If the voltages are sinusoidal, they can be represented by phasors.
Thus, we can apply KVL directly to phasors.
Procedure for steady-state analysis of circuits with sinusoidal
sources:
Example 5.4: Steady-state AC analysis of a Series Circuit
Equivalent impedance of three series elements:
Phasor current:
Current as a function of time
Phasor voltage across each element:
Equivalent impedance of three series elements:
Phasor current:
Phasor diagram:
Current as a function of time
Phasor voltage across each element:
Example 5.5 : Series and Parallel combination of complex Impedance
Impedance of parallel RC:
Converting to rectangular form
Now, using voltage division principal:
Converting phasor to a time function:
Current in each element:
Phasor diagram:
Example 5.6: Steady-state AC Node-Voltage Analysis
-j1/(100 × 2000 × 10-6 )
= -j5
j100 × 0.1 = j10
Replace the time
description with
corresponding phasor
Applying KCL at node 1:
Applying KCL at node 2:
Equations in standard form:
Solving for V1
Converting phasor to a time function:
Power in AC circuit
A voltage source delivering power to a load impedance: Z= R+jX
Consider
And
The phasor of the current:
where
a. Power for a Pure Resistive load
Z=R
Current is in phase with the voltage
( both reach their peak values at the
same time)
The value of power rises and falls with
the voltage and current magnitude
Average power absorbed by R
P(t) is positive at all time: energy flows
continually in the direction form the
source to the load (converted to heat).
b. Power for a Pure Inductive load
90˚
Current lags the voltage by 90˚
Half of the time the power is positive, showing
the energy delivered to the inductance (where
it is stored in the magnetic filed)
In this case, we say that the reactive
power is positive and flows from the
source to the load
Half of the time the power is negative, showing
the inductance returns energy to the source
c. Power for a Pure Capacitive load
Z=
90˚
Current leads the voltage by 90˚
Power for the capacitance carries the
opposite sign as that for the
inductance. In this case, we say that
the reactive power is negative for the
capacitance
c. Power for a General RLC load
The phase θ can be any value form -90˚ to +90˚
Using identity:
Complex Power
The complex power, denoted as S delivered to the load impedance Z= R+jX
P is the power associated with the
resistive load (real part of impedance).
Q is called the reactive power and it is the power associated
with the energy storage elements (Inductive and Capacitive
loads : imaginary part of impedance).
Units : Watt
Units: The physical units of Q are watts. However
to emphasize the fact that Q does not represent
the flow of net energy*, it’s units are usually gives
as Volts Amperes Reactive (VARs)
* Energy flows in and out of energy storage elements, so although instantaneous power can be very large, the net energy
transferred per cycle is zero for ideal L and C
This term is called power factor (PF)
apparent power
Units : volt-amperes (VA)
Power Factor in general
Example 5.7 AC Power Calculations
Example 5.7 Solution
Example 5.7 Solution (cont.)
Example 5.7 Solution (cont.)
Reactive power (Q) for each element:
Reactive power for resistance is zero
QR = 0
Note that reactive power delivered by the source is equal to the sum of the reactive powers
absorbed by the L and C
Example 5.7 Solution (cont.)
Power (P) for each element:
Power in AC circuits
P5.63 Solution
Problem 5.66
Problem 5.66 – Solution
Problem 5.77
Problem 5.77 Solution