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Symbulator 9 9

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, and Thévenin and Norton equivalents.

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, and Thévenin and Norton equivalents.

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 gives 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

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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.

Circuit Description

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
s4,3,5

Choose Solve circuit and DC, then run it. The current through each short is given in:

is1 = -2 A, is2 = 3 A, is3 = -8 A and is4 = -0.5 A.

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

4.2Finding the equivalent resistance

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

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

AS2's Practice Problem 2.15

Solution

Circuit Description

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

Solve it in DC. Because a source sits between the two nodes we care about, the equivalent resistance is one of the answers already: the resistance seen by the source, re, is 40 Ω.

The current i is the current through r13: 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.

Set Type of analysis to Find equivalent, and a second menu appears, Type of equivalent, whose first entry is Resistance / impedance. Two node boxes appear with it, for the terminals you are measuring between. Symbulator does the 1 A trick internally and reports the answer.

B11's Example 8.29

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Calculate the equivalent resistance of the circuit shown.

B11's Example 8.29

Solution

My solution: I label the nodes, then describe the circuit.

Circuit Description

r4,0,a,4
r2,0,b,2
r6,a,b,6
rb,a,c,3
ra,b,c,3

Set 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 answer, headed equivalent resistance:

Req = 2.89 Ω

This is correct.

What counts as passive

That example was made of resistors only. Resistance / impedance 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 it is found in exactly the same way, provided the dependent sources are described properly.

4.3Finding a Thévenin / Norton equivalent

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, V_TH), then a second with a short between those nodes for the current through it (the Norton current, I_NO). R_EQ is then V_TH/I_NO. That is the manual way.

An easier way is to let Symbulator do exactly that automatically: set Type of analysis to Find equivalent and Type of equivalent to Thévenin / Norton, and give it the two nodes.

RM3's Practice Problem 9-4

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Find the Thévenin and Norton equivalents of the circuit.

RM3's Practice Problem 9-4

Solution

This is my circuit description:

Circuit Description

e,1,0,3.3
r1,1,2,66
r2,2,0,24

Set 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.

Results replaces the node-and-element listing with a single group headed Thévenin / Norton equivalent, holding four answers:

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

Here they read vth = 880 mV, ino = 50 mA, Req = 17.6 Ω and pmax = 11 mW. The first three are what you came for; the fourth is 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 R_L problem. A typical one goes like this: "First, reduce the circuit, as seen by resistor R_L, to its Thévenin or Norton equivalent. Then, find the value of the voltage drop, current and/or power consumed in the load resistor R_L if its value is (whatever) ohms."

Since this is such a typical problem, Symbulator makes provision for it. When the equivalent asked for is Thévenin / Norton, a question appears under the two node boxes: Are you running a problem with a load connected to this equivalent circuit? Tick it, and Results adds three answers under the four, each an expression in terms of the variable load, the value of the load resistor:

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

To read one of them at a given load, ask Evaluate for its name and put the load's value in the Conditions box, as in load = 2. Untick the question and the four answers are all you see.

B11's Example 9.6

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

B11's Example 9.6

Solution

First we find the Thévenin equivalent.

Circuit Description

e,1,0,9
r1,1,2,3
r2,2,0,6

Set 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. Tick the question about the load. Choose DC and run it.

Results gives:

Thévenin voltage

\[v_{th} = 6\ \mathrm{V}\]

equivalent resistance

\[R_{eq} = 2\ \Omega\]

and, because the load question is ticked:

current in load

\[i_{rl} = \dfrac{6}{load + 2}\]

Correct. Now we find the current in the load for each value.

Here we use Rounding — approx to n digits, with n = 2, in Settings, so the answers read as the book prints them; exact, the third would come back as 1/17. Then ask Evaluate for irl, with the load's value in the Conditions box:

Evaluate

irl

Conditions

load = 2

Then load = 10, then load = 100.

The answers are 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 R_L equals the R_EQ of the equivalent. Symbulator 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, once the load question is ticked.

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, Symbulator writes that circuit for you: with the load question ticked, a button under the equivalent's answers drops the Norton equivalent, load and all, into the input fields, ready to be added to. When a load is not the only thing attached covers it, with a worked example — optional reading, and nothing else in this lesson depends on it.

4.4Instructive solved examples

Practice problems for resistive circuits

B11's Example 8.29

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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.

Circuit Description

r4,0,a,4
r2,0,b,2
r6,a,b,6
rb,a,c,3
ra,b,c,3

Set 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

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Find the equivalent resistance of the circuit.

Circuit Description

rac,a,c,6
rad,a,0,9
rab,a,b,6
rcd,c,0,9
rbc,b,c,6
rbd,b,0,9

Find equivalent, Resistance / impedance, nodes a and c, in DC: req is 3.273 Ω.

The problem names the four nodes a, b, c and d. Symbulator needs one of them to be the reference node, so d is called 0 here. 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 pr.

Evaluate

4+pr(1+5,2+pr(6,3))+8

Evaluate gives 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 pr.

Evaluate

2+pr(6,3+pr(4,4+5+3))+1

Evaluate gives 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 pr.

Evaluate

1/(pr(1/6,1/5+pr(1/8,1/12)))

Evaluate gives 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 pr.

Evaluate

1/(pr(1/8,1/4)+pr(1/2,1/12+1/6))

Evaluate gives the equivalent conductance: 4 S.

Examples with dependent sources

Bo2's Drill Problem 3.14

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Find the equivalent resistance of the circuit.

Circuit Description

r4,a,0,4
ri,a,0,6
ji,a,0,iri/2

Find equivalent, Resistance / impedance, nodes a and 0, in DC: req is 2 Ω.

AS2's Example 4.10

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Find the equivalent resistance of the circuit.

Circuit Description

r4,a,0,4
rx,0,a,2
j,a,0,2*irx

Find 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

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Find the equivalent resistance of the circuit.

Circuit Description

r15,a,0,15
e,1,a,4*vrx
r10,1,x,10
rx,x,0,5

Find equivalent, Resistance / impedance, nodes a and 0, in DC: req is -7.500 Ω.

HK5's Figure 2-29

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Find the equivalent resistance of the circuit.

Circuit Description

e,3,0,1.5*is
r3,3,2,3
r2,2,0,2
s,2,1

Find equivalent, Resistance / impedance, nodes 1 and 0, in DC: req is 0.6000 Ω.

HK5's Drill Problem 2-9d

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Find the equivalent resistance of the circuit.

Circuit Description

r10,1,2,10
r5,2,3,5
r1,2,0,30
e,1,0,20*ir1

Find equivalent, Resistance / impedance, nodes 3 and 0, in DC: req is 20 Ω.

Bo2's Example 3.12

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Find the equivalent resistance of the circuit.

Circuit Description

r1,1,0,6
r4,a,0,4
e,a,1,6*ir1

Find equivalent, Resistance / impedance, nodes a and 0, in DC: req is 3 Ω.

B11's Example 9.7

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

Circuit Description

j,0,1,12
r1,1,0,4
r2,1,a,2

Set 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.

In DC, vth gives V_TH = 48 V and req gives R_EQ = 6 Ω.

B11's Example 9.11

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

Circuit Description

e,1,0,9
r1,1,2,3
r2,2,0,6

Set 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.

In DC, ino gives I_NO = 3 A and req gives R_EQ = 2 Ω.

AS2's Practice Problem 4.12

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Find the Norton equivalent of the circuit.

Circuit Description

r6,1,0,6
j,0,1,10
r2,x,0,2
e,1,x,2*vx

Set 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.

In DC, ino gives I_NO = 10 A and req gives R_EQ = 1 Ω.

B11's Example 9.12

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

Circuit Description

r2,1,0,4
r1,1,2,5
j,1,2,10

Set 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.

In DC, ino gives I_NO = 5.56 A and req gives R_EQ = 9 Ω.

HK5's Drill Problem 2-8b

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Find the Thévenin equivalent of the circuit.

Circuit Description

j,0,2,0.01*v1
r,0,2,20
e,1,2,100

Set 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.

In DC, vth gives V_TH = 125 V and req gives R_EQ = 25 Ω.

HK5's Figure 2-27

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Find the Thévenin equivalent of the circuit.

Circuit Description

e,1,0,4
r2,1,2,2'k
r3,2,x,3'k
j,0,2,vx/4000

Set 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.

In DC, vth gives V_TH = 8 V and req gives R_EQ = 10 kΩ.

B11's Example 9.8

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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.

Circuit Description

r1,2,0,6
r2,2,1,4
r3,1,0,2
e,0,1,8

Set 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.

In DC, vth gives V_TH= -4.8 V and req gives R_EQ= 2.4 Ω.

B11's Example 8.6

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Find the Norton equivalent of the circuit.

Circuit Description

j1,0,1,6
r1,1,0,3
j2,1,0,10
r2,1,0,6

Set 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.

In DC, ino gives I_NO = -4 A and req gives R_EQ = 2 Ω.

Bo2's Drill Exercise 3.12

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Find the Norton equivalent of the circuit.

Circuit Description

e,1,0,12
r6,1,2,6
j,0,2,3*ir6
r3,2,0,3

Set 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.

In DC, ino gives I_NO = 8 A and req gives R_EQ = 1 Ω.

Bo2's Drill Exercise 3.9

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Find the Thévenin equivalent of the circuit.

Circuit Description

j,b,0,10
r1,0,b,1
e,a,0,3*ir1
r6,a,b,6

Set 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.

In DC, vth gives V_TH = 24 V and req gives R_EQ = 2.4 Ω.

AS2's Example 4.12

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Find Norton equivalent of the circuit.

Circuit Description

rx,0,b,4
e,0,b,10
r5,0,a,5
j,0,a,2*irx

Set 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.

In DC, ino gives I_NO = 7 A and req gives R_EQ = 5 Ω.

B11's Example 9.10 (Hidden source)

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Find the Thévenin equivalent of the circuit.

My solution is below:

Circuit Description

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

Set 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.

In DC, vth gives V_TH = -3 V and req gives R_EQ = 2 kΩ.

B11's Example 9.9

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

Circuit Description

e,1,0,72
r1,1,b,6
r2,1,a,12
r3,0,b,3
r4,0,a,4

Set 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.

In DC, vth gives V_TH = 6 V and req gives R_EQ = 5 Ω.

AS2's Example 4.11

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Find the Norton equivalent of the circuit.

Circuit Description

j,0,2,2
e,4,0,12
r1,2,4,4
r2,2,3,8
r3,0,1,8
r4,3,1,5

Set 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.

In DC, ino gives I_NO = 1 A and req gives R_EQ = 4 Ω.

HK5's Drill Problem 2-8a

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Find the Thévenin equivalent of the circuit.

Circuit Description

e1,1,0,100
r2,1,2,20
j,0,2,4
r1,2,3,10
e5,3,4,50

Set 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.

In DC, vth gives V_TH = 130 V and req gives R_EQ = 30 Ω.

B11's Example 8.7

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Find Norton equivalent of the circuit.

Circuit Description

j7,0,1,7
j3,1,0,3
r1,1,0,4
j4,0,1,4

Set 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.

In DC, ino gives I_NO = 8 A and req gives R_EQ = 4 Ω.

AS2's Practice Problem 4.9

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Find the Thévenin equivalent of the circuit.

Circuit Description

e,1,0,6
r5,1,2,5
rx,2,3,3
j,0,2,1.5*irx
r4,3,0,4

Set 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.

In DC, vth gives V_TH = 5.33 V and req gives R_EQ = 0.44 Ω.

AS2's Example 4.9

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Find the Thévenin equivalent of the circuit.

Circuit Description

j,b,0,5
rx,0,b,4
r1,0,1,2
r2,1,b,6
r3,1,a,2
e,1,0,2*vrx

Set 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.

In DC, vth gives V_TH = 20 V and req gives R_EQ = 6 Ω.

Bo2's Drill Exercise 3.8

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Find the Thévenin equivalent of the circuit.

Circuit Description

j,0,1,3
r1,0,1,1
r6,1,2,6
r10,1,3,10
r8,2,0,8
r2,2,3,2

Set 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.

In DC, vth gives V_TH = 2 V and req gives R_EQ = 4 Ω.

B11's Example 9.13

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Find Norton equivalent of the shaded part of the circuit.

Circuit Description

e,1,0,7
r1,1,a,4
j,a,0,8
r2,a,0,6

Set 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.

In DC, ino gives I_NO = -6.25 A and req gives R_EQ = 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)

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Find the Norton equivalent of the circuit.

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

Circuit Description

r2,0,b,2
r8,0,2,8
r3,2,a,3
r1,1,a,1
e,1,0,1
j,2,b,3*ir8

Set 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.

Circuit Description

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

Tricky, with a symbolOpen in app ↗Open in split view

Set 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,

equivalent resistance

\[R_{eq} = \dfrac{9x - 35}{4(x - 3)}\]

will divide by zero at \(x = 3\) — the very value we replaced.

The equivalent is in terms of x, so nothing needs running again. Put ino in Evaluate with x = 3 in its Conditions box: I_NO = 1 A. Ask for req the same way: infinite, the denominator being zero.

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

Bo2's Drill Exercise 3.13 (Tricky)

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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.

Circuit Description

ei,1,b,6
ed,b,0,2*x*ir2
r1,1,0,6
r2,0,a,2
jd,a,0,3*x*ir1

Set 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.

In the answers, the denominator of the expression for req, (x-1), divides by zero at x = 1.

Put ino in the Evaluate card with x = 1 in its Conditions box: I_NO = -3 A. Ask for req the same way and the answer is infinite, so R_EQ is undefined at that value.

TR5's Exercise 4-6 (Symbolic)

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

Circuit Description

ei,3,0,vs
rf,3,2,rf
ro,2,t,ro
ed,2,0,μ*vrf

The answer you want is rei, in Results.

It reads 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 R_L.

Circuit Description

ei,3,0,vs
rf,3,2,rf
ro,2,t,ro
ed,2,0,μ*vrf

Symbolic, output equivalentOpen in app ↗Open in split view

Set 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.

Results gives

Thévenin voltage

\[v_{th} = \dfrac{vs\,\mu}{\mu + 1}\]

equivalent resistance

\[R_{eq} = ro\]

which is correct, as can be seen in the textbook's answers for v_T and R_T, 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, written straight into the value as va-vb, as Lesson 3 describes.

Now I find the Thévenin equivalent of this circuit description:

Circuit Description

ei,a,0,vs
ed,1,0,μ*(va-vb)
ro,b,1,ro

Set 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 in Results,

Thévenin voltage

\[v_{th} = \dfrac{vs\,\mu}{\mu + 1}\]

equivalent resistance

\[R_{eq} = \dfrac{ro}{\mu + 1}\]

are correct, as can be seen by comparing them to those in the book:

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 R_L = 6, 16 and 36Ω.

My solution:

Circuit Description

e,1,0,32.
r4,1,2,4
r12,2,0,12
j,0,2,2
r1,2,3,1

Set 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. Tick the question about the load, choose DC and run it.

The answers you want are vth and req, in Results, and under them the load answers.

Ask Evaluate for irl, with the load in the Conditions box:

Evaluate

irl

Conditions

load = 6

Then load = 16, then load = 36.

The answer is vth = 30 V and req = 4 Ω, and irl is 3 A for 6 Ω, 1.5 A for 16 Ω and .75 A for 36 Ω. This 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 solution:

Circuit Description

e,3,0,3
r31,3,1,1/2
r10,1,0,1/2
r12,1,2,1/4
j,2,0,v1/2

Set 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. Tick the question about the load, choose DC and run it.

The answers you want are ino and req, in Results, and under them the load answers.

Ask Evaluate for vrl, then for irl, with the load in the Conditions box each time:

Evaluate

vrl

Conditions

load = 1/4

The book gives the answers as fractions. We get it right: ino = 21/8 A and req = 4/9 Ω, and at that load vrl = 21/50 V and irl = 42/25 A.

RM3's Example 9-7

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Find the Norton equivalent of the circuit external to R_L. Then determine the load current I_L when R_L = 0 Ω, 2 kΩ and 5 kΩ. My solution:

Circuit Description

e,1,0,15.
r1,1,2,6'k
j,0,2,5'm
r2,2,0,2'k

Set 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. Tick the question about the load, choose DC and run it.

The answers you want are ino and req, in Results, and under them the load answers.

Ask Evaluate for irl, with the load in the Conditions box:

Evaluate

irl

Conditions

load = 0

Then load = 2000, then load = 5000.

The answer is ino = .0075 A and req = 1500 Ω, and irl is .0075 A for 0 Ω, .00321 A for 2 kΩ and .00173 A for 5 kΩ. This 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:

Circuit Description

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

Set 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. Tick the question about the load, choose DC and run it.

The answers you want are ino and req, in Results, and under them the load answers.

Ask Evaluate for vrl, with the load in the Conditions box:

Evaluate

vrl

Conditions

load = 1

The answer is ino = -9/7 A and req = 7/2 Ω, and vrl in the 1 Ω load is -1 V. This 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 R_L.

I don't know Millman's Theorem, but we can find the Thévenin equivalent with Symbulator.

Circuit Description

r1,0,1,240.
e1,2,1,96
r2,0,3,200
e2,3,2,40
r3,0,4,800
e3,2,4,80

Set 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. Tick the question about the load, choose DC and run it.

The answers you want are vth and req, in Results, and under them the load answers.

Ask Evaluate for irl, with the load in the Conditions box:

Evaluate

irl

Conditions

load = 192

The answer is vth = 28.8 V and req = 96 Ω, and irl in the 192 Ω load is .1 A. This 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:

Circuit Description

e,1,0,20
r6,1,2,6
r1,1,3,1
r2,3,2,2
j32,3,2,15
j30,3,0,15

Set 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. Tick the question about the load, choose DC and run it.

The answers you want are vth and req, in Results, and under them the load answers.

Ask Evaluate for vrl, with the load in the Conditions box:

Evaluate

vrl

Conditions

load = 3

Then load = 6..

The answer is vth = 30 V and req = 2 Ω, and vrl is 18 V for 3 Ω and 22.5 V for 6 Ω. This 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:

Circuit Description

e,1,0,8
r3,1,2,3
r12,1,3,12
r6,2,0,6
r2,2,3,2

Set 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. Tick the question about the load, choose DC and run it.

The answers you want are vth and req, in Results, and under them the load answers.

Ask Evaluate for vrl, with the load in the Conditions box:

Evaluate

vrl

Conditions

load = 5.

The answer is vth = 6 V and req = 3 Ω, and vrl in the 5 Ω load is 3.75 V. This is correct.

RM3's Practice Problem 9.5

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My solution:

Circuit Description

e1,1,0,35.
r1,1,2,15'k
r2,2,3,60'k
e2,3,0,70
r3,2,4,30'k

Set 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. Tick the question about the load, choose DC and run it.

The answers you want are ino and req, in Results, and under them the load answers.

Ask Evaluate for irl, with the load in the Conditions box:

Evaluate

irl

Conditions

load = 0

Then load = 1e4, then load = 5e4, then load = 1e5.

The answer is ino = .001 A and req = 42000 Ω, and irl is .001 A for 0 Ω, .000808 A for 10 kΩ, .000457 A for 50 kΩ and .000296 A for 100 kΩ. This is correct.

Power transfer problems

AS2's Example 4.13

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

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

Circuit Description

e,1,0,12
r6,1,2,6
r12,2,0,12
r3,2,3,3
j,0,3,2
r2,3,4,2

Set 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.

Here we use Rounding — approx to n digits, with n = 3.

Choose DC. The maximum transfer of power occurs when the load is req = 9Ω, and the power transferred is then pmax = 13.44W. Now let's solve another one.

AS2's Practice Problem 4.13

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

Circuit Description

ei,1,0,9
rx,1,2,2
r1,2,4,1
ed,4,0,3*vrx
r4,2,3,4

Set 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.

Here we use Rounding — approx to n digits, with n = 3.

Choose DC. The logic of this problem is identical to the previous one. The answer is req = 4.22 Ω and pmax = 2.901 W.

B11's Example 9.15

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My solution to all three questions is presented below.

Circuit Description

j,0,1,10'm
rs,1,0,40'k

Set 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. Tick the question about the load, choose DC and run it.

The answers you want are req and pmax, in Results, and under them the load answers.

Ask Evaluate for prl, with the load in the Conditions box:

Evaluate

prl

Conditions

load = 68000.

Then load = 8200..

The answers we obtain, req = 40000 Ω and pmax = 1 W, then prl = .93 W and .57 W, are 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 prl, the power delivered by the equivalent to the load as a function of load, 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 R_L value for maximum power transfer and the maximum power transferred.

Circuit Description

j,2,0,6
r2,2,0,10
r1,2,3,3
r3,0,1,2
e,3,4,68

Set 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.

Here we use Rounding — approx to n digits, with n = 3.

Choose DC. The answer is req = 15 Ω and pmax = 273.07 W. Let's now see one that is a little different.