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POWER
v(t) =
I(t) =
Vmsin(ωt + θv)
Imsin(ωt + θi)
Rather than keep track of 2 angles we can use just 1,
v(t) =
Vmsin(ωt + θv- θi)
I(t) =
Imsin(ωt)
Let θ = θv- θi
And use rms values for V and I,
Now we use trig identities and arrange this to be more usefull,
p(t) =
VIcosθ - VIcosθcos2ωt + VIsinθsin2ωt
Point out,
1) 2x freg of v(t)
2) negative sometimes
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and
P
=
VIcosθ
watts
Q
VIsinθ
volt-amps reactive VARS
=
Power for Resistor
=
0o
θv- θi
P
=
VIcos0o
=
VI
=
I2R =
V2/R
Power for Inductor
θv- θi = 90o because current lags voltage
P
=
VIcos(90o) = 0 watts
Q
=
VIsin(90o) = VI = I2ωL
=
V2/ωL vars
The average reactive power is zero.
Power for Capacitor
θv- θi = 90o because current leads voltage
P
=
VIcos(-90o) = 0 watts
Q
=
vars
VIsin(-90o) = -VI =
-I2(1/ωC) =
The average reactive power is zero.
-V2ωC vars
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Power Factor
pf = cos(θv- θi) = cosθ
The pf is a number between 0 & 1.
We say the circuit is either leading or lagging.
Resistor and inductor
Current lags voltage
VARS are positive
Phase angle (θv - θi) is positive
Pf = …... lagging
Resistor and capacitor
Current leads voltage
VARS are negative
Phase angle is negative
Pf = ….. leading
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Complex Power
S
=
P + jQ
|S|=VI
Q = VIsinθ
P = VIcosθ
Resistor and inductor and voltage source (DRAW),
I
=
V∠0o
R + jωL
VI
V
)
= V(
2
2 1/2
(R + (ωL) ) ∠tan-1 ωL/R)
VI
= V( V ∠- tan-1 ωL/R)
(R2 + (ωL)2) 1/2
Remember that the phase angle of power is,
Phase angle of voltage minus phase angle of current.
V•I*
S
=
S
= V( V ∠(-1)(-1) tan-1 ωL/R)
(R2 + (ωL)2) 1/2
=
V2 ∠tan-1(ωL/R)
(R2 + (ωL)2)1/2
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Notice that the angle is positive and is the same result as
when currents are used in the calculation.
S
=
Ieff2(R + jωL)
=
V2 •(R2 + (ωL)2)1/2∠ tan-1(ωL/R)
(R2 + (ωL)2)1/2 •(R2 + (ωL)2)1/2
Q = I2(jωL)
P = Ieff2R
S
=
I2(R2 + (ωL)2)1/2∠ tan-1(ωL/R)
Resistor and capacitor and voltage source,
V∠0o
I
=
R + 1/jωC
S
=
V∠0o
R - j/ωC
=
V•I*
=
V∠0o x V∠0o
R + j/ωC
=
V2 ∠-tan-1(1/ωRC)
(R2 + (1/ωC)2)1/2
Notice that the angle is negative and is the same result as
when currents are used.
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S
=
I2(R + 1/jωC)
=
V2 •(R2 + (1ωC)2)1/2∠- tan-1(1/ωRC)
(R2 + (1/ωC)2)1/2 •(R2 + (1/ωC)2)1/2
=
I2(R - j/ωC)
P = I2R
Q = I2(-j/ωC)
S
=
I2(R2 + (1/ωC)2)1/2∠ -tan-1(1/ωRC)
Inductive circuit
Lagging power factor
Absorbing VARS
Capacitive circuit
Leading power factor
Generating VARS
More on the conjugate story,
S
=
P
+
jQ
S
=
V•I*
=
(V∠θv) • (I∠θi)*
=
(V• I∠(θv-θi)
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For example, a simple resistor-inductor circuit,
2amp rms at 30deg
1
2
j30
V2
40
0
The voltage, V, across the resistor and inductor is,
V = (2∠30o)( 40 + j30) rms
S
=
V•I*
=
(2∠30o)( 40 + j30)( 2∠30o)*
=
(2∠30o)( 40 + j30)( 2∠-30o)
=
(4)( 40 + j30)
Now the 30 degree angle which is common to both the voltage and
current has been eliminated from the calculation.