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BASIC ELECTRICAL ENGINEERING

ELE290

2
...
In order to understand ac systems, it is necessary
to be familiar with the generation of sinusoidal varying ac voltage, which is discussed in this
chapter
...

The concept of average and root mean square values are introduced
...
The analysis of series and parallel ac circuits containing resistances,
inductances and capacitances and the analytical techniques that greatly simplify the solution of
the circuit with sinusoidal excitation are also presented
...
0
...
The coil is mounted on a spindle DD’ as shown
in Figure 2 below

Fig
...
The ends of the coil are connected to two slip
rings S and S’ which are rigidly fixed to an insulated shaft DD’ which is used to rotate the coil
...


BASIC ELECTRICAL ENGINEERING

ELE290

AC can be defined as an ‘Alternating current form which have positive and negative portion and
is usually depicted in a sinusoidal form signal’
...
3
From Figure 3 above, V M sin ω(t – Ø) is a signal that lags the original signal equation V M sin ωt
by an angle of Ø
...
4

2
...


1
...
For a leading value of Ø:-

I



V

BASIC ELECTRICAL ENGINEERING

ELE290

2
...
This term is very important because in actual
practise, the rms voltage is the effective voltage that is utilised in practical or in theoretical
analysis
...
dt
2

For a sinusoidal signal

V RMS 

1
2

0 VM sin t  dt
2

2

The equation will yield the result as below:-

VRMS 

VM
2

BASIC ELECTRICAL ENGINEERING

ELE290

2
...

Calculation of currents in ac circuit will be using V M because V RMS will only be used to
calculate power related question
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1 – SADIKU ELEC TRIC AL CIRCUIT p age 363

BASIC ELECTRICAL ENGINEERING
2
...
1
-

ELE290

Purely Resistive Circuit

A purely resistive circuit only consists of resistor, R
...
f
...
f
...
f
...
3
...


powerfacto r , p
...
 cos 
powerfacto r , p
...
 cos   
powerfacto r , p
...
 value  lagging

BASIC ELECTRICAL ENGINEERING
2
...
3
-

-

ELE290

Capacitive Circuit

A capacitive circuit consists of
resistor , R connected in series
with capacitor, C
The impedances for capacitive
circuit

V

Z  R  jX C
-

1
2f  C
Circuit diagram of an inductive circuit

Current calculation

I
-

-jXC

Inductance reactance calculation

jX C  
-

R

VM 0
V 0
 M
 I(0   )
R  jX C Z   

Impedance triangle

R



 jX L

Z

-

Phasor diagram representation

I   


VM 0
-

Since angle between voltage and current is positive, this means that current in a
capacitor circuit leads the voltage
...
f
...
f
...
f
...
4 POWER IN AC CIRCUIT
There are three types of power in an AC circuit
1
...
Real Power, P (W)
3
...


S  (V0)  ( I   ) *
S  (V0)  ( I   )
S  VRMS I RMS cos   jVRMS I RMS sin 
S  P( watts )  jQ (var s )
Other names for the power in AC circuit:1
...
Q (vars) = reactive power/imaginary power
3
...

Power triangle for power in ac circuit is as shown below:-

P


S

 jQ


 jQ
S

P
Inductive

Capacitive

BASIC ELECTRICAL ENGINEERING

2
...
Pay-back period for an equipment
purchase including installation cost may be less than six months to a year
...
Other ancillary
benefits to be gained by correcting power factor are, lower energy losses, better voltage
regulation and released system capacity
...
These inductive characteristics are more pronounced in motors and
transformers, and therefore, can be quite significant in industrial facilities
...

Some utilities charge for these vars in the form of a penalty, or KVA demand charge, to
justify the cost for lost energy and the additional conductor and transformer capacity
required to carry the vars
...


ELE290

BASIC ELECTRICAL ENGINEERING

ELE290

Advantages of power factor correction done on a system:



A reduction in the overall cost of electricity can be achieved by improving the
power factor to a more economic level
...

Reducing the load on distribution network components by power factor
improvement will result in an extension of their useful life
...

Nuisance tripping of circuit breakers and other protective devises
...


Usually, the correction of power factor is done by connecting a capacitor bank in parallel
with the load
...


I

I

IL

IL
V

V

LOAD

Without capacitor bank

IC

LOAD

C

With capacitor bank connected

I SUPPLY  I LOAD  I CAPACITOR

I SUPPLY  I LOAD
IC
V

 OLD

 NEW

 OLD
IL

V

I L  IC

IL

BASIC ELECTRICAL ENGINEERING

ELE290

From the phasor diagram above, we can see that there is a difference if a capacitor bank is
connected in parallel with the load and if there is no capacitor bank connected with the load
...
The phasor
diagram above applies the lagging power factor characteristic to the diagram
...
5
...
95 p
...
The
frequency of the supply is set at 50 Hz
...
8  30 A
ZTOTAL 2530

Step 2: Get the Real power and Reactive power using old power factor (Ø is taken from the angle
of current)

P  VRMS I RMS cos 
1
P  (120VRMS )  (4
...
83W
1
QOLD  VRMS I RMS sin 
QOLD  (120VRMS )  (4
...
83  j 288)VA
Step 3: Compare parameters between old p
...
f

old
...
f  30
new
...
f  cos 1 0
...
p
...
19

BASIC ELECTRICAL ENGINEERING

ELE290

We can see that there is a difference between the angle of old power factor and new power factor

IC
S

18
...
19

IL

jQNEW
P1  P2

Real Power at old power factor = Real Power at new power factor

P1  P2  498
...
19˚

V RMS 1 I RMS1 cos  OLD  V RMS 2 I RMS 2 cos  NEW
V RMS1 I RMS1 cos   498
...
07VA
cos  
0
...
07VA sin 18
...
41uF
(V RMS ) 2 2f (120V ) 2  2    50 Hz

jQOLD

BASIC ELECTRICAL ENGINEERING

ELE290

TUTORIAL

1) How is AC different from DC?
2) Draw the phasor diagram for a purely resistive, inductive and capacitive AC circuit
3) List down the 3 main components that are involved in an AC circuit

4) Figure below shows a simple single phase AC power system with 2 loads
...


Assume that the voltage applied to these loads is V = 208<0˚ Volt
i) Calculate the current flowing to these loads, I
ii) Calculate the complex power, S consumed by these loads,
iii) Calculate the voltage for each of the load
...
Do the loads consume
reactive power from the source or supply it to the source?
5) The AC circuit shown in the Figure (a) is connected to an AC supply
...
Calculate the followings:
i)
The RMS voltage source
...
The equation for the
voltage supply is 150V cos(2500t + 30˚)V and the corresponding circuit parameters are
R1=100Ω,R2=250Ω, 10mH and 250μF
...

ii)
The phase shift
iii)
The maximum current, I a in the circuit
iv)
The voltage across capacitor,VC and the voltage across the inductor,VL

250

Ia



100

150 cos(2500t  30)

VL



VC

10mH



250F


Figure 1(a)

7) A simple RLC circuit is shown in Figure 1(b) is connected to an AC supply
...
Calculate the followings:
i)
ii)
iii)
iv)
v)
vi)

The frequency in Hz
...

Phase angle of the voltage
...

The phasor voltage of V C1 and V C2
...


IT

10

10

5mH

+

VC 2

VS

+

VC1

30F

-

Figure Q1b

50F

BASIC ELECTRICAL ENGINEERING

ELE290

8) The sinusoidal steady state voltage source for the circuit shown in Figure Q1a is 16cos(10t),
calculate;
i)
The total impedance and power factor seen by the voltage source
ii)
The phasor current I s, I1 , and I2
iii)
The average, reactive and complex power of the source
...
02F
Is

I1
0
...
The
corresponding circuit parameters are R1 = 250Ω, R2 = 300Ω, L1 = L2 = 20mH and C1 = 200uF
...


10) When connected to a 120-V (rms), 60-Hz power line, a load absorbs 4kW at a lagging power
factor of 0
...
Find the value of capacitance necessery to raise the power factor to 0
...


11) A 240V (rms), 50Hz power line is connected to a load
...
85
...
90
...
The connected load absorbs 5kW at a
lagging power factor of 0
...
Find the value of capacitance needed to raise the power factor to
0
...

13) Three loads A, B and C are connected in parallel across a 1400 V rms 50 Hz line
...
Load B consumes 10 kW with
a 0
...
Determine
the following:
i)
ii)
iii)

Total apparent power in the system
...

Calculate the capacitance necessary to establish 0
...




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