POWER IN AC CIRCUITS
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There
are several different definitions of power in AC circuits; all, however, have
dimension of V*A or W (watts).
1.
Instantaneous
power: p(t)
is the time function of the power, p(t)
= u(t)*i(t). It is the product of the time functions of the voltage and current.
This definition of instantaneous power is valid for signals of any waveform. The
unit for instantaneous power is VA.
2.
Complex
power: S
Complex
power is the product of the complex effective voltage and the complex effective
conjugate current. In our notation here, the conjugate is indicated by an
asterisk (*).Complex power can also be computed using the peak values of the
complex voltage and current, but then the result must be divided by 2. Note that
complex power is applicable only to circuits with sinusoidal excitation because
complex effective or peak values exist and are defined only for sinusoidal
signals. The unit for complex power is
VA.
3. Real or average power: P can be defined in two ways: as the real part of the complex power or as the simple average of the instantaneous power. The second definition is more general because with it we can define the instantaneous power for any signal waveform, not just for sinusoids. It is given explicitly in the following expression
The unit for real or average power
is watts (W), just as for power in DC circuits. Real power is dissipated
as heat in resistances.
4.
Reactive
power: Q
is
the imaginary part of the complex power. It is given in units of volt-amperes
reactive (VAR). Reactive power is positive
in an inductive circuit and negative in a capacitive
circuit. This power is defined only for sinusoidal excitation. The
reactive power doesn't do any useful work or heat and it
is the
power returned to the source by the reactive components (inductors, capacitors)
of the circuit
5.
Apparent
power: S
is
the product of the rms values of the voltage and the current, S = U*I. The unit
of apparent power is VA. The apparent
power is the absolute value of the complex
power, so it is defined only for sinusoidal excitation.
Power Factor
(cos
φ)
The power factor is very important in power systems because it indicates how
closely the effective power equals the apparent power. Power factors near one
are desirable. The definition:
TINA’s
power-measuring instrument also measures the power factor.
In
our first example, we calculate the powers in a simple circuit.
Example
1
Find
the average (dissipated) and reactive powers of the resistor and the capacitor.
Find the average and reactive powers provided by the source.
Check
to see if the powers provided by the source equal those in the components.
First
calculate the network current.
=
3.9 ej38.7B°
mA
Where
you see division by 2, remember that where the peak value is used for the source
voltage and the power definition, the power calculation requires the rms value.
Checking
the results, you can see that the sum of all three powers is zero, confirming
that the power from the source appears at the two components.
The
instantaneous power of the voltage source:
pV(t)
= -vS(t)*i(t) = -10 cos ωt
* 3.9 cos( ωt+38.7°
) =
Next,
we demonstrate how easy it is to obtain these results using a schematic and
instruments in TINA. Note that in the TINA schematics we use TINA’s jumpers to
connect the power meters.
You
can obtain the above tables by selecting Analysis/AC Analysis/Calculate nodal
voltages from the menu and then clicking the power meters with the probe.
We
can conveniently determine the apparent power of the voltage source using
TINA’s Interpreter:
S = VS*I = 10*3.9/2 = 19.5 VA
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{Solution
by TINA's Interpreter} |
You can see that there are ways other than the definitions themselves to
calculate the power in two-pole networks. The following table summarizes this:
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P
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Q
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S
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Z = R + jX
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R*I2 |
X*I2 |
½Z½*I2 |
Z*I2 |
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Y
= G + jB |
G*V2 |
-B*V2 |
½Y½*V2 |
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In
this table, we have rows for circuits characterized by either their impedance or
their admittance. Be careful using the formulas. When considering the impedance
form, think of the impedance as
representing a series circuit, for
which you need the current. When considering the admittance form, think of the admittance as
representing a parallel circuit, for
which you need the voltage. And don’t forget that although Y = 1/Z, in general
G ≠ 1/R. Except for the special case X = 0 (pure resistance), G = R/( R2
+ X2 ).
Example 2
Find the average power, the reactive power, p(t), and the power factor of the
two-pole network connected to the current source.
iS(t)=(100 * cosω t)mA w = 1 krad/s
Refer
to the table above and, since the two-pole network is a parallel circuit, use
the equations in the row for the admittance case.
Working
with an admittance, we must first find the admittance itself. Fortunately, our
two-pole network is a purely parallel one.
Yeq
= 1/R + jωC
+ 1/jωL
= 1/5 + j250*10-6103 + 1/(j*20*10-3103)
= 0.2+j0.2 S
We
need the absolute value of the voltage:
½V½
=½
Z½*I
= I/½Y½
= 0.1/ ê(0.2+j0.2)
ê=
0.3535 V
Q = -V2*B = - 0.125*0.2/2 = - 0.0125 var
= V2*
=0.125* (0.2-j0.2)/2 = ( 12.5 - j 12.5 ) mVA
S
= V2* Y = 0.125*ê0.2+j0.2ê/2
= 0.01768 VA
cos
φ
= P/S = 0.707
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{Solution
by TINA's Interpreter} om:=1000; Is:=0.1; V:=Is*(1/(1/R+j*om*C+1/(j*om*L))); V=[250m-250m*j] S:=V*Is/2; S=[12.5m-12.5m*j] P:=Re(S); Q:=Im(S); P=[12.5m] Q=[-12.5m] abs(S)=[17.6777m] |
Example 3
Find the average and reactive powers of the two-pole network connected to the voltage generator.
For
this example, we will dispense with manual solutions and show how to use
TINA’s measuring instruments and Interpreter to obtain the answers.
Selec
Analysis/AC Analysis/Calculate nodal voltages from the menu and then click the
power meter with the probe. The following table will appear:
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{Solution
by TINA's Interpreter!} |
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