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```AC power through resistor
10.0
1.0
5.0
0.5
Current, A
Voltage, V
Example: i = IM cos(ωt); IM=1 A; R = 10 Ω v = VM cos(ωt); VM = IM*R = 10 V
0.0
-5.0
0.0
-0.5
-10.0
-1.0
0
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12
0
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10
12
10
The power dissipated in the resistor:
At any point of time, p(t) = v(t)×i(t) =
VM cos (ωt)× IM cos (ωt) = VMIM cos2 (ωt)
Power, W
5
0
-5
p(t) is always positive
-10
0
2
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Instantaneous and Average (“rms”) powers
RMS = “root mean square” or the square root of the arithmetic mean (average)
10
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Power, W
5
Power, W
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-5
-10
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Instantaneous power:
p(t) = v(t)×i(t) = VMIM cos2 (ωt)
In terms of Joule heat dissipation,
instantaneous power is equivalent
to a certain DC power (i.e. timeindependent).
The equivalent DC power is called
Average or “rms” power
Average (“rms”) power dissipated in resistor
10
10
5
Power, W
Power, W
5
0
-5
-10
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VM I M
2
-5
-10
0
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p(t) = v(t)×i(t) = VMIM cos2 (ωt)
10
cos2 (ωt) = (1/2) [1+ cos(2 ωt)]
VM I M VM I M
+
cos (2ωt )
2
2
Prms = VMIM /2
Power, W
p (t ) =
5
0
-5
VM I M
cos ( 2ωt )
2
-10
0
Veff =VM / 2
I eff = I M / 2
2
4
6
8
10
12
Average (RMS) characteristics of sine waves
The sine voltage source with the voltage amplitude VM produces the same power as
the DC battery with the voltage VDC:
VDC =VM / 2 = 0.707VM
The equivalent DC voltage (producing the same Joule heat as the AC voltage) is called
RMS voltage (Root Mean Square) or effective value
Average (RMS) characteristics of sine waves
VRMS = VM
2 = 0.707VM
What is the amplitude of the AC source above?
VM = VRMS /0.707 = 10V/0.707 = 14.14 V
AC powers in R-L-C circuits
There is a phase angle between the current and voltage:
i = IM cos(ωt); v = VM cos(ωt + ϕ);
The instantaneous power:
p(t) = v(t)×i(t) = VM cos (ωt+ϕ) × IM cos (ωt) =
VMIM cos (ωt+ϕ) cos (ωt)
cos(a) cos(b) = (1/2) [cos(a+b) + cos(a-b)]
a = ωt+ϕ; b = ωt;
p (t ) =
VM I M
V I
cos (ϕ ) + M M cos ( 2ωt + ϕ )
2
2
cos(a+b) = cos(a) cos(b) - sin(a) sin(b)
a = ωt+ϕ; b = ωt;
cos(2ωt+ϕ) =
cos(2ωt) cos(ϕ) - sin(2ωt) sin(ϕ)
VM I M
VM I M
p (t ) =
cos (ϕ ) ⋅ [1 + cos(2ωt ) ] −
sin(ϕ )sin ( 2ωt )
2
2
AC powers in R-L-C circuits
In R-L-C circuits, the current and voltage may both have a phase angle
with respect to the source signal:
i = IM cos(ωt + ϕi); v = VM cos(ωt + ϕv);
We can offset the origin of the current and voltage waveforms by ϕi:
i = IM cos(ωt); v = VM cos(ωt + ϕv - ϕi) = VM cos(ωt + ϕ);
where ϕ = ϕv – ϕi; we can now use the previous expressions derived for ϕi = 0
p (t ) =
VM I M
V I
cos (ϕvi ) ⋅ [1 + cos(2ωt )] − M M sin(ϕvi ) sin ( 2ωt )
2
2
where ϕvi = ϕv – ϕi is the phase angle
between the voltage and the current
Active (average) and reactive powers in R-L-C circuits
p (t ) =
VM I M
V I
cos (ϕvi ) ⋅ [1 + cos(2ωt )] − M M sin(ϕvi ) sin ( 2ωt )
2
2
10
Power, arb. units
Power, arb.units
10
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0
VM I M
cos(ϕvi )
2
-5
-10
0
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-5
VM I M
cos(ϕvi ) cos ( 2ωt )
2
-10
0
2
4
Active (average) power:
VM I M
p A (t ) = P =
cos (ϕvi ) = Veff I eff cos (ϕvi )
2
6
8
10
12
Active (average) and reactive powers in R-L-C circuits
p (t ) =
VM I M
V I
cos (ϕvi ) ⋅ [1 + cos(2ωt )] − M M sin(ϕvi ) sin ( 2ωt )
2
2
Power, arb. units
10
5
0
-5
VM I M
sin(ϕ )sin ( 2ωt )
2
-10
0
2
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10
12
Reactive power:
pR (t ) = Q =
Average reactive power = 0
VM I M
sin(ϕvi ) = Veff I eff sin(ϕvi )
2
Active (average) and reactive powers in R-L-C circuits
p (t ) = P ⋅ [1 + cos(2ωt ) ] − Q sin(ϕvi ) sin ( 2ωt )
P=
VM I M
cos (ϕvi )
2
Q=
Examples:
1. Resistive circuit: ϕvi = 0;
P = VMIM/2; Q = 0;
2. Inductance: ϕvi = -π/2;
P = 0; Q = -VMIM/2;
2. Capacitance: ϕvi = π/2;
P = 0; Q = +VMIM/2;
VM I M
sin(ϕvi )
2
AC powers and complex amplitudes
Given the complex current amplitude:
and the complex voltage amplitude:
IˆM = I M e jϕi
VˆM =VM e jϕv
Complex Power is defined as
*
S =Vˆeff ⋅ Iˆeff
where
Veff =VM / 2; I eff = I M / 2
Iˆ *eff = I eff e − jφi is a complex conjugate of the Ieff
VM I M j (ϕv −ϕi ) VM I M
VM I M
*
ˆ
ˆ
S = Veff ⋅ I eff =
e
=
cos(ϕvi ) + j
sin(ϕvi )
2
2
2
Re(S) = P
Im(S) = Q
```
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