Lesson 8
Power in AC circuits
Learn to solve average power and complex power problems in AC. Learn how you can use RMS values in Symbulator with the rms flag. Solve power factor problems using the pf tool.
Last updated 2023-07-08
In this lesson you will learn how to run more advanced alternating current simulations. You will learn how to solve average power problems, when and how to use the rms flag, and how to solve complex power problems.
8.1Average power
AS7's Example 11.5
Open in app ↗Open in split view
Determine the load impedance ZL that maximises the average power drawn from the circuit. What is the maximum average power?
AS7's Example 11.5
Solution
Symbulator's th tool helps us find all the answers we need. Notice this circuit description does not include the load, because it is not needed as input to th.
e,1,0,10
r1,1,2,4
r2,2,0,8-6j
r3,2,3,5jFind equivalent, Thévenin / Norton, nodes 3 and 0, in AC. In AC the equivalent impedance is called zeq rather than req.
Thévenin / Norton saves the equivalent impedance in zeq, which evaluates approximately to 2.933 + j4.467. The load that will deliver the maximum power is its conjugate, 2.933 − j4.467.
The average maximum power delivered by the circuit is in pmax, which evaluates to 2.3674 W. There is no separate apmax in Symbulator 9: pmax looks at the domain you asked for, and in the AC domain it is already the average maximum power, computed from the real part of the equivalent impedance.
Now, here's the cool part. Symbulator 9 carries no load expressions on the result — the tool reports vth, ino, zeq and pmax, and nothing else — but you do not have to build anything: the maximum average power is one of the four answers it already gives you, in pmax.
Results shows pmax = 2.367 W, alongside vth, ino and zeq. That is the power delivered when the load is the conjugate of the equivalent impedance, which is the whole point of the exercise — Symbulator has done the conjugating and the halving for you.
8.2RMS and PF
In the previous lesson we mentioned that after an AC simulation you also get the complex power consumed. How to read these values depends on a flag called the RMS flag.
The RMS setting
You will learn in your circuits course that phasor analysis can be done in RMS values. There are advantages to it, and you will have to solve AC circuits both in normal values (also called magnitude values) and in RMS. Working in RMS simply means every current and voltage in the analysis is taken to be RMS. That's it.
Where the calculator versions used a flag variable that persisted between runs, Symbulator 9 puts it in Settings, under AC power convention: tick RMS phasors. Off means peak amplitude, the convention with the divide-by-two, and the setting affects AC power only. Being a setting rather than a stored variable, it is always in view — it cannot be left set from a problem you finished an hour ago.
AS7's Example 11.10
Open in app ↗Open in split view
Determine the power factor of the entire circuit as seen by the source. Calculate the average power delivered by the source.
AS7's Example 11.10
Solution
The source value in this problem is given in RMS values, so we must say so.
e,1,0,30
r1,1,2,6
r2,2,0,-2j
r3,2,0,4AC, with omega typed into the ω box, and RMS phasors ticked in Settings.
Or, if you want to simplify the impedances, the resistors can be collapsed into a single parallel combination.
The average power consumed by the source is in pe. To get the average power delivered, we ask for the negative of it, which gives us 125.4 W.
The pf tool
Symbulator 9 has the same tool, in the Mini-Tools card. Choose pf — power factor; it asks for the voltage and the current separately rather than for a single complex power or an element name:
ve-ieIt answers 0.97342 leading.
It returns the value and the verbal description together, in one string, just as the calculator prints them.
8.3Complex power
AS7's Example 11.13
Open in app ↗Open in split view
The figure shows a load being fed by a voltage source through a transmission line. The impedance of the line is represented by the (4+j2) Ω impedance and a return path. Find the real power and reactive power delivered by the source, and absorbed by the line and load.
AS7's Example 11.13
Solution
The value of this circuit's source is given in RMS. Since the problem asks for answers regarding the line and the load, we should have separate impedances, one for each.
e,1,0,220
r1,1,2,4+2j
r2,2,0,15-10jAgain AC with omega and RMS phasors ticked.
The complex power absorbed in the source, line and load are in -se, sr1 and sr2:
We look in the results and see:
-se= 2163.8 − 911.1𝐢 VAsr1= 455.5 + 227.8𝐢 VAsr2= 1708.2 − 1138.8𝐢 VA
These are correct.
AS7's Practice Problem 11.10
Open in app ↗Open in split view
Calculate the power factor of the entire circuit as seen by the source. What is the average power supplied by the source?
AS7's Practice Problem 11.10
Solution
The values are RMS, so tick RMS phasors in Settings.
e,1,0,165
r1,1,2,10
r2,2,0,4j
r3,2,3,8
r4,3,0,-6jWritten out as four elements rather than one, which is clearer and costs nothing here. The average power supplied is the opposite of the power the source consumes — -pe in Evaluate gives 2007.1 W.
For the power factor, use Mini-Tools with pf, giving it the source's voltage and the current it delivers — which is the opposite of the current through the source element:
ve-ieIt reads 0.93595 lagging.
AS7's Problem 11.97
Open in app ↗Open in split view
A power transmission system is modelled as shown. If Vs = 240 V rms, find the average power absorbed by the load.
AS7's Problem 11.97
Solution
Three impedances in series: the outgoing line, the load, and the return path.
evs,1,0,240
rl1,1,2,.1+1j
rl,2,3,100+1j
rl2,3,0,.1+1jAC with RMS phasors ticked. The rl block's power consumed line reads 573.2 W.
Correct.
AS7's Problem 11.75
a and bOpen in app ↗Open in split viewc, trying a capacitorOpen in app ↗Open in split viewc, an inductor insteadOpen in app ↗Open in split view
Consider the power system shown. Calculate the total complex power, the power factor, and the parallel capacitance needed for a unity power factor.
AS7's Problem 11.75
Solution
Three loads in parallel across a 240 V rms source. The first two parts need no frequency at all, so leave omega as anything.
e,1,0,240
r1,1,0,80-50j
r2,1,0,120+70j
r3,1,0,60(a) The complex power delivered is the opposite of the power the source consumes: -se in Evaluate gives 1835.9 − 114.7j VA.
(b) The power factor, from pf in Mini-Tools with ve and -ie, is 0.99805 leading.
(c) This one needs a frequency, because it needs a capacitor. Add one in parallel with a symbolic value (let's say x), and run at the stated 50 Hz — (2π)(50) in rad/s.
e,1,0,240
r1,1,0,80-50j
r2,1,0,120+70j
r3,1,0,60
c,1,0,xPut 2*pi*50 in the ω — angular frequency box — it takes an expression. Then ask the Solve card for the value of x that leaves no reactive power:
im(se) = 0xTick real solutions only.
It answers -6.34 µF.
A negative capacitance, and no positive value satisfies the equation — which is probably why the textbook gives no number for this part. It is not a failure of the method: the load is already leading, as part (b) said, so no capacitor can bring it to unity. What it needs is the opposite.
Try an inductor instead:
e,1,0,240
r1,1,0,80-50j
r2,1,0,120+70j
r3,1,0,60
l,1,0,xSolving the same way gives 1.5987 H, which is positive and therefore the real answer.
Check it by putting that number back in as the inductor's value: the power factor comes out 1. An inductor of about 1.6 H in parallel is what brings this system to unity.