Power Introduction - MavDISK

v(t) = Vmsin(ωt + θv)
I(t) = Imsin(ωt + θi)
POWER
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
θv- θi = 0 o
P = VIcos0 o
= VI = I 2 R = V 2 /R
Power for Inductor
θv- θi = 90o because current lags voltage
P = VIcos(90 o ) = 0 watts
Q = VIsin(90 o ) = VI = I 2 ωL = V 2 /ωL vars
The average reactive power is zero.
Power for Capacitor
θv- θi = 90o because current leads voltage
P = VIcos(-90 o ) = 0 watts
Q = VIsin(-90 o ) = -VI = -I 2 (1/ωC) = -V 2 ωC vars
vars
The average reactive power is zero.
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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
P = VIcosθ
Q = VIsinθ
Resistor and inductor and voltage source (DRAW),
I = V∠0 o
R + jωL
VI = V( V )
(R 2 + (ωL) 2 ) 1/2 ∠tan-1 ωL/R)
VI = V( V ∠- tan-1 ωL/R)
(R 2 + (ωL) 2 ) 1/2
Remember that the phase angle of power is,
Phase angle of voltage minus phase angle of current.
S = V•I *
S = V( V ∠(-1)(-1) tan-1 ωL/R)
(R 2 + (ωL) 2 ) 1/2
= V 2 ∠tan -1 (ωL/R)
(R 2 + (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 = Ieff 2 (R + jωL)
= V 2 •(R 2 + (ωL) 2 ) 1/2 ∠ tan -1 (ωL/R)
(R 2 + (ωL) 2 ) 1/2 •(R 2 + (ωL) 2 ) 1/2
P = Ieff 2 R
Q = I 2 (jωL)
S = I 2 (R 2 + (ωL) 2 ) 1/2 ∠ tan -1 (ωL/R)
Resistor and capacitor and voltage source,
I = V∠0 o
R + 1/jωC
= V∠0 o
R - j/ωC
S = V•I *
= V∠0 o x V∠0 o
R + j/ωC
= V 2 ∠-tan -1 (1/ωRC)
(R 2 + (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 = I 2 (R + 1/jωC) = I 2 (R - j/ωC)
= V 2 •(R 2 + (1ωC) 2 ) 1/2 ∠- tan -1 (1/ωRC)
(R 2 + (1/ωC) 2 ) 1/2 •(R 2 + (1/ωC) 2 ) 1/2
P = I 2 R
Q = I 2 (-j/ωC)
S = I 2 (R 2 + (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
V2
1 2
j30
0
The voltage, V, across the resistor and inductor is,
V = (2∠30 o) ( 40 + j30) rms
S = V•I *
= (2∠30 o )( 40 + j30)( 2∠30 o ) *
= (2∠30 o )( 40 + j30)( 2∠-30 o )
= (4)( 40 + j30)
40
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Now the 30 degree angle which is common to both the voltage and
current has been eliminated from the calculation.