Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Components
Revision Lecture 2: Phasors
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 1 / 7
Basic Concepts
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
• Phasors and Complex impedances are only relevant to sinusoidal
sources.
◦ A DC source is a special case of a cosine wave with ω = 0.
Components
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 2 / 7
Basic Concepts
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Components
• Phasors and Complex impedances are only relevant to sinusoidal
sources.
◦ A DC source is a special case of a cosine wave with ω = 0.
• For two sine waves, the leading one reaches its peak first, the lagging
one reaches its peak second. So sin ωt lags cos ωt.
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 2 / 7
Basic Concepts
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Components
• Phasors and Complex impedances are only relevant to sinusoidal
sources.
◦ A DC source is a special case of a cosine wave with ω = 0.
• For two sine waves, the leading one reaches its peak first, the lagging
one reaches its peak second. So sin ωt lags cos ωt.
• If A cos (ωt + θ) = F cos ωt − G sin ωt, then
√
◦ A = F 2 + G2 , θ = tan−1 G
F.
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 2 / 7
Basic Concepts
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Components
• Phasors and Complex impedances are only relevant to sinusoidal
sources.
◦ A DC source is a special case of a cosine wave with ω = 0.
• For two sine waves, the leading one reaches its peak first, the lagging
one reaches its peak second. So sin ωt lags cos ωt.
• If A cos (ωt + θ) = F cos ωt − G sin ωt, then
√
◦ A = F 2 + G2 , θ = tan−1 G
F.
◦ F = A cos θ ,
E1.1 Analysis of Circuits (2014-4472)
G = A sin θ .
Revision Lecture 2 – 2 / 7
Basic Concepts
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Components
• Phasors and Complex impedances are only relevant to sinusoidal
sources.
◦ A DC source is a special case of a cosine wave with ω = 0.
• For two sine waves, the leading one reaches its peak first, the lagging
one reaches its peak second. So sin ωt lags cos ωt.
• If A cos (ωt + θ) = F cos ωt − G sin ωt, then
√
◦ A = F 2 + G2 , θ = tan−1 G
F.
◦ F = A cos θ ,
G = A sin θ .
◦ In CMPLX mode, Casio fx-991ES can do complex arithmetic and
can switch between the two forms with SHIFT,CMPLX,3 or
SHIFT,CMPLX,4
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 2 / 7
Reactive Components
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
1
jωC
=
−j
ωC .
1
1
,
R jωL
=
−j
ωL ,
• Impedances: R, jωL,
◦ Admittances:
jωC
Components
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 3 / 7
Reactive Components
Revision Lecture 2: Phasors
1
jωC
=
−j
ωC .
1
1
,
R jωL
=
−j
ωL ,
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
• Impedances: R, jωL,
Components
• In a capacitor or inductor, the Current and Voltage are 90◦ apart :
◦ Admittances:
jωC
◦ CIVIL: Capacitor - current leads voltage; Inductor - current lags
voltage
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 3 / 7
Reactive Components
Revision Lecture 2: Phasors
1
jωC
=
−j
ωC .
1
1
,
R jωL
=
−j
ωL ,
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
• Impedances: R, jωL,
Components
• In a capacitor or inductor, the Current and Voltage are 90◦ apart :
◦ Admittances:
jωC
◦ CIVIL: Capacitor - current leads voltage; Inductor - current lags
voltage
• Average current (or DC current) through a capacitor is always zero
• Average voltage across an inductor is always zero
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 3 / 7
Reactive Components
Revision Lecture 2: Phasors
1
jωC
=
−j
ωC .
1
1
,
R jωL
=
−j
ωL ,
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
• Impedances: R, jωL,
Components
• In a capacitor or inductor, the Current and Voltage are 90◦ apart :
◦ Admittances:
jωC
◦ CIVIL: Capacitor - current leads voltage; Inductor - current lags
voltage
• Average current (or DC current) through a capacitor is always zero
• Average voltage across an inductor is always zero
• Average power absorbed by a capacitor or inductor is always zero
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 3 / 7
Phasors
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
A phasor represents a time-varying sinusoidal waveform by a fixed complex
number.
Components
Waveform
Phasor
x(t) = F cos ωt − G sin ωt
X = F + jG
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 4 / 7
Phasors
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
A phasor represents a time-varying sinusoidal waveform by a fixed complex
number.
Components
Waveform
Phasor
x(t) = F cos ωt − G sin ωt
X = F + jG
E1.1 Analysis of Circuits (2014-4472)
[Note minus sign]
Revision Lecture 2 – 4 / 7
Phasors
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
A phasor represents a time-varying sinusoidal waveform by a fixed complex
number.
Components
Waveform
Phasor
x(t) = F cos ωt − G sin ωt
x(t) = A cos (ωt + θ)
X = F + jG
X = Aejθ = A∠θ
E1.1 Analysis of Circuits (2014-4472)
[Note minus sign]
Revision Lecture 2 – 4 / 7
Phasors
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
A phasor represents a time-varying sinusoidal waveform by a fixed complex
number.
Components
Waveform
Phasor
x(t) = F cos ωt − G sin ωt
x(t) = A cos (ωt + θ)
max (x(t)) = A
X = F + jG
X = Aejθ = A∠θ
|X| = A
E1.1 Analysis of Circuits (2014-4472)
[Note minus sign]
Revision Lecture 2 – 4 / 7
Phasors
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
A phasor represents a time-varying sinusoidal waveform by a fixed complex
number.
Components
Waveform
Phasor
x(t) = F cos ωt − G sin ωt
x(t) = A cos (ωt + θ)
max (x(t)) = A
X = F + jG
X = Aejθ = A∠θ
|X| = A
[Note minus sign]
x(t) is the projection of a rotating rod onto
the real (horizontal) axis.
X = F + jG is its starting position at t = 0.
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 4 / 7
Phasor Diagram
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Draw phasors as vectors. Join vectors end-to-end to show how voltages in
series add up (or currents in parallel).
Find y(t) if x(t) = cos 300t .
Components
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 5 / 7
Phasor Diagram
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Draw phasors as vectors. Join vectors end-to-end to show how voltages in
series add up (or currents in parallel).
Find y(t) if x(t) = cos 300t .
Components
Y
X
=
E1.1 Analysis of Circuits (2014-4472)
1
jωC
1
R+ jωC
=
1
jωRC+1
Revision Lecture 2 – 5 / 7
Phasor Diagram
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Draw phasors as vectors. Join vectors end-to-end to show how voltages in
series add up (or currents in parallel).
Find y(t) if x(t) = cos 300t .
Components
Y
X
1
jωC
1
R+ jωC
=
=
=
1
jωRC+1
1
1+3j
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 5 / 7
Phasor Diagram
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Draw phasors as vectors. Join vectors end-to-end to show how voltages in
series add up (or currents in parallel).
Find y(t) if x(t) = cos 300t .
Components
Y
X
1
jωC
1
R+ jωC
=
=
=
1
jωRC+1
1
1+3j
= 0.1 − 0.3j = 0.32∠ − 72◦
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 5 / 7
Phasor Diagram
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Draw phasors as vectors. Join vectors end-to-end to show how voltages in
series add up (or currents in parallel).
Find y(t) if x(t) = cos 300t .
Components
Y
X
1
jωC
1
R+ jωC
=
=
=
1
jωRC+1
1
1+3j
= 0.1 − 0.3j = 0.32∠ − 72◦
x(t) = cos 300t ⇒ X = 1
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 5 / 7
Phasor Diagram
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Draw phasors as vectors. Join vectors end-to-end to show how voltages in
series add up (or currents in parallel).
Find y(t) if x(t) = cos 300t .
Components
Y
X
1
jωC
1
R+ jωC
=
=
=
1
jωRC+1
1
1+3j
= 0.1 − 0.3j = 0.32∠ − 72◦
x(t) = cos 300t ⇒ X = 1
Y =X×
E1.1 Analysis of Circuits (2014-4472)
Y
X
Revision Lecture 2 – 5 / 7
Phasor Diagram
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Draw phasors as vectors. Join vectors end-to-end to show how voltages in
series add up (or currents in parallel).
Find y(t) if x(t) = cos 300t .
Components
Y
X
1
jωC
1
R+ jωC
=
=
=
1
jωRC+1
1
1+3j
= 0.1 − 0.3j = 0.32∠ − 72◦
x(t) = cos 300t ⇒ X = 1
Y =X×
Y
X
= 0.1 − 0.3j = 0.32∠ − 72◦
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 5 / 7
Phasor Diagram
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Draw phasors as vectors. Join vectors end-to-end to show how voltages in
series add up (or currents in parallel).
Find y(t) if x(t) = cos 300t .
Components
Y
X
1
jωC
1
R+ jωC
=
=
=
1
jωRC+1
1
1+3j
= 0.1 − 0.3j = 0.32∠ − 72◦
x(t) = cos 300t ⇒ X = 1
Y =X×
Y
X
0
-0.2
-0.4
0
0.5
Real
1
= 0.1 − 0.3j = 0.32∠ − 72◦
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 5 / 7
Phasor Diagram
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Draw phasors as vectors. Join vectors end-to-end to show how voltages in
series add up (or currents in parallel).
Find y(t) if x(t) = cos 300t .
Components
Y
X
1
jωC
1
R+ jωC
=
=
=
1
jωRC+1
1
1+3j
= 0.1 − 0.3j = 0.32∠ − 72◦
x(t) = cos 300t ⇒ X = 1
Y =X×
Y
X
0
-0.2
-0.4
0
0.5
Real
1
= 0.1 − 0.3j = 0.32∠ − 72◦
y(t) = 0.1 cos 300t + 0.3 sin 300t
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 5 / 7
Phasor Diagram
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Draw phasors as vectors. Join vectors end-to-end to show how voltages in
series add up (or currents in parallel).
Find y(t) if x(t) = cos 300t .
Components
Y
X
1
jωC
1
R+ jωC
=
=
=
1
jωRC+1
1
1+3j
= 0.1 − 0.3j = 0.32∠ − 72◦
x(t) = cos 300t ⇒ X = 1
Y =X×
Y
X
0
-0.2
-0.4
0
0.5
Real
1
= 0.1 − 0.3j = 0.32∠ − 72◦
y(t) = 0.1 cos 300t + 0.3 sin 300t
= 0.32 cos (300t − 1.25)
= 0.32 cos (300 (t − 4.2ms))
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 5 / 7
Phasor Diagram
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Draw phasors as vectors. Join vectors end-to-end to show how voltages in
series add up (or currents in parallel).
Find y(t) if x(t) = cos 300t .
Components
Y
X
1
jωC
1
R+ jωC
=
=
=
1
jωRC+1
1
1+3j
= 0.1 − 0.3j = 0.32∠ − 72◦
0
-0.2
x(t) = cos 300t ⇒ X = 1
Y =X×
-0.4
Y
X
0
= 0.1 − 0.3j = 0.32∠ − 72◦
= 0.32 cos (300t − 1.25)
= 0.32 cos (300 (t − 4.2ms))
E1.1 Analysis of Circuits (2014-4472)
1
w = 300 rad/s, Gain = 0.32, Phase = -72°
1
x=blue, y=red
y(t) = 0.1 cos 300t + 0.3 sin 300t
0.5
Real
x
0.5
y
0
-0.5
-1
0
20
40
60
80
time (ms)
100
120
Revision Lecture 2 – 5 / 7
Complex Power
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
e = √1 × V to be the
If V = |V | ∠θV is a phasor, we define V
e
corresponding r.m.s. phasor. The r.m.s. voltage is V
=
2
√1
2
× |V |.
Components
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 6 / 7
Complex Power
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Components
e = √1 × V to be the
If V = |V | ∠θV is a phasor, we define V
e
corresponding r.m.s. phasor. The r.m.s. voltage is V
=
2
Power Factor
E1.1 Analysis of Circuits (2014-4472)
√1
2
× |V |.
cos φ = cos (θV − θI )
Revision Lecture 2 – 6 / 7
Complex Power
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Components
e = √1 × V to be the
If V = |V | ∠θV is a phasor, we define V
e
corresponding r.m.s. phasor. The r.m.s. voltage is V
=
2
Power Factor
Complex Power
E1.1 Analysis of Circuits (2014-4472)
cos φ = cos (θV − θI )
S = Ve × Ie∗ = P + jQ
√1
2
× |V |.
[∗ = complex conjugate]
Revision Lecture 2 – 6 / 7
Complex Power
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Components
e = √1 × V to be the
If V = |V | ∠θV is a phasor, we define V
e
corresponding r.m.s. phasor. The r.m.s. voltage is V
=
2
Power Factor
Complex Power
Apparent Power
E1.1 Analysis of Circuits (2014-4472)
cos φ = cos (θV − θI )
S = Ve ×Ie∗ = P + jQ
|S| = Ve × Ie
√1
2
× |V |.
[∗ = complex conjugate]
Revision Lecture 2 – 6 / 7
Complex Power
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Components
e = √1 × V to be the
If V = |V | ∠θV is a phasor, we define V
e
corresponding r.m.s. phasor. The r.m.s. voltage is V
=
2
Power Factor
Complex Power
Apparent Power
Average Power
E1.1 Analysis of Circuits (2014-4472)
cos φ = cos (θV − θI )
S = Ve ×Ie∗ = P + jQ
|S| = Ve × Ie
P = ℜ (S) = |S| cos φ
√1
2
× |V |.
[∗ = complex conjugate]
unit = Watts → heat
Revision Lecture 2 – 6 / 7
Complex Power
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Components
e = √1 × V to be the
If V = |V | ∠θV is a phasor, we define V
e
corresponding r.m.s. phasor. The r.m.s. voltage is V
=
2
Power Factor
Complex Power
Apparent Power
Average Power
Reactive Power
E1.1 Analysis of Circuits (2014-4472)
cos φ = cos (θV − θI )
S = Ve ×Ie∗ = P + jQ
|S| = Ve × Ie
P = ℜ (S) = |S| cos φ
Q = ℑ (S) = |S| sin φ
√1
2
× |V |.
[∗ = complex conjugate]
unit = VAs
unit = Watts → heat
unit = VARs
Revision Lecture 2 – 6 / 7
Complex Power
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Components
e = √1 × V to be the
If V = |V | ∠θV is a phasor, we define V
e
corresponding r.m.s. phasor. The r.m.s. voltage is V
=
2
Power Factor
Complex Power
Apparent Power
Average Power
Reactive Power
cos φ = cos (θV − θI )
S = Ve ×Ie∗ = P + jQ
|S| = Ve × Ie
P = ℜ (S) = |S| cos φ
Q = ℑ (S) = |S| sin φ
√1
2
× |V |.
[∗ = complex conjugate]
unit = VAs
unit = Watts → heat
unit = VARs
Conservation of power (Tellegen’s theorem): in any circuit the total complex
power absorbed by all components sums to zero
⇒ P = average power and Q = reactive power sum separately to zero.
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 6 / 7
Complex Power
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Components
e = √1 × V to be the
If V = |V | ∠θV is a phasor, we define V
e
corresponding r.m.s. phasor. The r.m.s. voltage is V
=
2
Power Factor
Complex Power
Apparent Power
Average Power
Reactive Power
cos φ = cos (θV − θI )
S = Ve ×Ie∗ = P + jQ
|S| = Ve × Ie
P = ℜ (S) = |S| cos φ
Q = ℑ (S) = |S| sin φ
√1
2
× |V |.
[∗ = complex conjugate]
unit = VAs
unit = Watts → heat
unit = VARs
Conservation of power (Tellegen’s theorem): in any circuit the total complex
power absorbed by all components sums to zero
⇒ P = average power and Q = reactive power sum separately to zero.
VARs are generated by capacitors and absorbed by inductors.
φ > 0 for inductive impedance.
φ < 0 for capacitive impedance.
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 6 / 7
Complex Power in Components
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
S = Ve Ie∗ =
|Ve |
2
Z∗
2
= Ie Z
Components
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 7 / 7
Complex Power in Components
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
S = Ve Ie∗ =
|Ve |
2
Z∗
2
= Ie Z ⇒ ∠S = ∠Z
Components
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 7 / 7
Complex Power in Components
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Components
S = Ve Ie∗ =
|Ve |
2
Z∗
2
= Ie Z ⇒ ∠S = ∠Z
Resistor: positive real (absorbs watts)
Inductor: positive imaginary (absorbs VARs)
Capacitor: negative imaginary (generates VARs)
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 7 / 7
Complex Power in Components
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Components
S = Ve Ie∗ =
|Ve |
2
Z∗
2
= Ie Z ⇒ ∠S = ∠Z
Resistor: positive real (absorbs watts)
Inductor: positive imaginary (absorbs VARs)
Capacitor: negative imaginary (generates VARs)
Power Factor Correction
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 7 / 7
Complex Power in Components
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Components
S = Ve Ie∗ =
|Ve |
2
Z∗
2
= Ie Z ⇒ ∠S = ∠Z
Resistor: positive real (absorbs watts)
Inductor: positive imaginary (absorbs VARs)
Capacitor: negative imaginary (generates VARs)
Power Factor Correction
Ve = 230. Motor is 5||7j Ω.
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 7 / 7
Complex Power in Components
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Components
S = Ve Ie∗ =
|Ve |
2
Z∗
2
= Ie Z ⇒ ∠S = ∠Z
Resistor: positive real (absorbs watts)
Inductor: positive imaginary (absorbs VARs)
Capacitor: negative imaginary (generates VARs)
Power Factor Correction
Ve = 230. Motor is 5||7j Ω.
Ie = 46 − 33j = 56.5∠ − 36◦
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 7 / 7
Complex Power in Components
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Components
S = Ve Ie∗ =
|Ve |
2
Z∗
2
= Ie Z ⇒ ∠S = ∠Z
Resistor: positive real (absorbs watts)
Inductor: positive imaginary (absorbs VARs)
Capacitor: negative imaginary (generates VARs)
Power Factor Correction
Ve = 230. Motor is 5||7j Ω.
Ie = 46 − 33j = 56.5∠ − 36◦
S = Ve Ie∗ = 13∠36◦ kVA
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 7 / 7
Complex Power in Components
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Components
S = Ve Ie∗ =
|Ve |
2
Z∗
2
= Ie Z ⇒ ∠S = ∠Z
Resistor: positive real (absorbs watts)
Inductor: positive imaginary (absorbs VARs)
Capacitor: negative imaginary (generates VARs)
Power Factor Correction
Ve = 230. Motor is 5||7j Ω.
Ie = 46 − 33j = 56.5∠ − 36◦
S = Ve Ie∗ = 13∠36◦ kVA
P
cos φ = |S|
= cos 36◦ = 0.81
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 7 / 7
Complex Power in Components
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Components
S = Ve Ie∗ =
|Ve |
2
Z∗
2
= Ie Z ⇒ ∠S = ∠Z
Resistor: positive real (absorbs watts)
Inductor: positive imaginary (absorbs VARs)
Capacitor: negative imaginary (generates VARs)
Power Factor Correction
Ve = 230. Motor is 5||7j Ω.
Ie = 46 − 33j = 56.5∠ − 36◦
S = Ve Ie∗ = 13∠36◦ kVA
P
cos φ = |S|
= cos 36◦ = 0.81
Add parallel capacitor: 300 µF
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 7 / 7
Complex Power in Components
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Components
S = Ve Ie∗ =
|Ve |
2
Z∗
2
= Ie Z ⇒ ∠S = ∠Z
Resistor: positive real (absorbs watts)
Inductor: positive imaginary (absorbs VARs)
Capacitor: negative imaginary (generates VARs)
Power Factor Correction
Ve = 230. Motor is 5||7j Ω.
Ie = 46 − 33j = 56.5∠ − 36◦
S = Ve Ie∗ = 13∠36◦ kVA
P
cos φ = |S|
= cos 36◦ = 0.81
Add parallel capacitor: 300 µF
Ie = 46 − 11j = 47∠ − 14◦
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 7 / 7
Complex Power in Components
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Components
S = Ve Ie∗ =
|Ve |
2
Z∗
2
= Ie Z ⇒ ∠S = ∠Z
Resistor: positive real (absorbs watts)
Inductor: positive imaginary (absorbs VARs)
Capacitor: negative imaginary (generates VARs)
Power Factor Correction
Ve = 230. Motor is 5||7j Ω.
Ie = 46 − 33j = 56.5∠ − 36◦
S = Ve Ie∗ = 13∠36◦ kVA
P
cos φ = |S|
= cos 36◦ = 0.81
Add parallel capacitor: 300 µF
Ie = 46 − 11j = 47∠ − 14◦
S = Ve Ie∗ = 10.9∠14◦ kVA
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 7 / 7
Complex Power in Components
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Components
S = Ve Ie∗ =
|Ve |
2
Z∗
2
= Ie Z ⇒ ∠S = ∠Z
Resistor: positive real (absorbs watts)
Inductor: positive imaginary (absorbs VARs)
Capacitor: negative imaginary (generates VARs)
Power Factor Correction
Ve = 230. Motor is 5||7j Ω.
Ie = 46 − 33j = 56.5∠ − 36◦
S = Ve Ie∗ = 13∠36◦ kVA
P
cos φ = |S|
= cos 36◦ = 0.81
Add parallel capacitor: 300 µF
Ie = 46 − 11j = 47∠ − 14◦
S = Ve Ie∗ = 10.9∠14◦ kVA
P
cos φ = |S|
= cos 14◦ = 0.97
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 7 / 7
Complex Power in Components
Revision Lecture 2: Phasors
• Basic Concepts
• Reactive Components
• Phasors
• Phasor Diagram
• Complex Power
• Complex Power in
Components
S = Ve Ie∗ =
|Ve |
2
Z∗
2
= Ie Z ⇒ ∠S = ∠Z
Resistor: positive real (absorbs watts)
Inductor: positive imaginary (absorbs VARs)
Capacitor: negative imaginary (generates VARs)
Power Factor Correction
Ve = 230. Motor is 5||7j Ω.
Ie = 46 − 33j = 56.5∠ − 36◦
S = Ve Ie∗ = 13∠36◦ kVA
P
cos φ = |S|
= cos 36◦ = 0.81
Add parallel capacitor: 300 µF
Ie = 46 − 11j = 47∠ − 14◦
S = Ve Ie∗ = 10.9∠14◦ kVA
P
cos φ = |S|
= cos 14◦ = 0.97
0.81
. Lower power transmission losses.
Current decreases by factor of 0.97
E1.1 Analysis of Circuits (2014-4472)
Revision Lecture 2 – 7 / 7