Concept of ac circuit for electrical engineering and this help to study about ac circuit

Power in Single Phase AC Circuits
Let us consider the following circuit.
−
+ v(t)
i(t)
Load
V
I
θV θI
Let
v(t) =
√
2V sin(ωt + θV )
i(t) =
√
2I sin(ωt + θI )

The instantaneous power delivered to the load is
p(t) = v(t)i(t)
p(t) = Vm sin(ωt + θV )Im sin(ωt + θI )
p(t) =
VmIm
2
(cos(θV − θI ) − cos(2ωt + θV + θI ))
p(t) =
VmIm
2
cos(θV − θI ) −
VmIm
2
cos(2ωt + θV + θI )

ωt
p, v, i
v
i
Vm
Im
θI
θV
φ
p
Figure: Voltage, current and power in RL circuit

Let θV − θI be φ.
p(t) = VI cos φ − VI cos(2ωt + θV + θI )
p(t) = VI cos φ − VI cos(2ωt + θV − θI + 2θI )
p(t) = VI cos φ − VI cos(2ωt + 2θI − φ)
p(t) = VI cos φ(1 − cos(2ωt + 2θI ))
| {z }
pI
− VI sin φ sin(2ωt + 2θI )
| {z }
pII
pI has an average value of VI cos φ which is called the average
power.
pII does not have an average. But it’s maximum value is VI sin φ
which is called reactive power.

ωt
p
p
pI
pII
P
Q
Figure: Power in RL circuit

Power
The average power P is
P =
VmIm
2
cos(θV − θI ) = VI cos(φ)
where φ = θV − θI . Its unit is watts (W).
The reactive power Q is
Q = VI sin φ
The apparent power S is
|S| = VI
Its unit is volt-ampere (VA).
The ratio of real power (P) to apparent power is called as the
power factor (pf).
pf =
VI cos φ
VI
= cos φ
Since cos φ can never be greater than unity, P ≤ |S|.

Complex Power
Let us define voltage phasor and current phasor.
V = V ∠θV , I = I∠θI
The complex power S is
S = VI∗
S = V ∠θV I∠ − θI
= VI∠(θV − θI )
S = VI cos φ + VI sin φ
The real part of S is called the average power (P). The imaginary
part of S is called the reactive power (Q).
S = P + Q

Re
Im
V
I
θV θI
φ
Re
Im
P
Q
S
φ
Figure: RL load
If V leads I (φ > 0), power factor is lagging.

Re
Im
V
I
θV = θI
Re
Im
P=S
Figure: Resisitive Load
If V and I are in phase (φ = 0), power factor is unity.

Re
Im
I
V
θI θV
φ
Re
Im
P
Q
S
−φ
Figure: RC load
If I leads V (φ < 0), power factor is leading.

For two loads (inductive and capacitive) in parallel,
P1
Q1
S1 P2
Q2
S2
ST QT
PT
PT = P1 + P2; QT = Q1 + Q2
But
|ST | 6= |S1| + |S2|

Power Factor Control
I If pf decreases, the current will increase to supply the same
real power.
I This will increase the line loss. (It is an additional cost to a
utility.)
I Capacitors which supply reactive power are connected in
parallel to improve the power factor.
Load C
P
Q
S
−Qc
Snew
Qnew
φnew

Example 1 : A single-phase inductive load draws to 1 kW at 0.6
power-factor lagging from a 230 V AC supply.
1. Find the current it draws.
2. Find the value of a capacitor to be connected in parallel with
the load to raise the power factor to 0.9 lagging. Determine
the current under this condition.
1.
I =
1000
230 × 0.6
= 7.24 A
Q = 230 × 7.24 × 0.8 = 1.332 kVAr
I
Load
1 kW
1.332 kVAR
1.6652 kVA
53.13◦

2.
pfnew = 0.9; φnew = 25.84◦
Qnew = QL − Qc
Qnew = P × tan 25.84◦
= 484.32 VAr
Qc = 847.68 VAr
Qc = V 2
ωC
C = 51 µF
I =
1000
230 × 0.9
= 4.83 A
I
Load C
1 kW
QL
S
−Qc
Snew
Qnew
25.84◦

Power in Balanced Three Phase Circuits
Let va, vb and vc be the instantaneous voltages of a balanced three
phase source.
va =
√
2V sin(ωt + θV )
vb =
√
2V sin(ωt + θV − 120◦
)
vc =
√
2V sin(ωt + θV − 240◦
)
When it supplies a balanced load,
ia =
√
2I sin(ωt + θI )
ib =
√
2I sin(ωt + θI − 120◦
)
ic =
√
2I sin(ωt + θI − 240◦
)

The instantaneous power is
p = vaia + vbib + vcic
p =
√
2Vp sin(ωt + θV ) ×
√
2Ip sin(ωt + θI )
√
2Vp sin(ωt + θV − 120◦
) ×
√
2Ip sin(ωt + θI − 120◦
)
√
2Vp sin(ωt + θV − 240◦
) ×
√
2Ip sin(ωt + θI − 240◦
)
p =VpIp cos(θV − θI ) + VpIp cos(2ωt + θV + θI )
VpIp cos(θV − θI ) + VpIp cos(2ωt + θV + θI − 120◦
)
VpIp cos(θV − θI ) + VpIp cos(2ωt + θV + θI − 240◦
)
p = 3VpIp cos φ
where φ = θV − θI .
The instantaneous power in a 3 phase balanced system is constant.

ωt
p, v, i
p
Figure: Voltage, current and power in a R-L load

The average/real power in a 3-phase system is
P = 3VpIp cos φ Watts
In a Y connected load, VL =
√
3Vp and IL = Ip,
P =
√
3VLIL cos φ
In a ∆ connected load, VL = Vp and IL
√
3Ip,
P =
√
3VLIL cos φ
Therefore, the three phase real power is
P = 3VpIp cos φ =
√
3VLIL cos φ

Since the instantaneous power in a 3-phase balanced system is
constant, it does not mean that there is no reactive power. Still
the instantaneous power of individual phases is pulsating.
The 3-phase reactive power is
Q = 3VpIP sin φ =
√
3VLIL sin φ VAr
The apparent power is
|S| =
p
P2 + Q2 = 3VpIp =
√
3VLIL VA

Per Phase Analysis
If a three phase system is balanced and there is no mutual
inductance between phases, it is enough to analyze it on per phase
basis.
1. Convert all ∆ connected sources and loads into equivalent Y
connections.
2. Solve for phase a variables using the phase a circuit with
neutrals connected.
3. Other phase variables can be found from the phase a variables
using the symmetry.
4. If necessary, find line-line variables from the original circuit.

Synchronous Machine Model
The per phase equivalent circuit of a synchronous machine is
+
−
E δ
Xs Ia
Ra
+
−
Vt 0◦

Transformer Model
The per phase equivalent circuit of a transformer is
Req Xeq
Rc Xm
I Since the impedance of the shunt path is larger, Rc and Xm
are neglected.
I Since Req is much smaller than Xeq, Req can be eliminated.
Xeq

Example 2: Consider a system where a three phase 440 V, 50 Hz
source is supplying power to two loads. Load 1 is a ∆ connected
load with a phase impedance of 10∠30◦ Ω and load 2 is a Y
connected load with a phase impedance of 5∠36.87◦ Ω.
1. Find the line current and the overall power factor of the
system.
2. Determine the capacitance per phase in µF of a three phase
bank of delta connected capacitors to be added in parallel to
the load to improve the overall power factor unity. Find the
line current under this condition.

1. The per phase equivalent circuit after using ∆ to Y
transformation,
−
+
440
√
3
IL
10 30◦
3
5 36.87◦
IL =
440/
√
3
10/3 30◦ +
440/
√
3
5 36.87◦
IL = 106.64 − 68.6A
IL = 126.8 −32.75◦ A
pf = cos(−32.75◦
) = 0.84 lag

2. The per phase equivalent circuit with a capacitor bank
−
+
440
√
3
IL
10 30◦
3
5 36.87◦ −
Xc
3
To make overall power factor unity, IL must be in phase with
the voltage.
∴ Ic = 68.6A
Xc =
3 × 440
√
3 × 68.6
= 11.11 Ω
C = 286.52 µF

Example 3: Consider a system where a three phase 400 V, 50 Hz
source is supplying power to two loads. Load 1 draws 5 kW at 0.8
pf lagging and load 2 draws 5 kW at unity power factor. The
voltage across the loads is 400 V.
1. Find the line current and the overall power factor.
2. Find the value of kVAR required from a bank of capacitors
connected across the loads to improve the overall power factor
to unity. Determine the line current under this condition.
1.
IL1 =
5000
√
3 × 400 × 0.8
= 9 A
IL2 =
5000
√
3 × 400 × 1
= 7.2; A
IL = 9 −36.87◦ + 7.2 0◦ = 15.38 −20.56◦ A
pf = 0.9363 lag
QT = Q1 + Q2 = 3.75 + 0 = 3.75 kVAR

2. To make overall power factor unity,
QC + QT = 0
QC = −3.75 kVAR
(-ve indicates that the capacitor supplies reactive power.)
∴ QC = 3.75 kVAR
When pf is unity, S = P.
IL =
PT
√
3VL
=
10 × 103
√
3 × 400
= 14.4 A
IL = 14.4 0◦ A

per unit
The per unit is defined as
per unit =
actual value in any unit
base value in the same unit
There are normally four quantities associated with a power system.
S, V , I, Z
How to find base quantities?
I Choose any two. Normally Sbase and Vbase are chosen.
I Find the remaining two using their relations.

Let us start with single phase.
Sb = S1φ MVA; Vb = V1φ kV
Ib =
Sb(MVA)
Vb(kV)
kA
Zb =
Vb(kV)
Ib(kA)
Ω
Substituting Ib in Zb,
Zb =
V 2
b ( in kV)
Sb( in MVA)
Ω
Zp.u. =
Zactual(Ω)
Zb(Ω)
∴ Zp.u. =
Zactual(Ω) × Sb( 1φ MVA)
V 2
b ( L- N in kV)

For three phase.
Sb = S3φ MVA; Vb = VL−L kV
Ib =
Sb(MVA)
√
3Vb(kV)
kA
Zb =
Vb(kV)
√
3 × Ib(kA)
Ω
Substituting Ib in Zb,
Zb =
V 2
b ( in kV)
Sb( in MVA)
Ω
Zp.u. =
Zactual(Ω)
Zb(Ω)
∴ Zp.u. =
Zactual(Ω) × Sb( 3φ MVA)
V 2
b (L-L in kV)

S3φ
b = 3S1φ
b ; VbL−L
=
√
3VbL−N
S3φ
p.u. =
S3φ
S3φ
b
=
3 × S1φ
3 × S1φ
b
= Sp.u.
Vp.u. =
VL−L
VbL−L
=
√
3VL−N
√
3VbL−N
= Vp.u.
I If the voltage magnitude is 1 p.u., the line-line voltage is 1
p.u. and the line-neutral voltage is also 1 p.u.
I Similarly, three phase power in p.u. and the single phase
power in p.u. are the same.
Zb =
(VbL−L
)2
S3φ
b
=
(
√
3VbL−N
)2
3 × S1φ
b
=
(VbL−N
)2
S1φ
b

Impedance values of a component when given in per unit without
specified bases are generally understood to be based on the MVA
and kV ratings of the component.
To change p.u. from one base to new base:
Zp.u. ∝
Sb
V 2
b
Zp.u. (new) = Zp.u. (given) ×
Sb(new in MVA)
Sb(given in MVA)
×
V 2
b (given in kV)
V 2
b (new in kV)
Advantages:
I The per unit values of impedance, voltage and current of a
transformer are the same regardless of whether they are
referred to the HV side or LV side. This is possible by
choosing base voltages on either side of the transformer using
the voltage ratio of the transformer.
I The factors
√
3 and 3 get eliminated in the per unit power
and voltage the equations.

Example 3 : Let us do the same example in p.u.
1.
Sb = 5 kVA; Vb = 400 V
P1p.u. =
5
5
= 1; P2p.u. =
5
5
= 1
Since Ip.u. =
Pp.u.
V p.u. × pf
IL1p.u. =
P1
V × pf
=
1
1 × 0.8
= 1.25
IL2p.u. =
P2
V × pf
=
1
1 × 1
= 1
IL = IL1 + IL2 = 1.25 −36.87◦ + 1 0◦ = 2.14 −20.56◦ p.u.
Ib =
Sb
√
3 × Vb
=
5 × 103
√
3 × 400
= 7.22 A.
IL = 2.14 −20.56◦ × 7.22 = 15.44 −20.56◦ A

2. To make unity pf, QC = QT .
QT p.u. = PT p.u. × tan φ = 2 × tan(20.56◦
) = 0.75 p.u.
QC = QC p.u. × Sb = 0.75 × 5 = 3.75 kVAr
Capacitors supply reactive power.
When the problems to be solved are more complex, and
particularly when transformers are involved, the advantages of
calculations in per unit are more apparent .

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