AC circuits and phasors
Fabrizio Sossan
fabrizio.sossan@mines-paristech.fr
Associate Professor, PERSEE Center MINES ParisTech - PSL University
Slides available at https://bit.ly/3lLhmjY - (second letter of the last portion is a lowercase ”L”)
(Version compiled on September 21, 2021)
Outline
I Review of Ohm’s law for DC circuits I Alternating current (AC)
I Physical evidence of active and reactive power I Phasors
I Three-phase systems
I Power flow on an transmission line
Ohm’s law
Figure:Direct current (DC) source with a series resistance
Current in the resistance:
i = v R
Quantity Symbol Unit
Voltage v Volt (V)
Current i Ampere (A)
Resistance R Ohm (Ω)
(Instantaneous) Power
Instantaneous power of the generator:
p = vi
Quantity Symbol Unit
Power p Watt (W)
Energy Joule, or more commonly, kilowatt-hours
Alternating current (AC) source
Figure: AC source with a series resistance
v (t)
Time
i (t)
Time v (t)/R
Alternating current (AC) source
Figure: AC source with a series resistance
v (t)
Time
i (t)
Time v (t)/R
Alternating current (AC) source
Figure: AC source with a series resistance
v (t)
Time
i (t)
Time v (t)/R
Inductor and capacitor
Inductor
vL= LdiL
dt
Capacitor
ic = Cdvc dt
Instantaneous power in a resistive load
v (t)
Time
i (t)
Time
Instantaneous power in a resistive load
v (t)
Time i (t)
Time
Instantaneous power in a resistive load (cont’d)
A resistive load consumes power that has an average larger than zero. The power it consumes is called active (or real) power.
Instantaneous power in an inductive load
v (t)
Time
i (t)
Time
Instantaneous power in an inductive load
v (t)
Time i (t)
Time
Instantaneous power in an inductive load (cont’d)
An inductive load absorbs (zero average) instantaneous power with the grid. This power bouncing back and forth between the grid and the load refers to the notion of
reactive power.
Real load
Actual loads are something in between a resistor and an inductor (e.g., electric motors) or a capacitor: their current and voltage are never aligned exactly. The phase
displacement between voltage and current,φ, is key to characterize a load.
Root mean square (RMS) values in AC circuits
Consider an AC voltage source v (t) = Vpsin(ωt) with peak voltage Vp, pulsation ω and period T (T = 2π
ω ) connected to a resistive load, R.
The average instantaneous power of the circuit is (double angle formulae): P¯ac = 1
R 1 T
Z T 0
v2(t) = 1 R
1 T
Z T 0
Vp2sin2(ωt) =
= 1 R
1 T
Z T 0
Vp2
2 (1 − cos 2ωt) = RVp2 2T
Z T 0
(1 − cos 2ωt) = 1 R
Vp2 2 .
Root mean square (RMS) values in AC circuits
Consider an AC voltage source v (t) = Vpsin(ωt) with peak voltage Vp, pulsation ω and period T (T = 2π
ω ) connected to a resistive load, R.
The average instantaneous power of the circuit is (double angle formulae):
P¯ac = 1 R
1 T
Z T 0
v2(t) = 1 R
1 T
Z T 0
Vp2sin2(ωt) =
= 1 R
1 T
Z T 0
Vp2
2 (1 − cos 2ωt) = RVp2 2T
Z T 0
(1 − cos 2ωt) = 1 R
Vp2 2 .
Root mean square (RMS) values (cont’d)
In a DC circuit, the instantaneous power is Pdc = 1
RVdc2 .
Equating the AC average power and DC power yields: Vdc2 = Vp2
2 → Vdc = Vp
√ 2 Vp
√2 is the root-mean-square (RMS) value of the AC voltage. The RMS value of an AC source is the voltage level that a DC should have to produce the same thermal
dissipation. Unless specified otherwise, AC voltage levels always refer to RMS values.
Root mean square (RMS) values (cont’d)
In a DC circuit, the instantaneous power is Pdc = 1
RVdc2 .
Equating the AC average power and DC power yields:
Vdc2 = Vp2
2 → Vdc = Vp
√ 2 Vp
√2 is the root-mean-square (RMS) value of the AC voltage. The RMS value of an AC source is the voltage level that a DC should have to produce the same thermal
dissipation. Unless specified otherwise, AC voltage levels always refer to RMS values.
Active and reactive power
Active power P = VI cosφ [W]
Reactive power Q = VI sinφ [VAr]
Apparent power S = VI [VA]
Power factor cosφ [-]
Complex power S = P + jQ˜
V and I are the RMS values.
Active and reactive power
Active power P = VI cosφ [W]
Reactive power Q = VI sinφ [VAr]
Apparent power S = VI [VA]
Power factor cosφ [-]
Complex power S = P + jQ˜
V and I are the RMS values.
Active and reactive power (cont’d)
Applying Pitagora’s theorem to the power triangle yields:
S2 = P2+ Q2
i.e., the so-called capability curve of ideal electrical machines and power converters. Such an electrical apparatus could provide any combination of active and reactive power as far it falls in the capability curve.
Active and reactive power (cont’d)
Applying Pitagora’s theorem to the power triangle yields:
S2 = P2+ Q2
i.e., the so-called capability curve of ideal electrical machines and power converters.
Such an electrical apparatus could provide any combination of active and reactive power as far it falls in the capability curve.
Example of apparent power rating
Figure:Small converter for photo-voltaic (PV) panels.
Phasors
To handle linear operations with AC quantities, it is convenient to adopt the phasor notation. Phasor notation is derived from Euler’s formula:
Aej (ωt+θ) = A cos(ωt + θ) + jA sin(ωt + θ) = Aej θ
|{z}
Phasor
·ej ωt
The portion Aej θ can be used to perform calculations with sinusoidal waveforms considering magnitude, A, and phase,θ, information only. For example:
cos(ωt + φ) + cos(ωt + ψ) =
= Real[cos(ωt + φ) + jA sin(ωt) + cos(ωt + ψ) + jA sin(ωt] =
= Real[ej φej ωt + ej ψej ωt] = Real[(ej φ+ ej ψ)ej ωt].
One can compute (ej φ+ ej ψ), determine the new phase angle, and retrieve the result. Phasor is a steady-state representation of electrical quantities and assumes the underlying frequency is constant (and the same).
Phasors
To handle linear operations with AC quantities, it is convenient to adopt the phasor notation. Phasor notation is derived from Euler’s formula:
Aej (ωt+θ) = A cos(ωt + θ) + jA sin(ωt + θ) = Aej θ
|{z}
Phasor
·ej ωt
The portion Aej θ can be used to perform calculations with sinusoidal waveforms considering magnitude, A, and phase,θ, information only. For example:
cos(ωt + φ) + cos(ωt + ψ) =
= Real[cos(ωt + φ) + jA sin(ωt) + cos(ωt + ψ) + jA sin(ωt] =
= Real[ej φej ωt+ ej ψej ωt] = Real[(ej φ+ ej ψ)ej ωt].
One can compute (ej φ+ ej ψ), determine the new phase angle, and retrieve the result. Phasor is a steady-state representation of electrical quantities and assumes the underlying frequency is constant (and the same).
Phasors
To handle linear operations with AC quantities, it is convenient to adopt the phasor notation. Phasor notation is derived from Euler’s formula:
Aej (ωt+θ) = A cos(ωt + θ) + jA sin(ωt + θ) = Aej θ
|{z}
Phasor
·ej ωt
The portion Aej θ can be used to perform calculations with sinusoidal waveforms considering magnitude, A, and phase,θ, information only. For example:
cos(ωt + φ) + cos(ωt + ψ) =
= Real[cos(ωt + φ) + jA sin(ωt) + cos(ωt + ψ) + jA sin(ωt] =
= Real[ej φej ωt+ ej ψej ωt] = Real[(ej φ+ ej ψ)ej ωt].
One can compute (ej φ+ ej ψ), determine the new phase angle, and retrieve the result.
Phasor is a steady-state representation of electrical quantities and assumes the underlying frequency is constant (and the same).
Phasors
To handle linear operations with AC quantities, it is convenient to adopt the phasor notation. Phasor notation is derived from Euler’s formula:
Aej (ωt+θ) = A cos(ωt + θ) + jA sin(ωt + θ) = Aej θ
|{z}
Phasor
·ej ωt
The portion Aej θ can be used to perform calculations with sinusoidal waveforms considering magnitude, A, and phase,θ, information only. For example:
cos(ωt + φ) + cos(ωt + ψ) =
= Real[cos(ωt + φ) + jA sin(ωt) + cos(ωt + ψ) + jA sin(ωt] =
= Real[ej φej ωt+ ej ψej ωt] = Real[(ej φ+ ej ψ)ej ωt].
One can compute (ej φ+ ej ψ), determine the new phase angle, and retrieve the result.
Phasor is a steady-state representation of electrical quantities and assumes the
Phasors (cont’d)
A numerical example:
3 cos(ωt) + cos(ωt − 100◦) In phasor notation
3ej 0◦+ 1e−j100◦ Converting from polar to rectangular coordinates yields:
3 cos(0◦) + j 3 sin(0◦) + cos(−100◦) + j sin(−100◦) =
=3 + cos(−100◦) + j sin(−100◦) = 2.83 − j0.98 Back to polar
(2.832+ 0.982)ej arctan(−0.98/2.83) = 2.99ej −19.1◦ times ej ωt and taking the real part:
2.99 cos(ωt − 19.1◦).
Complex power from voltage and current phasors
Given with the following voltage and current phasor:
V = v φ (1)
I = i ψ, (2)
the associated complex power, ˜S , is given by
VI∗= vi φ − ψ = (3)
= vi (cos(φ − ψ) + jsin(φ − ψ)) = (4)
= P + jQ := ˜S (5)
Phasors – Exercise
Calculate
cos(ωt) − cos(ωt − 120◦) (6)
Three-phase systems
Electricity is transmitted, delivered and distributed in a three-phase form.
I It is produced in such a way I More power with fewer conductors
I It creates a revolving magnetic field in generators
Power in three-phase balanced systems: Y-load
a
v6 0 n
v6 120 b v6 240
c
Figure: 3 lines feeding 3 phases of a balanced load.
I The line current is the same as the phase current.
I The line voltage is the phase voltage times√
3.
The total power of the load is:
P = 3VphIphcosφ
= 3
√3VLIL=√ 3VLIL.
Power in three-phase balanced systems: Y-load
a
v6 0 n
v6 120 b v6 240
c
Figure: 3 lines feeding 3 phases of a balanced load.
I The line current is the same as the phase current.
I The line voltage is the phase voltage times√
3.
The total power of the load is:
P = 3VphIphcosφ
= 3
√ VLIL=√ 3VLIL.
Power in three-phase balanced systems: ∆-load
a
v6 0 n
v6 120 b v6 240
c
I The line voltage is the phase voltage.
I The line current is NOT the same as the phase current. It is the phase current time√
3.
Power:
P = 3VLIphcosφ =
= 3VL IL
√3cosφ =√
3VLILcosφ.
The same load (ie, same impedances) absorb the same power irrespectively of the configuration of the three phases.
Power in three-phase balanced systems: ∆-load
a
v6 0 n
v6 120 b v6 240
c
I The line voltage is the phase voltage.
I The line current is NOT the same as the phase current. It is the phase current time√
3.
Power:
P = 3VLIphcosφ =
= 3VL IL
√3cosφ =√
3VLILcosφ.
The same load (ie, same impedances) absorb the same power irrespectively of the configuration of the three phases.
Power in three-phase balanced systems: ∆-load
a
v6 0 n
v6 120 b v6 240
c
I The line voltage is the phase voltage.
I The line current is NOT the same as the phase current. It is the phase current time√
3.
Power:
P = 3VLIphcosφ =
= 3VL IL
√3cosφ =√
3VLILcosφ.
The same load (ie, same impedances) absorb the same power irrespectively of the configuration of the three phases.
Active and reactive power transfer in a transmission line (
1)
← A sending and a receiving voltage source connected by an inductive line.
Power at the receiving end:
SR = PR + jQR = ERI∗=
= ER ES − ER jX
∗
= eR eScosδ + jeSsinδ − eR jX
∗
=
= eReS
X sinδ + j esercosδ − er2 X
.
In summary:
PR = eSeR X sinδ QR = esercosδ − er2
X Similarly, for the sending end:
PS = eReS X sinδ QS = es2− eSeRcosδ
X
Active and reactive power transfer in a transmission line (
1)
← A sending and a receiving voltage source connected by an inductive line.
Power at the receiving end:
SR = PR + jQR = ERI∗=
= ER ES − ER jX
∗
= eR eScosδ + jeSsinδ − eR jX
∗
=
= eReS
X sinδ + j esercosδ − er2 X
.
In summary:
PR = eSeR X sinδ QR = esercosδ − er2
X Similarly, for the sending end:
PS = eReS X sinδ QS = es2− eSeRcosδ
X
Active and reactive power transfer in a transmission line (cont’d)
PR = eSeR X sinδ QR = esercosδ − er2
X PS = eReS
X sinδ QS = es2− eSeRcosδ
X
For δ = 0:
PR = PS = 0 QR = eR(eS− eR)
X
For δ 6= 0 and eS = eR: PR = PS = e2
X sinδ QS = −QR = e2
X (1 − cosδ) =1 2Xi2
Active and reactive power transfer in a transmission line (cont’d)
PR = eSeR X sinδ QR = esercosδ − er2
X PS = eReS
X sinδ QS = es2− eSeRcosδ
X
For δ = 0:
PR = PS = 0 QR = eR(eS− eR)
X For δ 6= 0 and eS = eR:
PR = PS = e2 X sinδ QS = −QR = e2
X(1 − cosδ) = 1 2Xi2
Active and reactive power transfer in a transmission line – Conclusions
I Active power transfer depends mainly on the angle by which the sending source leads the receiving sink.
I Reactive power transfer depends mainly on voltage magnitudes; it is transmitted from the higher- to the lower-voltage end.
I Reactive power cannot be transmitted over long distance because this would require a large voltage gradient.
I In real lines with non-negligible resistance, reactive power causes active power losses too.