Symbulator 7

the best portable symbolic simulator of linear circuits

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Lesson 4

Shorts, equivalent resistance and Thévenin/Norton

Learn how to describe a short circuit using the s element. Learn how to find equivalent resistances using the er script, and Thevenin and Norton equivalents using the th script.

Last updated 2023-07-08

In this lesson you will learn how to describe a short circuit with the s element, how to find equivalent resistances with the er script, and Thévenin and Norton equivalents with the th script.

4.1How to describe a short circuit

Shorts are used mostly to find a current in a part of the circuit with no element in it already. Otherwise we would just make it a single node.

What answers do you get

No power is consumed, and no voltage is dropped, in a short circuit. For each short in a circuit, Symbulator stores only the current through it, flowing from the first node towards the second. For a short called sx, that is isx.

HK5's Drill Problem 1-13

Find i1, i2, i3 and i4.

HK5's Drill Problem 1-13

Solution

My solution: I define the shorts in the same direction as the arrows in the schematic.

type
s\dc("r1,1,0,25:jd,0,2,.2v1:r2,2,3,10:ji,4,3,2.5:r3,4,5,100:s1,1,2:s2,2,4:s3,0,3:s4,3,5")

We ask for the values of the variables:

type
approx({is1,is2,is3,is4})
returns
{–2., 3., –8., –.5}

These are correct, and they can only be found using short circuits.

4.2The equivalent resistance script: er

As we saw, Symbulator gives the equivalent resistance of a circuit as seen from any source. That lets us solve problems like this one.

AS2's Practice Problem 2.15

Find the equivalent resistance as seen by the 100 V source, and the value of current i.

AS2's Practice Problem 2.15

Solution

type
s\dc("e,a,0,100:r13,a,1,13:r24,1,2,24:r10,1,3,10:r20,2,3,20:r30,2,0,30:r50,3,0,50"):re

Evaluating re gives the equivalent resistance as seen by the source e: 40 Ω. Evaluating ir13 gives 2.5 A for current i. The answers came easily, because a source sat between the two nodes we wanted.

What if there is no independent source?

But how do we find the equivalent resistance of a passive circuit, one with no independent source in it? One way is to connect a 1 A current source between the two nodes and read the voltage drop across it. That is the manual way.

An easier way is to let Symbulator do that for us.

Just run the er script and it does the same thing automatically. It finds the equivalent resistance of a passive circuit and stores it — for a DC analysis — in req. It takes three arguments: the circuit description as a string, and the two nodes to measure between.

B11's Example 8.29

Calculate the equivalent resistance of the circuit shown.

B11's Example 8.29

Solution

Let me solve this problem step by step. After I label the nodes, I describe the circuit and store it in a variable.

type
"r4,0,a,4:r2,0,b,2:r6,a,b,6:rb,a,c,3:ra,b,c,3"→cir
s\er(cir,0,c)

When prompted, choose DC as analysis type. Once it is done, evaluate:

type
approx(req)
returns
2.89

The value is 2.89 Ω. This is correct.

What counts as passive

That example was made of resistors only. The er script also handles a second kind of passive circuit: one with resistors and dependent sources, but no independent sources. Such a circuit can only be reduced to an equivalent resistance, not to a Thévenin or Norton equivalent, so er is the script to use — in exactly the same way, provided the dependent sources are described properly.

4.3The Thévenin / Norton script: th

Just as a passive circuit can be reduced to an equivalent resistance, an active circuit — one with independent sources — can be reduced to a Thévenin or Norton equivalent.

One way is to run a first simulation for the voltage between the two nodes where we want the equivalent (the Thévenin voltage, VTH), then a second with a short between those nodes for the current through it (the Norton current, INO). REQ is then VTH/INO. That is the manual way.

An easier way is the th script, which does exactly that automatically. It takes three arguments: the circuit description, the first node and the second node.

RM3's Practice Problem 9-4

Find the Thévenin and Norton equivalents of the circuit.

RM3's Practice Problem 9-4

Solution

This is my circuit description, and this is how we run the th script:

type
"e,1,0,3.3:r1,1,2,66:r2,2,0,24"→cir
s\th(cir,2,0)

We could also have passed the circuit description directly:

type
s\th("e,1,0,3.3:r1,1,2,66:r2,2,0,24",2,0)

When prompted to select a type of analysis, choose DC. Symbulator tells you what it is doing: one simulation for the Thévenin voltage, then a second for the Norton current and the equivalent resistance. In this case, VTH = 0.88 V, INO = 0.05 A and REQ = 17.6 Ω.

When you press ENTER, Symbulator asks whether you are running a problem with a load connected to this equivalent circuit. For now, say No. The script then stores these variables:

  • vth has the Thévenin voltage
  • ino has the Norton current
  • req has the equivalent resistance
  • pmax has the maximum power that the equivalent can deliver to a hypothetical load

Problems with a load

One type of problem books and professors like when teaching the Thévenin / Norton equivalents is what I call an RL problem. A typical one goes like this: "First, reduce the circuit, as seen by resistor RL, to its Thévenin or Norton equivalent. Then, find the value of the voltage drop, current and/or power consumed in the load resistor RL if its value is (whatever) ohms."

Since this is such a typical problem, I've made some provisions in Symbulator to help you solve them. Right after a Thévenin or Norton equivalent is found, the th script asks whether you are planning to connect a load. The default is No, but if you select Yes, Symbulator will save some special answers for that very typical case, as functions of the load variable:

  • irl has the current in the load
  • vrl has the voltage drop in the load
  • prl has the power consumed in the load

B11's Example 9.6

Find the Thévenin equivalent circuit for the network in the shaded area. Then find the current through RL for RL values of 2 Ω, 10 Ω and 100 Ω.

B11's Example 9.6

Solution

First we find the Thévenin equivalent.

type
s\th("e1,1,0,9:r1,1,2,3:r2,2,0,6",2,0):{vth,req}
returns
{6, 2}

Correct. Now we find the values of the current in the load for the different values. We can do this in a single push, or separately. Here I find them in one go:

type
{irL|Load=2.,irL|Load=10.,irL|Load=100.}
returns
{1.5, .5, .059}

Where | is the "given" operator. The answers are correct.

Power transfer problems

Another problem often associated with the Thévenin / Norton equivalents is power transfer to a load, particularly the maximum possible. Maximum power is transferred when the load RL equals the REQ of the equivalent. Symbulator's th script gives you the maximum power that can be delivered in pmax, and the power transferred to the load as a function of its value in prl.

What if it's more than a load?

The formulas for the current, voltage and power in the load apply only when a load is the only thing connected to the equivalent circuit. If the problem you want to solve includes something more complicated, you will have to run your own simulation.

To help you in those cases, let me show you one more goody of the th script, at risk of promoting vagrancy among EE students: once it has found the Norton equivalent of a circuit, it will automatically write for you the circuit description of that Norton equivalent connected to a load, and store it in a string called eqcir. You can use it as a starting point. It has the Norton equivalent connected, between nodes n and 0, to a load called rl with a symbolic value of load, in ohms.

RM3's Example 9-8

Find the Norton equivalent of the circuit left of a-b; then find the current through RL.

RM3's Example 9-8

Solution

Let's first find the circuit equivalent:

type
s\th("e,1,0,24:r1,1,2,120:r2,2,0,280:j,2,0,560'm",2,0):{ino,req}
returns
{.36, 84.}

Correct. Now to the second part of the question. In order to find the current through RL, we cannot use the load expressions, because now the load is not the only thing connected to the terminals of the equivalent: there is also a current source. We have to run a new simulation.

The fastest way is to start from the equivalent circuit description:

returns
"jN,0,n,iNo:rE,n,0,rEq:rL,n,0,L"

We change the value of the load to 168 Ω, and add the 180 mA source flowing from node 0 to node n. Then we run a dc simulation and ask for the current in the load:

type
s\dc("jN,0,n,iNo:rE,n,0,rEq:rL,n,0,168:j,0,n,180'm"):irL
returns
–.06

Correct: there is a current of 60 mA flowing through RL from 0 to n.

Using the equivalent circuit description is meant to save you time. If you find it confusing to use, just don't use it.

4.4Instructive solved examples

Practice problems for resistive circuits

B11's Example 8.29

Calculate the equivalent resistance of the circuit shown.

You saw this circuit solved step by step in the walkthrough above; here it is again, in the compact form the rest of these problems use.

After I label the nodes, I describe the circuit. In this case, I store it in a variable.

type
"r4,0,a,4:r2,0,b,2:r6,a,b,6:rb,a,c,3:ra,b,c,3"→cir

Run the er script, giving it as arguments the circuit and the nodes: s\er(cir,0,c) When prompted, choose DC as analysis type. Once it is done, evaluate approx(req) The value is 2.89 Ω.

This is correct. Below are many practice examples of this type.

B11's Example 8.30

Find the equivalent resistance of the circuit.

type
s\er("rac,a,c,6:rad,a,d,9:rab,a,b,6:rcd,c,d,9:rbc,b,c,6:rbd,b,d,9",a,c)

Choose DC. When Done, use approx(req) to find the equivalent resistance is 3.27 Ω.

AS2's Example 2.9

Find the equivalent resistance of the circuit.

We don't need to run a simulation for this. We can reduce it using s\pr.

type
4+s\pr({1+5,2+s\pr({6,3})})+8

Evaluating approximately gives us the equivalent resistance: 14.4 Ω.

AS2's Practice Problem 2.9

Find the equivalent resistance of the circuit.

We don't need to run a simulation for this. We can reduce it using s\pr.

type
2+s\pr({6,3+s\pr({4,4+5+3})})+1

Evaluating approximately gives us the equivalent resistance: 6 Ω.

AS2's Example 2.11

Find the equivalent conductance of the circuit.

We don't need to run a simulation for this. We can reduce it using s\pr.

type
1/(s\pr({1/6,1/5+s\pr({1/8,1/12})}))

Evaluating approximately gives us the equivalent conductance: 10 S.

AS2's Practice Problem 2.11

Find the equivalent conductance of the circuit.

We don't need to run a simulation for this. We can reduce it using s\pr.

type
1/(s\pr({1/8,1/4})+s\pr({1/2,1/12+1/6}))

Evaluating approximately gives us the equivalent conductance: 4 S.

Examples with dependent sources

Bo2's Drill Problem 3.14

Find the equivalent resistance of the circuit.

type
s\er("r4,a,0,4:ri,a,0,6:ji,a,0,iri/2",a,0)

Choose DC. Wait for Done. Evaluate req to find the equivalent resistance is 2 Ω.

AS2's Example 4.10

Find the equivalent resistance of the circuit.

type
s\er("r4,a,0,4:rx,0,a,2:j,a,0,2irx",a,0)

Choose DC. Wait for Done. Evaluate req. The equivalent resistance is -4 Ω. It may be surprising to have a negative resistance. This is the result of the dependent sources.

AS2's Practice Problem 4.10

Find the equivalent resistance of the circuit.

type
s\er("r15,a,0,15:e,1,a,4vrx:r10,1,x,10:rx,x,0,5",a,0)

Choose DC. Wait for Done. Evaluating req approximately, we find the equivalent resistance is -7.5 Ω.

HK5's Figure 2-29

Find the equivalent resistance of the circuit.

type
s\er("e,3,0,1.5is:r3,3,2,3:r2,2,0,2:s,2,1",1,0)

Choose DC. Wait for Done. Evaluate req. The equivalent resistance is 0.6 Ω.

HK5's Drill Problem 2-9d

Find the equivalent resistance of the circuit.

type
s\er("r10,1,2,10:r5,2,3,5:r1,2,0,30:e,1,0,20ir1",3,0)

Choose DC. Wait for Done. Evaluate req. The equivalent resistance is 20 Ω.

Bo2's Example 3.12

Find the equivalent resistance of the circuit.

type
s\er("r1,1,0,6:r4,a,0,4:e,a,1,6ir1",a,0)

Choose DC. Wait for Done. Evaluate req. The equivalent resistance is 3 Ω.

B11's Example 9.7

Find the Thévenin equivalent of the circuit, as seen from the R3 resistor.

type
s\th("j,0,1,12:r1,1,0,4:r2,1,a,2",a,0)

Choose DC. Via vth we find VTH = 48 V. Via req we find REQ = 6 Ω.

B11's Example 9.11

Find the Norton equivalent of the circuit, as seen from the RL resistor.

type
s\th("e,1,0,9:r1,1,2,3:r2,2,0,6",2,0)

Choose DC. Via ino we find INO = 3 A. Via req we find REQ = 2 Ω.

AS2's Practice Problem 4.12

Find the Norton equivalent of the circuit.

type
s\th("r6,1,0,6:j,0,1,10:r2,x,0,2:e,1,x,2vx",x,0)

Choose DC. Via ino we find INO = 10 A. Via req we find REQ = 1 Ω.

B11's Example 9.12

Find the Norton equivalent of the circuit, as seen from the RL resistor.

type
s\th("r2,1,0,4:r1,1,2,5:j,1,2,10",2,0)

Choose DC. Via ino we find INO = 5.56 A. Via req we find REQ = 9 Ω.

HK5's Drill Problem 2-8b

Find the Thévenin equivalent of the circuit.

type
s\th("j,0,2,0.01v1:r,0,2,20:e,1,2,100",1,0)

Choose DC. Via vth we find VTH = 125 V. Via req we find REQ = 25 Ω.

HK5's Figure 2-27

Find the Thévenin equivalent of the circuit.

type
s\th("e,1,0,4:r2,1,2,2'k:r3,2,x,3'k:j,0,2,vx/4000",x,0)

Via vth we find VTH = 8 V. Via req we find REQ = 10 kΩ.

B11's Example 9.8

Find the Thévenin equivalent of the circuit, as seen from the R4 resistor.

I ignore the textbook's decision to call the nodes a and b, since b is ground anyway.

type
s\th("r1,2,0,6:r2,2,1,4:r3,1,0,2:e,0,1,8",2,0)

Choose DC. Via vth we get VTH= -4.8 V. Via req we get REQ= 2.4 Ω.

B11's Example 8.6

Find the Norton equivalent of the circuit.

type
s\th("j1,0,1,6:r1,1,0,3:j2,1,0,10:r2,1,0,6",1,0)

Choose DC. Via ino we find INO = -4 A. Via req we find REQ = 2 Ω.

Bo2's Drill Exercise 3.12

Find the Norton equivalent of the circuit.

type
s\th("e,1,0,12:r6,1,2,6:j,0,2,3ir6:r3,2,0,3",2,0)

Choose DC. Via ino we find INO = 8 A. Via req we find REQ = 1 Ω.

Bo2's Drill Exercise 3.9

Find the Thévenin equivalent of the circuit.

type
s\th("j,b,0,10:r1,0,b,1:e,a,0,3ir1:r6,a,b,6",a,b)

Choose DC. Via vth we find VTH = 24 V. Via req we find REQ = 2.4 Ω.

AS2's Example 4.12

Find Norton equivalent of the circuit.

type
s\th("rx,0,b,4:e,0,b,10:r5,0,a,5:j,0,a,2irx",a,b)

Choose DC. Via ino we find INO = 7 A. Via req we find REQ = 5 Ω.

B11's Example 9.10 (Hidden source)

Find the Thévenin equivalent of the circuit.

My solution is below:

type
s\th("e1,3,0,-6:e2,4,0,10:r1,2,3,0.8'k:
r2,2,4,4'k:r3,2,0,6'k:r4,2,1,1.4'k",1,0)

Choose DC. Via vth we find VTH = -3 V. Via req we find REQ = 2 kΩ.

B11's Example 9.9

Find the Thévenin equivalent of the circuit, as seen from the RL resistor.

type
s\th("e,1,0,72:r1,1,b,6:r2,1,a,12:r3,0,b,3:r4,0,a,4",b,a)

Choose DC. Via vth we find VTH = 6 V. Via req we find REQ = 5 Ω.

AS2's Example 4.11

Find the Norton equivalent of the circuit.

type
s\th("j,0,2,2:e,4,0,12:r1,2,4,4:r2,2,3,8:r3,0,1,8:r4,3,1,5",3,1)

Choose DC. Via ino we find INO = 1 A. Via req we find REQ = 4 Ω.

HK5's Drill Problem 2-8a

Find the Thévenin equivalent of the circuit.

type
s\th("e1,1,0,100:r2,1,2,20:j,0,2,4:r1,2,3,10:e5,3,4,50",4,0)

Choose DC. Via vth we find VTH = 130 V. Via req we find REQ = 30 Ω.

B11's Example 8.7

Find Norton equivalent of the circuit.

type
s\th("j7,0,1,7:j3,1,0,3:r1,1,0,4:j4,0,1,4",1,0)

Choose DC. Via ino we find INO = 8 A. Via req we find REQ = 4 Ω.

AS2's Practice Problem 4.9

Find the Thévenin equivalent of the circuit.

type
s\th("e,1,0,6:r5,1,2,5:rx,2,3,3:j,0,2,1.5irx:r4,3,0,4",3,0)

Choose DC. Via vth we find VTH = 5.33 V. Via req we find REQ = 0.44 Ω.

AS2's Example 4.9

Find the Thévenin equivalent of the circuit.

type
s\th("j,b,0,5:rx,0,b,4:r1,0,1,2:r2,1,b,6:r3,1,a,2:e,1,0,2vrx",a,b)

Choose DC. Via vth we find VTH = 20 V. Via req we find REQ = 6 Ω.

Bo2's Drill Exercise 3.8

Find the Thévenin equivalent of the circuit.

type
s\th("j,0,1,3:r1,0,1,1:r6,1,2,6:r10,1,3,10:r8,2,0,8:r2,2,3,2",3,0)

Choose DC. Via vth we find VTH = 2 V. Via req we find REQ = 4 Ω.

B11's Example 9.13

Find Norton equivalent of the shaded part of the circuit.

type
s\th("e1,1,0,7:r1,1,a,4:j,a,0,8:r2,a,0,6",a,0)

Choose DC. Via ino we find INO = -6.25 A. Via req we find REQ = 2.4 Ω.

Tricky problems

Some circuit theory books – and some professors – find it entertaining or instructive to surprise unsuspecting students with tricky problems, like the two we solve below.

Bo2's Example 3.11 (Tricky)

Find the Norton equivalent of the circuit.

Since you do not know beforehand that this is a tricky problem, you go for the usual:

type
s\th("r2,0,b,2:r8,0,2,8:r3,2,a,3:r1,1,a,1:e,1,0,1:j,2,b,3ir8",a,b)

You choose DC, press Enter and wait. Symbulator reports the calculator was unable to solve the equations. This often means there is a division by zero somewhere.

Clean the variables from the MAIN folder and try again, this time – following Symbulator's advise –using a symbolic value. We chose to use x instead of 3 in the dependent source.

type
s\th("r2,0,b,2:r8,0,2,8:r3,2,a,3:r1,1,a,1:e,1,0,1:j,2,b,x*ir8",a,b)

Now it solves. The expression for ino is fine, but the one for req, \((9x-35)/(4(x-3))\), will divide by zero at \(x = 3\) — the very value we replaced.

Via Define x=3: {ino,req} we find that INO = 1 A, and REQ is undefined.

This means the equivalent resistance is, for practical purposes, infinite. Your idea of fun, right?

Bo2's Drill Exercise 3.13 (Tricky)

Another tricky one. Find the Norton equivalent of the circuit.

As in the previous example, we get an error if we simulate using 2 and 3 in the dependent sources. So, we use 2x and 3x instead, where x will be later defined as 1.

type
s\th("ei,1,b,6:ed,b,0,2*x*ir2:r1,1,0,6:r2,0,a,2:jd,a,0,3*x*ir1",a,b)

Exploring the answers, we see that the denominator of the expression for req, (x-1), results in a division by zero at x = 1.

Via 1→x:{ino,req} we find that INO = -3A, and REQ is undefined or, for practical purposes, infinite.

TR5's Exercise 4-6 (Symbolic)

First let's find the input resistance, RIN, i.e. the resistance as seen by the vS source. We use μ for the constant in the dependent source.

type
s\dc("ei,3,0,vs:rf,3,2,rf:ro,2,t,ro:ed,2,0,μ*vrf"):rei

We get rf*(µ+1), which is correct. The textbook's answers are shown right of the circuit schematic. Now we find the output Thévenin equivalent circuit as seen by RL.

type
s\th("ei,3,0,vs:rf,3,2,rf:ro,2,t,ro:ed,2,0,μ*vrf",t,0):{vth,req}

We get {vs*µ/(µ+1),ro}, which is correct, as can be seen in the textbook's answers for vT and RT, shown right of the circuit schematic above.

TR5's Example 4-8 (Symbolic)

Find the Thévenin equivalent as seen by the load.

My answer starts by defining vx as va-vb. This is done thus:

type
Define vx=va-vb

Now I run the th script, with this circuit description:

type
s\th("ei,a,0,vs:ed,1,0,μ*(vx):ro,b,1,ro",b,0):
{vth,req}

The answers we get, {vs*µ/(µ+1),ro/(µ+1)}, are correct, as can be seen by comparing them to those in the book:

I am not sure there is any other circuit simulator for calculators that can do this.

AS2's Example 4.8

Find the Thévenin equivalent of the circuit shown to the left of terminals a-b. Then find the current through RL = 6, 16 and 36Ω.

My solution:

type
s\th("e,1,0,32.:r4,1,2,4:r12,2,0,12:j,0,2,2:r1,2,3,1",3,0):
{vth,req,irL|Load=6,irL|Load=16,irL|Load=36}

Choose DC. Answer Y when offered the load formulas.

The answer, {30.,4.,3.,1.5,.75}, is correct.

Bo2's Example 3.10

Find the Norton equivalent of the circuit left of the a-b terminals, and then find the voltage drop and the current through the ¼ Ω resistor. My one-line solution:

type
s\th("e,3,0,3:r31,3,1,1/2:r10,1,0,1/2:r12,1,2,1/4:j,2,0,v1/2",2,0):
{ino,req,vrL|Load=1/4,irL|Load=1/4}

Choose DC. Answer Y when offered the load formulas.

The book gives the answers as fractions. We get it right: {21/8, 4/9, 21/50, 42/25}.

RM3's Example 9-7

Find the Norton equivalent of the circuit external to RL. Then determine the load current IL when RL = 0 Ω, 2 kΩ and 5 kΩ. My one-line solution:

type
s\th("e,1,0,15.:r1,1,2,6'k:j,0,2,5'm:r2,2,0,2'k",2,0):
{ino,req, irL|Load=0,irL|Load=2000,irL|Load=5000}

Choose DC. Answer Y when offered the load formulas.

The answer, {.0075, 1500., .0075, .00321, .00173}, is correct.

Bo2's Example 3.5

Find the Norton equivalent of the circuit external to the 1Ω resistor. Then determine the voltage drop across this 1Ω resistor.

My solution:

type
s\th("e,1,0,24:r12,1,2,12:r20,2,0,4:r23,2,3,4:r34,3,4,2:
j,4,0,3:r40,4,0,5",3,0):{ino,req,vrL|Load=1}

Choose DC. Answer Y when offered the load formulas.

The answer, {-9/7,7/2,-1}, is correct.

RM3's Example 9-13

Use Millman's Theorem to simplify the circuit left of a-b so that there is only one voltage and one resistor. Then find the current in the load resistor RL.

I don't know Millman's Theorem, but in my book this is called the Thévenin equivalent.

type
s\th("r1,0,1,240.:e1,2,1,96:r2,0,3,200:e2,3,2,40:
r3,0,4,800:e3,2,4,80",2,0):{vth,req,irL|Load=192}

Choose DC. Answer Y when offered the load formulas.

The answer, {28.8,96.,.1}, is correct.

Bo2's Example 3.7

Find the Thévenin equivalent for the circuit left of a-b. Then find the voltage across the 3 Ω resistor, and also if it was 6 Ω. My answer:

type
s\th("e,1,0,20:r6,1,2,6:r1,1,3,1:r2,3,2,2:j32,3,2,15:j30,3,0,15",2,0):
{vth,req,vrL|Load=3,vrL|Load=6.}

Choose DC. Answer Y when offered the load formulas.

The answer we find, {30,2,18,22.5}, is correct.

Bo2's Drill Exercise 3.7

Find the Thévenin equivalent for the circuit left of a-b. Then find the voltage v.

My answer:

type
s\th("e,1,0,8:r3,1,2,3:r12,1,3,12:r6,2,0,6:r2,2,3,2",3,0):
{vth,req,vrL|Load=5.}

Choose DC. Answer Y when offered the load formulas.

The answer, {6,3,3.75}, is correct.

RM3's Practice Problem 9.5

My solution:

type
s\th("e1,1,0,35.:r1,1,2,15'k:r2,2,3,60'k:e2,3,0,70:r3,2,4,30'k",4,0):
{ino,req,irL|Load=0,irL|Load=1e4,irL|Load=5e4,irL|Load=1e5}

Choose DC. Answer Y when offered the load formulas.

The answer, {.001,42000,.001,.000808,.000457,.000296}, is correct.

Power transfer problems

AS2's Example 4.13

Find the RL value for maximum power transfer and the maximum power transferred.

RL for maximum transfer is req. The maximum power is in pmax. My solution:

type
s\th("e,1,0,12:r6,1,2,6:r12,2,0,12:r3,2,3,3:j,0,3,2:r2,3,4,2",4,0):
approx({req,pmax})

Choose DC. You can answer N when asked about the load equations. The answer, {9.,13.44}, is correct: the maximum transfer of power occurs when the load is 9Ω. At this point, the power transferred is 13.44W. Now let's solve another one.

AS2's Practice Problem 4.13

Find the RL value for maximum power transfer and the maximum power transferred.

type
s\th("ei,1,0,9:rx,1,2,2:r1,2,4,1:ed,4,0,3vrx:r4,2,3,4",3,0):
approx({req,pmax})

Choose DC. You can answer N. The logic of this problem is identical to the previous one. The answer is {4.22,2.901}.

B11's Example 9.15

My one-line solution to all three questions is presented below.

type
s\th("j,0,1,10'm:rs,1,0,40'k",1,0):
{req,pmax,prl|load=68000.,prl|load=8200.}

Choose DC. Answer Y about the load formulas. The answer we obtain, {40000, 1, .93, .57}, is correct. Let's deconstruct it.

Part (a) is answered by the first two values: a 40 kΩ resistor as load would receive 1W power. Since this is the maximum – this is the most that any load could receive ever.

Part (b) is answered by the third value. Making use of the variable prL, which contains the power delivered by the circuit equivalent to the load, as a function of the load value L, we find that a load of 68 kΩ receives .93W, which is less than the maximum.

Part (c) is answered in similar manner by the fourth value. A load of 8.2 kΩ receives .57W, which is less than the maximum. Any resistance other than 40 kΩ gets less power.

B11's Example 9.17

Find the RL value for maximum power transfer and the maximum power transferred.

type
s\th("j,2,0,6:r2,2,0,10:r1,2,3,3:r3,0,1,2:e,3,4,68",4,1):
approx({req,pmax})

Choose DC. You can answer N. The answer is {15., 273.07}. Let's now see one that is a little different.