Introduction to AC Circuits

Dr. Lakshmi Varaha Iyer, PhD

Significant Features of Three-Phase AC Circuits
• Almost all AC power generation and transmission is in the form of 3-phase AC circuits
• AC power systems have a great advantage over DC systems in that their voltage levels can be
changed with transformers to reduce transmission losses.
• Three-phase (3f) AC power system consists of
• 3f AC generators
• 3f transmission lines
• 3f loads
• Advantages of having 3f power systems over 1f ones:
• More power per kilogram of metal from a 3f electric machine.
• Power delivered to a 3f load is constant, instead of pulsating as it does in a 1f system.
• 3-phase systems also make the use of induction motors easier by allowing them to start
without special auxiliary starting windings.
Generation of 3f Voltages and Currents
A 3f generator consists of three 1f generators:

- voltage of all phases are equal in magnitude

- differing in phase angle from each other by 120o.
Generator Connected to Three identical Loads
• Each of these three
generators could be
connected to one of the
VC
three identical loads by a
pair of wires and the
resulting power system
would be as shown in the
figure VA
• Current flowing to each
load can be found from
the equation:
VB

Phasor diagram showing the

voltages in each phase
Currents in the Three Phases and the Neutral
Currents flowing in the three phases:
V 0 0
Ia = = I − 
Z 
V − 1200
Ib = = I − 1200 − 
Z 
V − 2400
Ic = = I − 2400 − 
Z 

It is possible to connect the negative ends of these three single phase generators and the loads
together, so that they share a common return line, called neutral.
I N = Ia + Ib + Ic
( ) (
= I −  + I − 120 0 −  + I − 240 0 −  )
= I
 ( ) ( ) 
cos(−  ) + j sin(−  ) + cos − 120 0 −  + j sin − 120 0 −  As long as the three loads are equal,
  ( ) (
+ cos − 240 0 −  + j sin − 240 0 − ) the return current in the neutral is zero.
=0
Balanced Power Systems
• In a balanced power system:
• Three generators have same voltage magnitude and phase difference is 120o.
• Three loads are equal and magnitude and angle.
• Neutral is actually unnecessary and we could get away with three wires instead of six

• abc phase sequence: the voltages in the three phases peak in the order a, b and c. It is
possible to have acb phase sequence.

abc phase sequence acb phase sequence

Y and Δ Connections
Ib

+ Va Ia Z
+
A connection of this sort - - Vb In Z
is called Wye-connection. - Z
Vc
+ Ic

Ia

Va + -
Another possible connection Vb Z Z
is delta-connection, in which - +
the generators are connected Z
head to tail. + - Ib
Vc
Ic
Voltages and Currents in a Y-Connected 3f Circuit
Phase quantities: voltages and currents in a given phase (Van, Vbn, Vcn)
Line quantities: voltages between lines and current in the lines (Vab, Vbc, Vca)
A
If +
- Ib
Vab
+
+ Van + V - Ia (=IL)
Resistive
bc
- - Vbn Load
n V
- Vca
Vcn
+ + Ic
-

Van = Vf 0 0 I a = I f 0 0
Vbn = Vf  − 1200 I b = I f  − 1200
Vcn = Vf  − 1200 I c = I f  − 2400

Vab = Van − Vbn = Vf00 − Vf − 120 0 = 3Vf30 0

Voltages and Currents in a Y-Connected 3f Circuit (cont’d)
• The relationship between the magnitude of the line-to-line voltage and the line-to-neutral
(phase) voltage in a Y-connected generator or load

VLL = 3Vf

• In a Y-connected generator or load, the current in any line is the same as the current in the
corresponding phase.
I L = If Vcn
Vca Vab

Van
• Line voltages are shifted 30 degrees with respect
to phase voltages.
Vbn

Vbc
Voltages and Currents in a Δ Connected 3f Circuit
• In a delta-connected generator or load, the line-to- Iab
line voltage between any two lines will be the
VA + I -
same as the voltage in the corresponding phase. a VB
- Ib +
VLL = Vf Ic
+ - Ibc
VC
• In a delta-connected generator or load, the Ica
relationship between the magnitudes of the line • Line currents are
and phase currents: shifted 30 degrees Ic
I L = 3I f with respect to Ica
phase currents.

Iab

Ib Ia
Ibc
Power Relationship in 3f Circuits
The 3f voltages applied to this load:
van (t ) = 2V sin(t )
vbn (t ) = 2V sin(t − 1200 )
vcn (t ) = 2V sin(t − 2400 )
A balanced Y-connected load.
The 3f currents flowing in this load:
ia (t ) = 2 I sin(t −  )
ib (t ) = 2 I sin(t − 1200 −  )
ic (t ) = 2 I sin(t − 2400 −  )
• Power supplied by each phase
consists of constant component
Instantaneous power supplied to each of the three phases: plus a pulsing component.
Pa (t ) = van (t )ia (t ) = 2VI sin(t )sin(t −  ) = VI cos − cos(2t −  )
• However, the pulsing components
Pb (t ) = vbn (t )ib (t ) = 2VI sin(t − 120 )sin(t − 120 −  ) = VI cos − cos(2t − 240 −  )
0 0 0
in the 3 phases cancel each other
Pc (t ) = vcn (t )ic (t ) = 2VI sin(t − 240 )sin(t − 240 −  ) = VI cos − cos(2t − 480 −  )
0 0 0
since they are 120 deg out of
phase with each other. Final power
Total power supplied to the 3f load: supplied by three phase power
Ptotal (t ) = P3f = Pa (t ) + Pb (t ) + Pc (t ) = 3VI cos = 3P1f system is a constant.
3f Power Equations Involving Phase Quantities

S1f = Vf I f
P1f = Vf I f cos 
Q1f = Vf I f sin 

S3f = 3Vf I f
P3f = 3Vf I f cos 
Q3f = 3Vf I f sin 
3f Power Equations Involving Line Quantities
For a Y-connected load: I L = I f and VLL = 3Vf

V 
P3f = 3Vf I f cos  = 3 LL  I L cos  = 3VLL I L cos 
 3
• It is important to realize that the
For a delta-connected load: I L = 3I f and VLL = Vf cos theta and sin theta are the
I  cosine and sine of the angle
P3f = 3Vf I f cos  = 3VLL  L  cos  = 3VLL I L cos  between the phase voltage and
 3 phase current, not angle
between the line-line voltage
Therefore, regardless of the connection of the load: and line current.
P3f = 3VLL I L cos 
Q3f = 3VLL I L sin 
S3f = 3VLL I L
Analysis of Balanced 3f Systems
If a three-phase power system is
balanced, it is possible to determine
voltages and currents at various points in
the circuit with a per phase equivalent
circuit.

• Neutral wire can be inserted, as no

current would be flowing through it,
thus, system is not affected.
• Three phases are identical except for
120o phase shift for each phases.
• It is thus possible to analyze circuit
consists of one phase and neutral.
• Results would be valid for other two
phases as well if 120o phase shift is
included.
Wye-Delta Transformation
• Above analysis if OK for Y-connected sources and loads, but no neutral can be connected
for delta-connected sources and loads.
• As a result, the standard approach is to transform the impedances by using the delta-wye
transform of elementary circuit theory.

Delta connected load consisting of

three equal impedances each of value
Z is totally equivalent to a Y connected
load consisting of three equally
impedances of value Z/3.
Numerical Problem - 1
A 3-phase star connected electric motor rated at 400V is consuming 50A at a
power factor of 0.8 when delivering a torque of 110Nm at a speed of 500 rpm.

Calculate :
1. Phase voltage in RMS quantity
2. Peak of phase voltage
3. Line voltage
4. Phase current
5. Line current
6. Input power
7. Output power
8. Efficiency of the motor
Numerical Problem - 2
A 3-phase delta connected electric motor rated at 275V is consuming 11A at a
power factor of 0.9 when delivering a torque of 70Nm at a speed of 575 rpm.

Calculate :
1. Phase voltage in RMS quantity
2. Peak of phase voltage
3. Line voltage
4. Phase current
5. Line current
6. Input power
7. Output power
8. Efficiency of the motor
Numerical Problem – 3 (pg. 695)
Numerical Problem – 3 (pg. 695)
Numerical Problem – 3 (pg. 695)
Reference
• Appendix A – Three Phase Circuits in Electric Machinery Fundamentals by Stephen J Chapman,
Fourth Edition, Mc Graw Hill