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 phasors setting. Solve power factor problems using the pf mini-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 phasors setting, and how to solve complex power problems.

8.1Average power

AS7's Example 11.5

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Determine the load impedance ZL that maximises the average power drawn from the circuit. What is the maximum average power?

Solution

Thévenin / Norton helps us find all the answers we need. Notice this circuit description does not include the load, because it is not needed.

Circuit Description

e,1,0,10
r1,1,2,4
r2,2,0,8-6j
r3,2,3,5j

Find equivalent, Thévenin / Norton, nodes 3 and 0, in AC. In AC the equivalent impedance is called zeq rather than req.

Thévenin / Norton reports the equivalent impedance, zeq, as 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 pmax = 2.3674 W.

In AC, pmax is the average maximum power, computed from the real part of the equivalent impedance: Results shows pmax = 2.367 W, the power delivered when the load is the conjugate of zeq.

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 the RMS phasors setting.

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.

The switch is in Settings, under AC power convention: tick RMS phasors.

With the tick off, Symbulator takes every phasor you type as a peak value — the amplitude the sinusoid rises to — and the average power in an element is computed with the factor of one half that peak values require:

\[P = \tfrac{1}{2}\,V_{\text{peak}}\,I_{\text{peak}}\cos\theta\]

With the tick on, it takes every phasor you type as an RMS value, and the half disappears:

\[P = V_{\text{rms}}\,I_{\text{rms}}\cos\theta\]

That is the whole difference. The currents and voltages you get back are in whatever convention you typed the sources in, unchanged; only the power answers — p, q and s — depend on the tick, and they keep their names whichever way it is set: pe is the average (real) power, qe the reactive power, se the complex power (the calculator named the average ap with peak phasors and p with RMS ones; ape still answers, as the same number as pe). With the tick on, the cards call every current and voltage effective, the RMS values being the effective values. So for the same numbers typed in, RMS on reports twice the power that RMS off does, and neither is wrong: they answer different questions. Match the book you are working from. The setting is always in view, so it is easier for you to notice whether it is on or off.

AS7's Example 11.10

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Determine the power factor of the entire circuit as seen by the source. Calculate the average power delivered by the source.

Solution

The source value in this problem is given in RMS values, so we must say so.

Circuit Description

e,1,0,30
r1,1,2,6
r2,2,0,-2j
r3,2,0,4

AC, 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 pe; the power delivered is its opposite, which the source's card reads as -pe, 125.4 W.

The pf mini-tool

pf is the second of the mini-tools Lesson 7 introduced, and it works as the calculator's did. In the Mini-Tools card choose pf — power factor; it asks for one value, and there are two things you can give it.

A complex power. Give it the name of a complex power answer, such as se, or any complex expression — a number in rectangular or polar form, an impedance, an expression with symbols in it:

Value

se

It answers 0.97342 lagging: the absolute value of the real part divided by the magnitude of the whole thing, and the word for the value as given — nothing more, because by now you know what se is. It is the power the source consumes, and the word belongs to that power.

The name of an element. Give it the name of an element of the circuit you have just solved in AC:

Value

It answers 0.97342 leading: the value and the word together, and under them this is the power factor for the power delivered by source e. This form works only when the element's voltage and current came out as numbers.

The same number, the opposite word. That is not a contradiction: se is the power the source consumes and e reads the power it delivers, and the two powers are opposite, so their words are. The question what is the power factor of the circuit as seen by the source? is answered by the name, e.

In a nutshell: given a complex power, or any complex expression, pf returns the power factor of that value as given, with its word and no explanation. Given the name of an element, right after an AC solve, it reads the right power for that element's kind and says in words which power that was. With symbols in the value there is no word, only the expression.

Note

Which power the reading is taken on

The two forms do not read the same power, and the difference is what makes the word come out right.

Given a variable such as se, sj or sr1, the calculation is done on the value as it stands — the complex power consumed, which is what Symbulator stores for every element, source or load alike — and the word is that value's own. For an impedance that is its own reading. For a source it is the reading of the power it consumes, which is the opposite of the power it delivers, so the word comes out opposite to the source's own. The tool cannot know better from a number alone: the same complex power is consumed by one side of a branch and delivered by the other.

Given a name, the tool reads the sign convention off the element's kind, the way the calculator did. For an impedance — r, and in version 9 l and c as well — the angle is taken between the element's voltage and the current it consumes, so the reading is the element's own: an inductive load reads lagging. For a source — e or j — the current is negated first, so the angle is taken on the power the source delivers, and the reading is that of the whole circuit the source sees. A source feeding an inductive load reads lagging, the same word as the load.

The value is the same either way; only the word depends on it. A source read on its consumed power, se, says the opposite word — which is why the question as seen by the source wants the name, and the tool does the negating itself.

8.3Complex power

AS7's Example 11.13

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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.

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.

Circuit Description

e,1,0,220
r1,1,2,4+2j
r2,2,0,15-10j

Again AC with omega and RMS phasors ticked.

The complex power delivered by the source is the opposite of the power it consumes, -se; the complex power absorbed in the line and in the load are sr1 and sr2:

Results shows:

-se = 2163.8 − 911.1𝐢 VA
sr1 = 455.5 + 227.8𝐢 VA
sr2 = 1708.2 − 1138.8𝐢 VA

These are correct.

AS7's Practice Problem 11.10

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Calculate the power factor of the entire circuit as seen by the source. What is the average power supplied by the source?

Solution

The values are RMS, so tick RMS phasors in Settings.

Circuit Description

e,1,0,165
r1,1,2,10
r2,2,0,4j
r3,2,3,8
r4,3,0,-6j

Written 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 reads -2007.1 W, so the source supplies 2007.1 W, which its card reads as -pe.

For the power factor, use Mini-Tools with pf, giving it the source's name, so that the reading is of the circuit the source sees:

Value

It reads 0.93595 lagging.

AS7's Problem 11.97

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A power transmission system is modelled as shown. If V_s = 240 V rms, find the average power absorbed by the load.

Solution

Three impedances in series: the outgoing line, the load, and the return path.

Circuit Description

evs,1,0,240
rl1,1,2,.1+1j
rl,2,3,100+1j
rl2,3,0,.1+1j

AC with RMS phasors ticked. The power consumed prl reads 573.2 W.

Correct.

AS7's Problem 11.75

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Consider the power system shown. Calculate the total complex power, the power factor, and the parallel capacitance needed for a unity power factor.

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.

Circuit Description

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: the opposite of se gives 1835.9 − 114.7j VA.

(b) The power factor, from pf in Mini-Tools given the source's name, e, 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.

Circuit Description

e,1,0,240
r1,1,0,80-50j
r2,1,0,120+70j
r3,1,0,60
c,1,0,x

c, trying a capacitorOpen in app ↗Open in split view

Put 2*pi*50 in the ω — angular frequency box — it takes an expression. Give pf the complex power se now and it answers with an expression in x: the power factor as a function of the capacitance, which is what the calculator solved for 1. In version 9, ask the Solve card instead for the value of x that leaves no reactive power, which is the same condition:

Equation(s) to solve in terms of the results

im(se) = 0

Unknown(s) to solve for

Tick 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:

Circuit Description

e,1,0,240
r1,1,0,80-50j
r2,1,0,120+70j
r3,1,0,60
l,1,0,x

c, an inductor insteadOpen in app ↗Open in split view

Solving 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.