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 the current through line of the sx block in Results.
HK5's Drill Problem 1-13
Open in app ↗Open in split view
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.
r1,1,0,25
jd,0,2,0.2*v1
r2,2,3,10
ji,4,3,2.5
r3,4,5,100
s1,1,2
s2,2,4
s3,0,3
s4,3,5Choose Solve circuit and DC, then run it. Each short has its own block in Results, and the number you want is on its current through line:
s1 reads -2 A, s2 3 A, s3 -8 A and s4 -0.5 A.
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
Open in app ↗Open in split view
Find the equivalent resistance as seen by the 100 V source, and the value of current i.
AS2's Practice Problem 2.15
Solution
e,a,0,100
r13,a,1,13
r24,1,2,24
r10,1,3,10
r20,2,3,20
r30,2,0,30
r50,3,0,50Solve it in DC. There is no special tool to reach for here: because a source sits between the two nodes we care about, the equivalent resistance is one of the answers Symbulator works out anyway. In the block for e, the resistance seen line reads re = 40 Ω.
The current i is the current through r13, and its block gives ir13 = 2.5 A.
Both answers came free, because there was a source connected between the two desired nodes. The rest of this lesson is about what to do when there is not.
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.
Version 9 puts this on the Type of analysis menu. Choose Find equivalent instead of Solve circuit, and a second menu appears — Type of equivalent — whose first entry is Resistance / impedance. Two node boxes appear with it, for the pair of terminals you are measuring between.
That is the whole of it: no source to invent, no voltage to divide. Symbulator does the 1 A trick internally and reports the answer.
B11's Example 8.29
Open in app ↗Open in split view
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.
r4,0,a,4
r2,0,b,2
r6,a,b,6
rb,a,c,3
ra,b,c,3Set Type of analysis to Find equivalent, Type of equivalent to Resistance / impedance, and the two node boxes to 0 and c. Leave Analysis on DC — direct current and press Run Symbulator.
Results looks different from a normal solve: instead of the node and element listing there is a single block headed Equivalent impedance, with one line in it.
Req = 2.89 Ω
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
Open in app ↗Open in split view
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:
e,1,0,3.3
r1,1,2,66
r2,2,0,24Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put 2 in the first and 0 in the second — the pair of terminals you are looking into. Leave Analysis on DC — direct current and press Run Symbulator.
The answers appear in Results under these names:
- 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
Results replaces the usual node-and-element listing with a single block headed Thévenin / Norton equivalent. There are no prompts and no second step; all four answers arrive together:
- Thevenin voltage, vth = 880 mV
- Norton current, ino = 50 mA
- Equivalent resistance, Req = 17.6 Ω
- Maximum deliverable power, pmax = 11 mW
The first three are what you came for. The fourth is Symbulator being generous — the most this circuit could deliver into a matched 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."
Symbulator 9 does not ask, and it does not carry the load quantities on the results either. Results shows four answers and no more: vth, ino, req and pmax.
That is no great loss, because each load quantity is one line in Evaluate, built from the two answers you already have. Writing R for the load resistance:
| To find | Type into Evaluate |
|---|---|
| the current in the load | vth/(req+R) |
| the voltage drop in the load | vth*R/(req+R) |
| the power consumed in the load | vth^2*R/(req+R)^2 |
Put the load's actual value where R is. The rest of this section reads exactly as it does on the calculator, with those three in place of the calculator's irL, vrL and prL.
B11's Example 9.6
Open in app ↗Open in split view
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.
e,1,0,9
r1,1,2,3
r2,2,0,6Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put 2 in the first and 0 in the second. Choose DC and run it.
Results gives vth = 6 V and req = 2 Ω.
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:
There is no single push here — you ask Evaluate three times, once per load, and the answers come back one at a time:
vth/(req+2)Then vth/(req+10), then vth/(req+100).
The three answers come back one at a time: 1.5 A, .5 A and .059 A.
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 the expression vth^2*R/(req+R)^2 derived above.
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.
Symbulator 9 writes no such string for you, and does not need to: the equivalent is three lines, and you already have both numbers on screen. Type them into a fresh Circuit Description, putting ino and req where they belong:
jn,0,n,ino
re1,n,0,req
rl,n,0,loadReplace ino and req with the numbers Results just gave you. The load is called rl and its value is the symbol load, so the answers come back in terms of it.
You can use it as a starting point for a new simulation.
RM3's Example 9-8
Open in app ↗Open in split viewwith the loadOpen in app ↗Open in split view
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:
e,1,0,24
r1,1,2,120
r2,2,0,280
j,2,0,560'mFind equivalent, Thévenin / Norton, nodes 2 and 0, in DC.
Results gives ino = -0.36 A and req = 84 Ω.
The sign differs from the calculator's, which prints 0.36 A: version 9 reports the Norton current in the direction it actually flows, from the first node to the second. Nothing else changes — carry the sign through and the load current below comes out the same.
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:
"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:
jn,0,n,-0.36
re1,n,0,84
rl,n,0,168
j,0,n,180'mRun it in DC. The rl block's current through line reads -0.06 A.
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
Open in app ↗Open in split view
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.
r4,0,a,4
r2,0,b,2
r6,a,b,6
rb,a,c,3
ra,b,c,3Set Type of analysis to Find equivalent and Type of equivalent to Resistance / impedance, with nodes 0 and c, in DC. The answer is req = 2.889 Ω.
This is correct. Below are many practice examples of this type.
B11's Example 8.30
Open in app ↗Open in split view
Find the equivalent resistance of the circuit.
rac,a,c,6
rad,a,0,9
rab,a,b,6
rcd,c,0,9
rbc,b,c,6
rbd,b,0,9Find equivalent, Resistance / impedance, nodes a and c, in DC: req is 3.273 Ω.
The calculator description names its four nodes a, b, c and d. Version 9 needs one of them to be the reference node, so d is called 0 here instead. Which node you ground makes no difference to the answer in a network with no sources — grounding b gives the same 3.273 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.
4+pr(1+5,2+pr(6,3))+8Evaluating 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.
2+pr(6,3+pr(4,4+5+3))+1Evaluating 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.
1/(pr(1/6,1/5+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.
1/(pr(1/8,1/4)+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
Open in app ↗Open in split view
Find the equivalent resistance of the circuit.
r4,a,0,4
ri,a,0,6
ji,a,0,iri/2Find equivalent, Resistance / impedance, nodes a and 0, in DC: req is 2 Ω.
AS2's Example 4.10
Open in app ↗Open in split view
Find the equivalent resistance of the circuit.
r4,a,0,4
rx,0,a,2
j,a,0,2*irxFind equivalent, Resistance / impedance, nodes a and 0, in DC: req 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
Open in app ↗Open in split view
Find the equivalent resistance of the circuit.
r15,a,0,15
e,1,a,4*vrx
r10,1,x,10
rx,x,0,5Find equivalent, Resistance / impedance, nodes a and 0, in DC: req is -7.500 Ω.
HK5's Figure 2-29
Open in app ↗Open in split view
Find the equivalent resistance of the circuit.
e,3,0,1.5*is1
r3,3,2,3
r2,2,0,2
s1,2,1Find equivalent, Resistance / impedance, nodes 1 and 0, in DC: req is 0.6000 Ω. Only the short is renamed, s to s1: a short named bare s would have its current spelled is, which already means something in Python, so version 9 refuses that one name and suggests s1. The dependent value follows it — is1.
HK5's Drill Problem 2-9d
Open in app ↗Open in split view
Find the equivalent resistance of the circuit.
r10,1,2,10
r5,2,3,5
r1,2,0,30
e,1,0,20*ir1Find equivalent, Resistance / impedance, nodes 3 and 0, in DC: req is 20 Ω.
Bo2's Example 3.12
Open in app ↗Open in split view
Find the equivalent resistance of the circuit.
r1,1,0,6
r4,a,0,4
e,a,1,6*ir1Find equivalent, Resistance / impedance, nodes a and 0, in DC: req is 3 Ω.
B11's Example 9.7
Open in app ↗Open in split view
Find the Thévenin equivalent of the circuit, as seen from the R3 resistor.
j,0,1,12
r1,1,0,4
r2,1,a,2Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put a in the first and 0 in the second — the pair of terminals you are looking into.
Choose DC. Via vth we find VTH = 48 V. Via req we find REQ = 6 Ω.
B11's Example 9.11
Open in app ↗Open in split view
Find the Norton equivalent of the circuit, as seen from the RL resistor.
e,1,0,9
r1,1,2,3
r2,2,0,6Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put 2 in the first and 0 in the second — the pair of terminals you are looking into.
Choose DC. Via ino we find INO = 3 A. Via req we find REQ = 2 Ω.
AS2's Practice Problem 4.12
Open in app ↗Open in split view
Find the Norton equivalent of the circuit.
r6,1,0,6
j,0,1,10
r2,x,0,2
e,1,x,2*vxSet Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put x in the first and 0 in the second — the pair of terminals you are looking into.
Choose DC. Via ino we find INO = 10 A. Via req we find REQ = 1 Ω.
B11's Example 9.12
Open in app ↗Open in split view
Find the Norton equivalent of the circuit, as seen from the RL resistor.
r2,1,0,4
r1,1,2,5
j,1,2,10Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put 2 in the first and 0 in the second — the pair of terminals you are looking into.
Choose DC. Via ino we find INO = 5.56 A. Via req we find REQ = 9 Ω.
HK5's Drill Problem 2-8b
Open in app ↗Open in split view
Find the Thévenin equivalent of the circuit.
j,0,2,0.01*v1
r,0,2,20
e,1,2,100Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put 1 in the first and 0 in the second — the pair of terminals you are looking into.
Choose DC. Via vth we find VTH = 125 V. Via req we find REQ = 25 Ω.
HK5's Figure 2-27
Open in app ↗Open in split view
Find the Thévenin equivalent of the circuit.
e,1,0,4
r2,1,2,2'k
r3,2,x,3'k
j,0,2,vx/4000Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put x in the first and 0 in the second — the pair of terminals you are looking into.
Via vth we find VTH = 8 V. Via req we find REQ = 10 kΩ.
B11's Example 9.8
Open in app ↗Open in split view
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.
r1,2,0,6
r2,2,1,4
r3,1,0,2
e,0,1,8Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put 2 in the first and 0 in the second — the pair of terminals you are looking into.
Choose DC. Via vth we get VTH= -4.8 V. Via req we get REQ= 2.4 Ω.
B11's Example 8.6
Open in app ↗Open in split view
Find the Norton equivalent of the circuit.
j1,0,1,6
r1,1,0,3
j2,1,0,10
r2,1,0,6Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put 1 in the first and 0 in the second — the pair of terminals you are looking into.
Choose DC. Via ino we find INO = -4 A. Via req we find REQ = 2 Ω.
Bo2's Drill Exercise 3.12
Open in app ↗Open in split view
Find the Norton equivalent of the circuit.
e,1,0,12
r6,1,2,6
j,0,2,3*ir6
r3,2,0,3Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put 2 in the first and 0 in the second — the pair of terminals you are looking into.
Choose DC. Via ino we find INO = 8 A. Via req we find REQ = 1 Ω.
Bo2's Drill Exercise 3.9
Open in app ↗Open in split view
Find the Thévenin equivalent of the circuit.
j,b,0,10
r1,0,b,1
e,a,0,3*ir1
r6,a,b,6Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put a in the first and b in the second — the pair of terminals you are looking into.
Choose DC. Via vth we find VTH = 24 V. Via req we find REQ = 2.4 Ω.
AS2's Example 4.12
Open in app ↗Open in split view
Find Norton equivalent of the circuit.
rx,0,b,4
e,0,b,10
r5,0,a,5
j,0,a,2*irxSet Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put a in the first and b in the second — the pair of terminals you are looking into.
Choose DC. Via ino we find INO = 7 A. Via req we find REQ = 5 Ω.
B11's Example 9.9
Open in app ↗Open in split view
Find the Thévenin equivalent of the circuit, as seen from the RL resistor.
e,1,0,72
r1,1,b,6
r2,1,a,12
r3,0,b,3
r4,0,a,4Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put b in the first and a in the second — the pair of terminals you are looking into.
Choose DC. Via vth we find VTH = 6 V. Via req we find REQ = 5 Ω.
AS2's Example 4.11
Open in app ↗Open in split view
Find the Norton equivalent of the circuit.
j,0,2,2
e,4,0,12
r1,2,4,4
r2,2,3,8
r3,0,1,8
r4,3,1,5Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put 3 in the first and 1 in the second — the pair of terminals you are looking into.
Choose DC. Via ino we find INO = 1 A. Via req we find REQ = 4 Ω.
HK5's Drill Problem 2-8a
Open in app ↗Open in split view
Find the Thévenin equivalent of the circuit.
e1,1,0,100
r2,1,2,20
j,0,2,4
r1,2,3,10
e5,3,4,50Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put 4 in the first and 0 in the second — the pair of terminals you are looking into.
Choose DC. Via vth we find VTH = 130 V. Via req we find REQ = 30 Ω.
B11's Example 8.7
Open in app ↗Open in split view
Find Norton equivalent of the circuit.
j7,0,1,7
j3,1,0,3
r1,1,0,4
j4,0,1,4Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put 1 in the first and 0 in the second — the pair of terminals you are looking into.
Choose DC. Via ino we find INO = 8 A. Via req we find REQ = 4 Ω.
AS2's Practice Problem 4.9
Open in app ↗Open in split view
Find the Thévenin equivalent of the circuit.
e,1,0,6
r5,1,2,5
rx,2,3,3
j,0,2,1.5*irx
r4,3,0,4Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put 3 in the first and 0 in the second — the pair of terminals you are looking into.
Choose DC. Via vth we find VTH = 5.33 V. Via req we find REQ = 0.44 Ω.
AS2's Example 4.9
Open in app ↗Open in split view
Find the Thévenin equivalent of the circuit.
j,b,0,5
rx,0,b,4
r1,0,1,2
r2,1,b,6
r3,1,a,2
e,1,0,2*vrxSet Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put a in the first and b in the second — the pair of terminals you are looking into.
Choose DC. Via vth we find VTH = 20 V. Via req we find REQ = 6 Ω.
Bo2's Drill Exercise 3.8
Open in app ↗Open in split view
Find the Thévenin equivalent of the circuit.
j,0,1,3
r1,0,1,1
r6,1,2,6
r10,1,3,10
r8,2,0,8
r2,2,3,2Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put 3 in the first and 0 in the second — the pair of terminals you are looking into.
Choose DC. Via vth we find VTH = 2 V. Via req we find REQ = 4 Ω.
B11's Example 9.13
Open in app ↗Open in split view
Find Norton equivalent of the shaded part of the circuit.
e,1,0,7
r1,1,a,4
j,a,0,8
r2,a,0,6Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put a in the first and 0 in the second — the pair of terminals you are looking into.
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)
as it comesOpen in app ↗Open in split viewwith a symbolOpen in app ↗Open in split view
Find the Norton equivalent of the circuit.
Since you do not know beforehand that this is a tricky problem, you go for the usual:
r2,0,b,2
r8,0,2,8
r3,2,a,3
r1,1,a,1
e,1,0,1
j,2,b,3*ir8Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put a in the first and b in the second — the pair of terminals you are looking into.
Run it in DC and it is refused: Could not solve the system of equations. This often means there is a division by zero somewhere.
Symbulator's own advice is to try again with a symbolic value, which is what we do. We chose to use x instead of 3 in the dependent source.
r2,0,b,2
r8,0,2,8
r3,2,a,3
r1,1,a,1
e,1,0,1
j,2,b,x*ir8Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put a in the first and b in the second — the pair of terminals you are looking into.
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.
The equivalent is already on screen, written in terms of x, so there is nothing to run again. Put ino in the Evaluate card with x = 3 in its Conditions box: INO = 1 A. Ask for req the same way and the answer is infinite, the denominator being zero at that value.
This means the equivalent resistance is, for practical purposes, infinite. Your idea of fun, right?
Bo2's Drill Exercise 3.13 (Tricky)
Open in app ↗Open in split view
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.
ei,1,b,6
ed,b,0,2*x*ir2
r1,1,0,6
r2,0,a,2
jd,a,0,3*x*ir1Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put a in the first and b in the second — the pair of terminals you are looking into.
Exploring the answers, we see that the denominator of the expression for req, (x-1), results in a division by zero at x = 1.
Put ino in the Evaluate card with x = 1 in its Conditions box: INO = -3 A. Ask for req the same way and the answer is infinite, so REQ is undefined at that value.
TR5's Exercise 4-6 (Symbolic)
input resistanceOpen in app ↗Open in split viewoutput equivalentOpen in app ↗Open in split view
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.
ei,3,0,vs
rf,3,2,rf
ro,2,t,ro
ed,2,0,μ*vrfThe answer you want is rei, in Results.
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.
ei,3,0,vs
rf,3,2,rf
ro,2,t,ro
ed,2,0,μ*vrfSet Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put t in the first and 0 in the second — the pair of terminals you are looking into.
The answers you want are vth and req, in Results.
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)
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Find the Thévenin equivalent as seen by the load.
The dependent source's value is the difference between two node voltages, so version 9 writes that difference straight into the value — va-vb, exactly as Lesson 3 describes. There is nothing to define beforehand.
Now I run the th script, with this circuit description:
ei,a,0,vs
ed,1,0,μ*(va-vb)
ro,b,1,roSet Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put b in the first and 0 in the second — the pair of terminals you are looking into.
The answers you want are vth and req, in Results.
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
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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:
e,1,0,32.
r4,1,2,4
r12,2,0,12
j,0,2,2
r1,2,3,1Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put 3 in the first and 0 in the second — the pair of terminals you are looking into.
The answers you want are vth and req, in Results.
Ask Evaluate for:
vth/(req+6)
vth/(req+16)
vth/(req+36)Choose DC.
The answer, {30.,4.,3.,1.5,.75}, is correct.
Bo2's Example 3.10
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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:
e,3,0,3
r31,3,1,1/2
r10,1,0,1/2
r12,1,2,1/4
j,2,0,v1/2Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put 2 in the first and 0 in the second — the pair of terminals you are looking into.
The answers you want are ino and req, in Results.
Ask Evaluate for:
vth*(1/4)/(req+1/4)
vth/(req+1/4)Choose DC.
The book gives the answers as fractions. We get it right: {21/8, 4/9, 21/50, 42/25}.
RM3's Example 9-7
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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:
e,1,0,15.
r1,1,2,6'k
j,0,2,5'm
r2,2,0,2'kSet Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put 2 in the first and 0 in the second — the pair of terminals you are looking into.
The answers you want are ino and req, in Results.
Ask Evaluate for:
vth/(req+0)
vth/(req+2000)
vth/(req+5000)Choose DC.
The answer, {.0075, 1500., .0075, .00321, .00173}, is correct.
Bo2's Example 3.5
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Find the Norton equivalent of the circuit external to the 1Ω resistor. Then determine the voltage drop across this 1Ω resistor.
My solution:
e,1,0,24
r12,1,2,12
r20,2,0,4
r23,2,3,4
r34,3,4,2
j,4,0,3
r40,4,0,5Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put 3 in the first and 0 in the second — the pair of terminals you are looking into.
The answers you want are ino and req, in Results.
Ask Evaluate for:
vth*1/(req+1)Choose DC.
The answer, {-9/7,7/2,-1}, is correct.
RM3's Example 9-13
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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.
r1,0,1,240.
e1,2,1,96
r2,0,3,200
e2,3,2,40
r3,0,4,800
e3,2,4,80Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put 2 in the first and 0 in the second — the pair of terminals you are looking into.
The answers you want are vth and req, in Results.
Ask Evaluate for:
vth/(req+192)Choose DC.
The answer, {28.8,96.,.1}, is correct.
Bo2's Example 3.7
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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:
e,1,0,20
r6,1,2,6
r1,1,3,1
r2,3,2,2
j32,3,2,15
j30,3,0,15Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put 2 in the first and 0 in the second — the pair of terminals you are looking into.
The answers you want are vth and req, in Results.
Ask Evaluate for:
vth*3/(req+3)
vth*6/(req+6.)Choose DC.
The answer we find, {30,2,18,22.5}, is correct.
Bo2's Drill Exercise 3.7
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Find the Thévenin equivalent for the circuit left of a-b. Then find the voltage v.
My answer:
e,1,0,8
r3,1,2,3
r12,1,3,12
r6,2,0,6
r2,2,3,2Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put 3 in the first and 0 in the second — the pair of terminals you are looking into.
The answers you want are vth and req, in Results.
Ask Evaluate for:
vth*5/(req+5)Choose DC.
The answer, {6,3,3.75}, is correct.
RM3's Practice Problem 9.5
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My solution:
e1,1,0,35.
r1,1,2,15'k
r2,2,3,60'k
e2,3,0,70
r3,2,4,30'kSet Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put 4 in the first and 0 in the second — the pair of terminals you are looking into.
The answers you want are ino and req, in Results.
Ask Evaluate for:
ino*req/(req+0)
ino*req/(req+1e4)
ino*req/(req+5e4)
ino*req/(req+1e5)One line at a time, as before.
Choose DC.
The answer, {.001,42000,.001,.000808,.000457,.000296}, is correct.
Power transfer problems
AS2's Example 4.13
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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:
e,1,0,12
r6,1,2,6
r12,2,0,12
r3,2,3,3
j,0,3,2
r2,3,4,2Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put 4 in the first and 0 in the second — the pair of terminals you are looking into.
The answers you want are req and pmax, in Results.
The calculator versions wrap this in approx to get a decimal; version 9 does it through Rounding — approx to n digits, with n = 3 here.
Choose DC. 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
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Find the RL value for maximum power transfer and the maximum power transferred.
ei,1,0,9
rx,1,2,2
r1,2,4,1
ed,4,0,3*vrx
r4,2,3,4Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put 3 in the first and 0 in the second — the pair of terminals you are looking into.
The answers you want are req and pmax, in Results.
The calculator versions wrap this in approx to get a decimal; version 9 does it through Rounding — approx to n digits, with n = 3 here.
Choose DC. The logic of this problem is identical to the previous one. The answer is {4.22,2.901}.
B11's Example 9.15
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My one-line solution to all three questions is presented below.
j,0,1,10'm
rs,1,0,40'kSet Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put 1 in the first and 0 in the second — the pair of terminals you are looking into.
The answers you want are req and pmax, in Results.
Ask Evaluate for:
vth^2*68000./(req+68000.)^2
vth^2*8200./(req+8200.)^2Choose DC. 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. 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
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Find the RL value for maximum power transfer and the maximum power transferred.
j,2,0,6
r2,2,0,10
r1,2,3,3
r3,0,1,2
e,3,4,68Set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton. Two node boxes appear: put 4 in the first and 1 in the second — the pair of terminals you are looking into.
The answers you want are req and pmax, in Results.
The calculator versions wrap this in approx to get a decimal; version 9 does it through Rounding — approx to n digits, with n = 3 here.
Choose DC. The answer is {15., 273.07}. Let's now see one that is a little different.