Instantaneous Power
Let us consider the following circuit.
R
−
+
v
i
L
if v = Vm sin(ωt + θ), the steady state current is
i = Im sin(ωt + φ)
Vm
ωL
where Im = p
and φ = − tan−1
.
2
2
R
R + (ωL)
The instantaneous power delivered to any load is the product of
the instantaneous voltage across the load and the current through
it.
p(t) = v (t)i(t)
p(t) = Vm sin(ωt + θ)Im sin(ωt + φ)
Vm Im
(cos(θ − φ) − cos(2ωt + θ + φ))
2
Vm Im
Vm Im
p(t) =
cos(θ − φ) −
cos(2ωt + θ + φ)
2
2
First component is constant. Second component varies with time
but twice the supply frequency. Since the second term is cosine
Vm Im
whose average value is zero, the average power is
cos(θ − φ).
2
p(t) =
p, v , i
Vm
v
Im
i p
θ−φ
Figure: Voltage, current and power in RL circuit
ωt
Average Power
1
P = Vm Im cos(θ − φ)
2
Average power P is product of one half of maximum voltage,
maximum current and the cosine of the phase angle difference
between voltage and current. Its unit is watts (W).
1. Resistor: Vm and Im are in phase.
1
1
P = Vm Im cos(0) = Vm Im
2
2
2. Ideal Inductor : Vm leads Im by 90◦ .
1
P = Vm Im cos(90◦ ) = 0
2
3. Ideal Capacitor : Im leads Vm by 90◦ .
1
P = Vm Im cos(90◦ ) = 0
2
Root Mean Square (RMS) Value
RMS value of any periodic current is the dc current that delivers
the same average power to a resistor as periodic current.
i(t)
I
−
+
v (t)
1
P=
T
R
Z
V
R
T
i 2 (t)R
P = I 2R
0
Z
1 T 2
I R=
i (t)R
T 0
s
Z
1 T 2
i (t)
I =
T 0
2
Similarly
s
V =
1
T
T
Z
v 2 (t)
0
Average power is
1
P = Vm Im cos(θ − φ)
2
For sinusoids,
Vm
V =√ ,
2
Im
I =√
2
where V and I are rms voltage and current respectively.
∴ P = VI cos(θ − φ)
Apparent Power and Power Factor
If the voltage and the current were dc quantities, the power
delivered to a load is
|S| = VI
As dc values and RMS are equal,
|S| = VRMS IRMS
where |S| is called as the apparent power. Its unit is
volt-ampere (VA).
The ratio of real power (P) to apparent power is called as the
power factor (pf).
pf =
VI cos(θ − φ)
= cos(θ − φ)
VI
Since cos(θ − φ) can never be greater than unity, P ≤ |S|.
1. Resistor:
θ − φ = 0◦
pf = cos(0◦ ) = 1
2. Ideal Inductor:
θ − φ = 90◦
pf = cos(90◦ ) = 0
3. Ideal Capacitor:
θ − φ = −90◦
pf = cos(−90◦ ) = 0
I
I
RL Circuit: Let (θ − φ) = 60◦ . pf = cos(60◦ ) = 0.5.
RC Circuit: Let (θ − φ) = −60◦ . pf = cos(−60◦ ) = 0.5
In order to differentiate these two cases, power factor in RL
circuit is mentioned lagging pf and in RC circuit as leading
pf.
Complex Power
Let us define voltage (rms) and current (rms) in frequency domain.
V = V ∠θ,
I = I ∠φ
The complex power S is
S = VI∗
S = V ∠θ I ∠ − φ
= VI ∠(θ − φ)
S = VI cos(θ − φ) + VI sin(θ − φ)
The real part of S is called the average power (P). The imaginary
part of S is called reactive power (Q).
S = P + Q
The unit of reactive power (Q) is volt ampere reactive (VAr).
Im
S Q
(θ − φ)
P
Re
Figure: Power triangle for an inductive load
If the power triangle lies in the first quadrant ((θ − φ) > 0), power
factor is lagging (inductive load).
Im
P
Re
−(θ − φ)
S
Q
Figure: Power triangle for a capacitive load
If the power triangle lies in the fourth quadrant ((θ − φ) < 0),
power factor is leading (capacitive load).
Example
Find the average power, reactive power and power factor of the
circuit.
I ∠φ =
−
+
230 ∠0
(4 + 3)Ω
230∠0
= 46∠ − 36.87◦ A
4 + 3
P = VI cos(φ) = 230 × 46 × cos(−36.87◦ ) ≈ 8.5 kW
Q = VI sin(φ) = 230 × 46 × sin(−36.87◦ ) ≈ 6.3 kVAr
pf = cos(φ) = cos(−36.87◦ ) = 0.8 lagging