIV

rmsrms

rmsrms

rmsrmsrmsrms

rmsrmsrmsrms

**

Q

P

jQPS

jS

IVVIS iviv

BEKG 2433 28

Example 1:

A coil having a resistance of 6Ω & an inductance of 0.03H is connected across a 50V, 60Hz supply. Calculate:

a) The current

b) The phase angle between the current & the applied voltage

c) The apparent power

d) The active power

Answer :

a) 3.91 (A) b) 62.05°

b) 195.5 (VA) c) 91.61 (W)

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BEKG 2433 29

Example 2: A coil of inductance 159.2mH and resistance 40 Ω is connected in

parallel with a 30 µF capacitor across a 240V, 50Hz supply.

Calculate:

(a) the current in the coil and its phase angle

(b) the current in the capacitor and its phase angle,

(c) the supply current and its phase angle,

(d) the circuit impedance,

(e) the power consumed,

(f) the apparent power, and

(g) the reactive power.

(h) draw the phasor diagram

Answer: a) 3.75/-51.35° b) 2.26 / 90° c) 2.43/-15.98°

d) 98.76Ω e)560.66 (W) f)582.3 (VA) g)160.56 (VAR)

BEKG 2433 30

• The power factor of an AC electric power system is defined as the ratio of the real power (P) flowing to the load to the apparent power (S) .

Power Factor, PF = P/S = cos θ

where θ is the angle between V & I

• In an electric power system, a load with low power factor draws more current than a load with a high power factor, for the same amount of useful power transferred. The higher currents increase the energy lost (I2R) in the distribution system, and require larger wires and other equipment

• Linear loads with low power factor (such as induction motors) can be corrected with a passive network of capacitors or inductors

15. Power factor

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BEKG 2433 31

PF can be categorized into: Unity, Lagging or Leading

If I lags V, the pf is lagging (Inductive)

If I leads V, the pf is leading (Capacitive)

Example pf = 0.85 lagging

Power factor

PF can be improve or corrected by adding the Capacitive or Inductive

loads.

BEKG 2433 32

Basically the load being supplied consists of resistance and inductance.

This will lead to the appearance of an active (P) and reactive (Q) power (S = P +jQ)

Our aim is to have a value of apparent power (S) , closes to the active power (P) so that the

excessive current drawn from the supply can be reduced.

This can be done by placing reactance of opposite type in parallel to the load so that the

positive Q can be cancelled by the negative Q and vice versa.

The reason to do this correction is to improve system loading, to reduce “copper loss” and

eliminate power factor penalty

Power factor correction

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BEKG 2433 33

,

+ QC

How????

old

P

Qold Sold

new

P

Qnew

Snew

Example

BEKG 2433 34

A load (L1) absorbs an average power of 8kW at a lagging power factor of 0.6 from 230V, 50Hz supply. Calculate:

a) the value of capacitance (at L2) to bring up the power factor to 0.85

b) the old and new supply current

V(t)

L1 L2

I

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BEKG 2433 35

,

PL1 = 8kW PFold = 0.6

old = cos-10.6 = 53.1

Qold = P tan old

= 8000 x tan(53.1)

= 10.667kVAR

P = V I cos , Q = V I sin = P tan 22 QPIVS

PL1 = 8kW PFnew = 0.85

new = cos-10.85 = 31.8

Qnew = P tan new

= 8000 x tan(31.8)

= 4.956kVAR

old

P

Qold Sold

new

P

Qnew

Snew

BEKG 2433 36

We need to add capacitance to reduce the Qold to Qnew

QC = Qold – Qnew

= (10.667 – 4.956)kVAR

= 5.711kVAR

But since c

cCX

VXIQ

22

Therefore 269107115

2303

22

.VAR.

)V(

Q

VX

c

c

F..HzX

Cc

75343269502

11

CX c

1Also since

Hence

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BEKG 2433 37

For the current, we know

cosVIP

Therefore

A.

.V

kW

cosV

PI

old

old

9757

60230

8

A.

.V

kW

cosV

PI

new

new

9240

850230

8

BEKG 2433 38

16. Measurement of Power

• What do we need to measure voltage?

• What do we need to measure Current?

Voltmeter / Multimeter

Ammeter / Multimeter

• What do we need to measure power? Wattmeter

• Like voltmeter & Ammeter, the wattmeter can be analog or digital

Digital:

Numerical

Readout

Analog:

Pointer

on a

scale

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BEKG 2433 39

• To help understand the concept of power measurement, consider the following figures:

• Instantaneous load power is the product of load voltage times load current, an average power is the average of this.

• Therefore, wattmeter should consist of:

Current sensing circuit Voltage sensing circuit

Multiplier circuit Averaging circuit

BEKG 2433 40

• The following figure shows a simplified representation of a wattmeter:

• Connection of a wattmeter:

• Basically there are 4 points, 2 for current measurement (in series), another 2 for voltage measurement (in parallel)

In series

In parallel

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BEKG 2433 41

Example:

For the circuit of the following figure, what does the wattmeter indicate if:

a. Vload = 100 0 V and Iload = 1560 A

Answer:

a. 750 W

b. 1409.5 W

WHY???

b. Vload = 100 10 V and Iload = 1530 A

BEKG 2433 42

Example:

Determine the wattmeter reading:

Answer:

600 W

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BEKG 2433 43

It should be noted that the wattmeter reads only for circuit elements on the load side of the meter. In addition, if the load consists of several elements, it reads the total Real/Active power.

Example:

Answer:

750 W