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Power in 1- φ AC Circuits
INTRODUCTION
Of the three circuit elements: resistors, inductors and capacitors, only resistors absorb energy from the source and dissipate this energy in the form of heat. The power dissipated by a resistor is given by:
PR = i^2 R
This power is always positive, since i^2 is always positive. Inductors and capacitors do not dissipate power but store energy in the magnetic and electrostatic fields respectively and release this energy back to the source. The energy stored or released by a magnetic and electrostatic field is given by:
2 2 W^1 Li Mag = 2 2 WEStat =^1 CV
POWER IN RESISTIVE CIRCUITS Consider the circuit shown in Figure 1. The supply voltage is sinusoidal and is given by: V = Vm sin ω t The current flowing in the circuit is given by:
i = VRm^ sin ω t
The instantaneous power is given by: P = i V and waveforms of V , i and P are drawn as functions of time and shown in Figure 1.
Figure 1
The power waveform in Figure 1 is fluctuating but its value is always positive. The average power dissipated by the resistor can be determined from the power waveform, whose frequency is twice that of the applied voltage. P = Vi
P T Vi dt
T av.
0 ∫
f (^) T ω = 2 π =^2 π ∴ ω T =^2^ π
= ∫
ωπ ωπ ω ω
2
0
Pav (^) 2 Vm sin t. Im sin t. dt
= ∫
ωπ πω^ ω
2
0
sin^2 .. 2 t dt P VmIm av
= × ∫ ( − )
ωπ πω^ ω
2
0
Pav V 2 m I 2 m 1 cos 2 t. dt
ω
π ω
ω π
ω
2 (^20)
sin 2 4
Pav = VmIm t − t
.^2
Pav = VmIm = Vm^ I^ m ω
π π
ω
av m m VrmsI rms
P V I.
Hence, the average power developed by a resistor is given by the product of the rms voltage and current. Since Vrms = IrmsR
Pav = Vrms. I (^) rms = Irms^2 R
given by the shaded area marked ‘+’. When the current is falling from its peak positive value to zero, the energy which was supplied to the field is now returned back to the source as the field collapses. This is represented by the shaded region marked ‘–‘. These are again two cycles of power for one cycle of voltage. The average power delivered to the circuit is given by:
P T Vi dt
T av.
ωπ ωπ ω ω
2
0
Pav (^) 2 Vm sin t. Im cos t. dt
ωπ πω ω^ ω
2
0
Pav Vm 2 I m sin t .cos t. dt
ωπ ω π
ω
2
0 2
sin 2. 2 dt P VmIm t av
2 2 [^ sin^2 ]^0
2 = − × 0 ω =
π Pav VmImπω^ ωt
Pav = 0
Hence the average power supplied to the inductance is zero and the inductor receives energy from the electrical source and returns this energy back to the source without dissipating any of this energy across it.
POWER IN A PURELY CAPACITIVE CIRCUIT When a sinusoidal voltage is applied to a pure capacitive circuit, the current through the circuit is given by:
X C
i = V 90 o
C
m X
V
I
∴ 90 o 2
C
m X
I V
∴ i = Im sin ( ω t + 90 o) = Im cos ωt
Waveforms of voltage, current and power are drawn in Figure 3.
Figure 3
The average power delivered to the capacitor is given by:
P T Vi dt
T av.
0 ∫
= ∫ ( )
ωπ ωπ ω ω
2
0
Pav (^) 2 Vm sin t. Im cos t. dt
= ∫
ωπ ω π
ω
2
0 2
sin 2. 2 dt P VmIm t av
2 2 cos 22 0
2 0
^ =
= ×
ωπ ω
ω π P VmImω^ t av Pav = 0
Therefore the average power supplied to the capacitor is zero. Energy is supplied to the electrostatic field when the capacitor is charged and this same quantity of energy is released from the field as the capacitor discharges and releases its energy back to the source. It must be noted that in the purely inductive and capacitive circuits, high rms values of current flow in the circuits but the average power delivered or work done is zero.
Therefore average power supplied to the R-L load over a cycle is given by:
2 cos. cos 2
0
T m m θ m m θ av
V I dt V I P = (^) T ∫ =
Pav = V 2 m.^ I^ m 2 cos θ
Pav = Vrms. Irms cos θ
The waveforms of voltage, current and power are drawn and shown in Figure 4.
Figure 4
The negative power in Figure 4 represents the energy returned from the magnetic field to the source and the positive power represents the energy supplied to the field. The difference between these areas represent the average power dissipated in the resistor.
The average power over a cycle is called the active power , measured in Watts. The product of the rms Voltage and rms Current is called the apparent power , measured in Volt-amperes (VA). The relationship between apparent and active power is given by:
rms rms
av V I cos θ = P = power factor
cos θ has a range between 1 and 0 for θ between 0 and 90º. When cos θ = 1, Pav = Papp And all the apparent power is converted to average power with no power being supplied and released from the field in a cycle of operation. When cos θ = 0, Pav = 0 and all the apparent power is supplied and and released to and from the magnetic field. For values of cos θ between 1 and 0, some of the apparent power is converted to average power loss in the resistor, while the remaining apparent power is stored and released in and from the magnetic field. If apparent power, VI = S Real or reactive power, VI cos θ = P Then the other side of the right angle triangle formed is named Q where Q = reactive power.
Figure 5: Power Triangle
∴ S^2 = P^2 + Q^2 where P = S cos θ = active power Q = S sin θ = reactive power
COMPLEX POWER
Given a voltage applied to a circuit in phasor form as:
V = V ∠ φ v r r
And the resulting current flowing in the circuit is given by:
I = I ∠ φ i
r r
These two phases are shown in Figure 6
Figure 6: Phasor Diagram
The phase angle between the voltage and current phases is given by:
φ = φi − φ v
The active power delivered is given by: P = VI cos φ The corresponding reactive power developed is given by: Q = VI sin φ The apparent power is given by: S = P + jQ S = VI cos φ + j VI sin φ S = VI (cos φ + j sin φ ) S = VIej^ φ^ = VIej (^ φi − φv^ ) S = VIejφ^ e − jφv^ )
S = Ve − j^ φIejφi S = V r^ * I r
but S = P + jQ
∴ Real ( S = V r* I r)= P
and Imaginary ( S = V r* I r)= Q
