1

Conservation of AC Power

Dr. Mohd Hafizi Ahmad

IVAT UTM
 2

Learning Outcome

After completing this lesson, you should able to:

 Understand the concept of conservation power

 determine the power factor of the AC system
 analyze power correction problem
 3

Conservation of AC Power

The principle applies to both ac and dc circuits

The complex, real, and reactive powers of the sources

equal the respective sums of the complex, real, and
reactive powers of the individual loads.

SkT = Sk1 + Sk2 + Sk3

= (P1 + jQ1) + (P2 + jQ2) + (P3 + jQ3)
= (P1 + P2 +P3) + j (Q1+ Q2 + Q3)
= PT + j Q T
 4

Conservation of AC Power

Parallel Circuit

The complex power supplied by the source equals the

complex powers delivered to loads Z1 and Z2
 
Sk  VI *  V I1*  I 2*  VI1*  VI2*  Sk1  Sk 2
I  I1  I 2
 5

Conservation of AC Power

Parallel Circuit

V  V1  V2

S  VI *  V1  V2 I * 
 V1 I *  V2 I *  S1  S 2

Whether the loads are connected in

series/parallel/general, the total power supplied by the
source equals the total power delivered to the load.
 6

Conservation of AC Power

This means that the total complex power in a network

is the sum of the complex powers of the individual
elements, BUT not true of apparent power.

SkT = Sk1 + Sk2 + Sk3

= (P1 + jQ1) + (P2 + jQ2) + (P3 + jQ3)
= (P1 + P2 +P3) + j (Q1+ Q2 + Q3)
= PT + j QT
 7

Example 4

i- Find average power and reactive power supplied by the source

ii- Find average power absorb by the resistor.
iii- Find the source power factor.
 8

Example 5

Find the average power generated by each source and average

power absorb by each passive element.
 9

Power Factor Correction

 Power factor is very important in power delivery. It is known that power

utility company charged the energy use by the consumer in terms of
kWh, which mean that only AVERAGE/REAL power is taken into
account. How about REACTIVE power?

 Reactive power contributes to the value of pf.

 If pf  1, excessive current is supplied by source. This is not efficient.

 Thus, pf must be improved (corrected) and the method to do this is

called power factor correction, (pfc).
 10

Power Factor Correction

Most all practical loads are inductive (motors, fluorescent

lights, etc.) and operate at a lagging power factor.

Power factor correction is necessary for economic reason.

 11

Power Factor Correction

The design of transmission system is very sensitive to the

magnitude of the current as determined by the applied load.

Increased currents result in increased power losses due to

the resistance of the lines

Heavier currents also require larger conductors, increasing

the amount of copper and obviously the require increased
generating capacities

Since the voltage is fixed, the apparent power is directly

related to the current level
 12

Power Factor Correction

The smaller the apparent power, the smaller is the current

drawn from the supply

Minimum current is therefore drawn from a supply when

S=P and Q=0.

PF angle approaches zero degrees and PF approaches 1,

revealing that the network is appearing more and more
resistive at the input terminals.
 13

Power Factor Correction

REMINDER :
Power factor correction is the process of increasing the
power factor without altering the voltage or current to
the original load.

Because most loads are inductive, power factor can be

corrected by installing a capacitor in parallel with the
load.
 14

Power Factor Correction

Consider a circuit with R and L load shown below

iS (t)
pf  0.707 lagging
10Ω
v(t)=240cos20t 0.5H io (t)=16.97cos(20t–450) A.
io(t)
is (t) = io(t)

The phasor diagram is Im

240
Re
 = 450
|IS| = 16.97
 15

Power Factor Correction

 If a capacitor is connected in parallel with the load,

is (t) io(t) 10Ω where in phasor,

Is = Io + Ic
C
v(t)
0.5H
ic (t)

V V
IC  
ZC 1/jC
Io is remaining the same   jCV
 CV900
 16

Power Factor Correction

Rearranging the current phasor diagram,

Im Im
Ic
Re
 = 45 Ic
Io = 16.97 Re
pfc
New/corrected power factor, pf = cos(pfc) Is Is = Io + Ic
Io
The value of the capacitor
V and 1
ZC  ZC 
IC j C
 17

Power Factor Correction

Using power triangle to solve for pfc

Im(S)
|S1|
+jQ1
|S2| S2 = S1 + SC
+jQ2
+pfc
P Re(S)
SC = –jQC

VI * V 2
The power at the capacitor SC  
2 2Z *
 18

Power Factor Correction

 Using power triangle to solve for pfc

VI C * V2
SC   Im(S)
2 2ZC * |S1|
+jQ1
 Rearranging the equation |S2|
+pfc +jQ2
V2 P Re(S)
ZC *  SC = –jQC
2SC
1
ZC 
j C
 19

Example 6

A circuit with R and L load consumes P=50kW real

power with pf=0.8 lagging. The input voltage is 230V,
50Hz frequency. A capacitor is connected in parallel
with the RL load to improve the power factor to 0.95.
Find the capacitor value.
 20

Example 7

A 4kW, 110Vrms load has a power factor of 0.82

lagging. Find the value of the parallel capacitor
that will correct the power factor to 0.95 lagging
when ω=377rad/s