Power In AC Circuits

Power In AC Circuits

        When an AC (sinusoidal) voltage is applied to an impedance, both the current and the voltage are sinusoidal function of time.  That presents problems when calculating power because the power fluctuates as the voltage and the current change.

        Consider the case of a resistor.  The current and the voltage are given by:

v(t) = Vmax cos(wt)

and

i(t) = Imax cos(wt) = (Vmax/R)cos(wt)

and the power is given by:

P(t) = (Vmax cos(wt))( Imax cos(wt))

We can manipulate the power time function to see things a little more clearly:

P(t) = (Vmax  Imax cos2(wt))

P(t) = (Vmax  Imax )(1 - cos(2wt))/2

This is an interesting result, because it really says that the average power is:

Pavg = (Vmax  Imax )/2

You can convince yourself that this is true if you consider that the average value of the cosine function over a period is zero.  What's left is the result above.  In the early days of municipal electrical distribution systems it was necessary to design light bulbs that worked on both AC and DC.  A 100 watt bulb that worked on a DC system had to work on an AC system as well.  There were many other reasons to worry about equivlance between AC and DC systems, so there had to be some way of making things equivalent - determining an equivalent AC voltage that produces the same power in a resistor as that value of DC voltage.  The convention since that time has often been to represent AC signals in terms of a heating equivalent.  For sinusoidal signals that equivalent is given by:

Vequiv = Vmax /(SQRT(2)

The effective value is then the peak value of the sine voltage divided by the square root of 2.

        Now, what happens if we have a general impedance, Z, instead of a resistor R.  Assume we have an impedance:

Z = |Z|/f
Then, we would have:

v(t) = Vmax cos(wt)

and:
i(t) = (Vmax /|Z|)cos(wt - f)

So, now the power - as a function of time - is given by:

P(t) = (Vmax cos(wt))( (Vmax /|Z|)cos(wt - f))

P(t) = (V2max/|Z|)cos(wt)cos(wt - f))

Example

        Here is a circuit with the following parameters.

The current and voltage are plotted in the plot below.  In this plot the red trace is the voltage and the blue trace is the current.  The black trace is the instantaneous power - which we have to talk more about.

        First, note that the voltage starts at zero, and has the look of a typical sine wave.  The current begins at a negative value in the plot, and peaks 45o after the voltage.  The peak value of the voltage is 10 v, and the peak value of the current is 5 A.

        Now, examine the power.  There are a number of interesting points in the plot of the instantaneous power.


 
 
 
 

        When a load is connected to a Thevinin Equivalent Circuit (TEC), two things happen - current flows from the source to the load, and the voltage across the output terminals of the TEC usually drops.  In the process, the load consumes energy.  There are many situations in which the rate of energy consumption - the power delivered to the load - needs to be as large as possible.  In this section, we will examine the conditions under which that will occur.
        Consider a TEC with an attached load as shown in the figure at the right.  The load draws power from the source.  The source is represented with a TEC that has an open circuit voltage phasor, V0, and an output impedance, Z0.   When you have a configuration like this, one question that arises is this:

How do you determine the load impedance, ZL, which will draw the most power from the source?

In order to answer this question, you need to account for the fact that the load impedance and the source impedance may both be complex.

        We will break both the output impedance, and the load impedance into real and imaginary parts.

Z0 = Z0,r + j Z0,i

ZL = ZL,r + j ZL,i

Now, the current flowing in the circuit is given by:

IL = V0/( Z0 + ZL)

and the power delivered to the load is given by:
UNDER CONSTRUCTION