circuit analysis with symbulator - volume 1

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Praise for Symbulator

“The masterpiece of TI-89 programming.” Alex

Astashyn. EE+CoE+CoS major / Polytechnic University in

Brooklyn / New York, USA (Born in Russia)

“Symbulator is and always will be the best program for

TI-89.” Charles Ware. EE major / Columbus, Ohio, USA

“Roberto, your program Symbulator is by itself a reason

to buy the calc. It's a masterpiece of programming.”

Pepe Iborra. Telecom / Spain

“The program is really a masterpiece. It gives you

everything you need, very accurate, very fast. It's really

what every Electrical Engineer dreams of. It's so easy to

use and it does everything! … I'm really amazed. Keep

up the good work.” Nikolaos Trichakis. EE major /

Aristoteles University of Thessaloniki / Greece

“Brillant work by Mr. Pérez-Franco! Check it out!” Arne

Harstad. EE major / Norges tekniske-

naturvitenskapelige Universitet / Norway

“Congratulations for your program: it's simply fantastic.

Symbulator is worth the cost of a TI-89!” Pier Giorgio

Raponi. Engineering student / Pisa, Italy

“I wish I had it while going through school.” Joe Riel.

BSEE & Creator of Syrup (a symbolic circuit simulator

for MapleV) / UC / Irvine CA, USA

“Symbulator is a wonderful tool.” Reinhard Willinski. EE

major / Germany

“It runs beautifully in TI-89.” Harold Martínez. CoE

major / City College / New York, USA

“I find Symbulator to be a phenomenal program. You

did a great job. The more I use it, the more I want to

use it.” Michael. EE/CoE student / San Diego State

University / USA

“Totally awesome! Thanks a million.” Chris Spencer. EE

major / University of Florida, USA

“You've made an excellent program. Symbulator is

really the best circuit simulator ever made for a

calculator.” Rui Sebastiao. EE major / University of

Coimbra / Portugal

“Your program is absolutely great, and I use it all the

time, not without enjoyment .” Carmel Nachmany. EE

major / Ben-Gurion University of the Negev / Israel

“Symbulator is a great tool to check answers.” Brad

Norman. EE junior / Montana State University /

Bozeman, USA

“Congratulations for your software Symbulator. It's

great!” Eduardo Silveira. EE major / Brazil

“Thanks for an exquisite, free program!” Chandler

Sorenson. Computer Engineering / University of

Arizona / Tucson, USA

“Symbulator is probably the most useful program ever

made for a TI calculator. I am an electronics student

and even paid for the EE*Pro software, yet I find myself

using Symbulator more because it is much better at

what it does.” Josh Cunningham. Electronics major

“Symbulator, by far, has been the best tool for EE I have

come across yet. ... It is truly an impressive piece of

work. You should be proud. It certainly is helping me to

focus on my studying of EE instead of trying to

remember all the math. Again, good job!” Jim Wachtel.

EE / Process Control & Electrical Co. / St. Louis,

Missouri, USA

“Your programs are the best thing I have ever seen on

the TI website. I can tell that you are making a lot of

engineering college students happy.” Ade George. EE

major / Morgan State University / USA

“I love this program! Trying to learn how to use it has

increased my knowledge of actual circuit analysis,

which helps me out in the long run. Your software is

great. I'm taking my first ELEN class and this software

not only confirms my work, but helps me understand

the circtuits as well.” Josh Earley. Computer

Engineering major / Texas A&M University / USA

“I will, one day in the near future, buy a TI-89 just to use

your Symbulator.” Ron Ross

“This program is the premier circuit solver for

calculators. Yes, there are others, but none are better.This program is similar to SPICE or PSpice in the way it

is used, and it is even easier when it comes to

dependent sources and biport simulations. Once a

circuit is solved, it has ALL the information the user

would want: node voltages, element currents, element

power usage, voltage drops across elements. It solves

symbolically, unlike almost any other circuit simulator

(even PC based). This program is, hands down, the best

program for circuit analysis that I have seen. … I could

not possibly praise this program well enough. It's hard



to believe that Mr. Perez-Franco is letting the world use

his fantastic program free of charge. Excellent job!” Jay

Myers. EE major / University of Texas at Arlington /

Texas, USA

“I have been using your wonderful program for about

four months now. The freedom it affords is amazing. ...

I use Symbulator extensively for analysis and design in

EE and find it easier to use (and really a better learningtool because you must understand what you are doing)

than PSpice. I am truly impressed with the efficiency of

the program and the simple fact that I have had no

problems with the program itself.” Douglas Fisher.

BioMedical Eng / Math Wright State University / Ohio,

USA

“I think your program is absolutely great. ... Your

Symbulator is the neatest thing to have around. ... Good

job, Roberto. ... What a great program by such a

gentleman.” Tim Hutcheson. BS in Comp Sci / Institute

for Human and Machine Cognition / Florida, USA

“Let me congratulate you on Symbulator. Amazing job!

I have EE200 and EE*Pro, but your app seems to far

surpass the capabilities of these two apps.” Ralph

McCarthy. EE major / Minneapolis / Minnesota, USA

“I think your program is one of the best programs out

there in the community, and in my personal Top 10

chart, it's in the top. It was the reason I bought the ti-89

for in the first place. I have distributed the Symbulator

to everybody in my class and they've been enjoying it a

lot when solving EE problems.” Melnic Atrav. EE major

“Hey, hats off to the programmer of this one, and for

the generosity to share it so freely. I guess it's too bad I

bought EE*Pro for $50 less than a week ago, huh?”

Brandon Newsom. High school student / Newport

News / Virginia, USA

“A million thanks for your program. You don't know

how fine I feel every time I solve a circuit with paper

and pencil, and I get the same answers I got with

Symbulator. It is much easier to learn this way. Truly, I

believe you deserve a mention in circuits books.”

Gabriel Florit. EE major / Dordt College, Sioux Center /

Iowa, USA

“Thank you for your brilliant program. ... I have used your program quite a bit throughout my studies. ... I

have to commend you on such a comprehensive and

impressive program. In my mind, it is the most

important program to have on a calculator for those

taking EE courses. ... Thanks for your time, and for your

Symbulator, Roberto.” Albert Meng. System Science

Mathematics major / Washington University / St. Louis,

USA

“It is a great way to check your answers in homework

and on a test. Try it out for yourself: that is the only

way you will figure out what Symbulator is capable of. I

am still amazed about how well Roberto's program

works...” Al Charpentier. EE major / Pennsylvania State

University / Pennsylvania, USA

“We all know about SPICE and what a lifesaver it is in

our design projects, but the learning curve has alwaysbeen a steep one. In my quest for the best EE

applications for design and analysis, I came across

Roberto Perez-Franco's Symbolic Circuit Simulator

(Symbulator) for the TI-89. I have been using

Symbulator for almost a year now, and believe me, it is

a must for every electrical engineering student. Perhaps

the most startling aspect of Symbulator is its fully

symbolic implementation. Answers are left in terms of

variables instead of being converted into numerical

approximations. This allows for easier analysis,

interpretation, and plotting. If you're taking Dr.

Gedney's Circuits II course, Symbulator will save you

every time. Dr. Gedney is infamous for assigning problems which do not have solutions in the textbook.

With Symbulator, the solution is only seconds away,

and a better understanding can be obtained by reverse-

engineering the problem so you know you understand

the correct solution. Since the interface to Symbulator is

similar to SPICE, once you know one, you can use both.

For EE 211 and EE 221, Symbulator actually works

better than SPICE because you get the results

symbolically. Once you use Symbulator you'll see what

everybody has been talking about. If you're not using

Symbulator, you're working too hard.” Chris Riegel.

EE+CoE major & President of the IEEE student chapterin his U. / University of Kentucky / Kentucky, USA

“I am really satisfied with the program. ... I have greatly

enjoyed using your program and have convinced most

of my classmates to use it also. ... I am on my third year

in EE and have used Symbulator since day one. I am still

using it on a regular basis.” Nevin McChesney. EE major

/ University of Kansas / Member of the US Navy / USA

“I would like to thank you for creating such a great tool

for EE analysis. I especially appreciate that you released

it for free.” Jeff Bach. EE & Music major / University of

New Mexico / New Mexico / USA

“I'd like to congratulate you on the success of

Symbulator for the TI-89... it seems to be used and

appreciated by many students.” Paul King. Texas

Instruments / Texas, USA

“Perez-Franco's Symbulator and Frederiksen's DiffEq

programs are two musts for any serious TI-89 user!”

Daniele Martini. EE major & author of S2S plug-in for

Symbulator / Florence, Italy



“The TI-89 opens up all kinds of opportunities for useful

programs, the Symbulator suite being a great

example.” Michael Rans. E+CSc major / Oxford

University / London, UK

“Just want to say that I admire your work. ... I enjoy

developing programs for my TI89 that help solving my

day by day problems. That's why I admire your work.”

Sergio Coelho. Engenharia Electrotecnica / Faculdadede Engenharia da Universidade do Porto, Portugal

“I and all of my friends at the University of Texas in

Arlington think the circuit simulator is great. Thank you

for all of your hard work on this program.” Jeffery Allen.

EE major / University of Texas at Arlington / Texas, USA

“I'm using Symbulator for only two weeks now and I

can say that I am in a way addicted to your program. It

is really the best software you can get for a calculator

on the field of Electrical Engineering. When I show it to

my colleagues at the faculty, I know that it will spread

in Slovenia too. ... Thanks again on this wonderful

program.” Matic Hibernik. EE major / University of

Ljubljanja / Slovenia

“Congratulations for Symbulator program. It's really

good!” Bernardo Merino Bermejo. EE major / Escuela

Universitaria de Ingenieria Tecnica / Madrid, Spain

“I really appreciate this program and I'm sure many

students will as well. I hope that now all other

programmers would try to create useful science

programs.” Vic Fischer. Physics+CoE+EE major /

Universite Jean Monnet / France

“I just loaded Symbulator onto my calc last night and Iam liking it a lot.” Samuel Nyall Stearley

“Your programs are great and useful for all students

with a good calculator.” Michael Preuhs. EE major /

Universität - Gesamthochschule Siegen / Germany

“Thanks for coming out with such a great program.”

Abhishek Shrestha. CoE Sophomore / Western

Michigan University / USA

“You can be really proud of Symbulator!” Kamil

Malinksi. EE major / Technical University / Vienna,

Austria

“Thank you very much for all your hard work on

Symbulator - nice job!” Doug Burkett. EE major / Eaton

/ Ohio, USA

“Thanks for carrying me through more than two years

of EE studies.” Thomas Schallhofer. Northest University

of Applied Sciences / Flensburg,Germany

“Thank's for doing this program!” Thomas Knoblauch.

EE major / ETH / Zürich, Switzerland

“Wow! Now I can do Bode Plots!” Damion Hadcroft. ME

major / Adelaide University / Australia

“I'm sure this program will prove to be quite an asset.”

Andrew Hockman. EE+CoE major / Missouri, USA

“It is really simple. At the same time, it's a very useful,

great program.” Andrej Bencúr. EE major / Technical

University of Ostrava / Czech Republic

“I love the software, it helps me out in the lab quite a

bit.” Eric. EE major / University of Dayton / Ohio, USA

“Thank you so much for letting me get this program. It

has been wonderful and I appreciate all of your hard

work.” Jared Whitaker. EE major / UAA

“I want to thank you and congratulate you as well for

such a nice and useful program.” Dan Safoory. EE major

“Thanks for your work. Symbulator is fantastic! ... God

has really blessed you with quite a talent for solving

difficult problems. ... I am not going to take another

exam without it.” Jake Adams. ME pro + EE student

“I love your program and I greatly appreciate all the

hours and hard work you have invested. ... I must give

you partial credit for placing 2nd in my circuits class

and getting a perfect score on my final exam. My

advisor is no longer telling me I may want to reconsider

my choice of major from electrical engineering to

something more suited to my advanced age of 41.”

Annie. EE major

“I must say that this program looks extremely tight and

well-tuned. You should be hugely commended for your

abilities! Great job and keep up the excellent work!”

Neel. EE major

“These have been great learning tools and helped

reemphasize what school teaches. I have to use these

tools to help me better understand rather than just

mindlessly answering the questions. Thanks to this

calculator, your programs and lots of hard work and

studying, I actually survived my “Advanced Circuit

Analysis” class, with an “A” to boot. It was a class from

hell!” Jackie. CoE+EE major

“Roberto, you are doing a good job for TI and EE

students that use TI89. Keep up the good work.” JLA

“Roberto, thank you for your diligence and hard work in

creating and perfecting a GOOD circuit simulator. (Not

to mention FREE). The use of Symbulator has improved

my learning of electrical circuits.” Nevin McChesney, on

the release of v.5. EE major / University of Kansas /

Member of the US Navy / USA



“Thanks, again. This is an excellent program and I am

very interested in seeing the book you are writing for

Symbulator Q.” Eric Wierwille, on the release of v.5

“I want to thank Roberto Perez-Franco for writing such

a wonderful program - the Symbulator. Spending so

much development time on the home page and

software, and not charging a nickel for it, makes you a

real hero. That kind of thing brings a tear to my eye!Thank you!” Bez, on the release of v.5

“Thanks for the program. It is a true gem, and a free

one at that...a true rarity!” Dorian, on the release of v.5

“I must say, Rob, that once again you've outdone

yourself. I hope that this isn't the last time, and look

forward to more genial programs from you. Thanks for

letting us enjoy the Symbulator for free. It's not often

that great programs come free. You, my friend, are an

example to follow.” Anonimous poster @ TI discussion

group, on the release of v.5

“And the Oscar for Best Program of the Year for a TIcalcs goes to... (drums) Symbulator Q, by Roberto Perez

Franco. Huray! Huray!” Anonimous poster @ TI

discussion group, on the release of v.5

“This program is one of the major TI programming

accomplishments, and in fact one the most significant

programs for any calculator. ... Until I started the

(Impala) testing, I did not realize just how powerful

Symbulator really is. ... Symbulator makes it possible to

solve every homework and test problem, without

knowing much more than how to enter the circuit in the

matrix form! ... I have been both playing with

calculators and using them professionally for over

twenty years. I am quite aware of their capabilities. And

I think that Symbulator is the most amazing end-user

calculator program that has ever been accomplished. I

am grateful to you for having done this work, and for

having improved Symbulator so enthusiastically. You

know that Symbulator is good enough to be a for-pay

program, yet you gave it to us for free. That means a lot

to me.” Doug Burkett, on the release of v.4. EE major /

Eaton / Ohio, USA

“I'd like to thank Roberto for his effort in releasing

Symbulator (Impala) It's really an incredible program.”

Farooq Ahmed Weshah, on the release of v.4. EE major

/ Yarmouk University / Jordan-Salt, Jordan

“I was going to sell my TI-89 because I didn't see any

use for it. But this new program by Roberto made me

want to try this program out. Congratulations for your

work!” Marcelo, on the release of v.4. EE senior /

Canada (born in Chile)

“Roberto, I've been enjoying your program for over a

year and use it regularly. It has made my EE studies

much easier over the years. ... Thanks again for a

superb program, which is the reason for which I

choosed TI-89.” Gordon, on the release of v.4

“Another triumph in BASIC programing for 68K

calculators. Awesome! I can't describe it any better.”

Captain Ginyu, on the release of v.4

“When I start my next circuits course and my first

Controls course in the fall, I know Symbulator will be my

best friend once again, especially with all of the

improvements you have made.” Charles Ware. EE major

/ Ohio State University / Columbus, Ohio, USA

“Congratulations on receiving the IEEE award - I believe

it is very much deserved. It makes me smile to see you

rewarded for your hard work and dedication to

Symbulator and the TI community.” Paul King, on the

IEEE Award. Texas Instruments / Texas, USA

“Many, many congratulations, Roberto! Passion and intelligence will always beat greed and narrow vision.”

Doug Burkett, on the IEEE Award. EE major / Eaton /

Ohio, USA

“Congratulations, Roberto. Symbulator is a masterpiece

worthy of the best programs out there, free or

otherwise. And the fact that it is free speaks volumes

about its creator. Peace be with you always.” Jake

Adams, on the IEEE Award. ME pro + EE student

“Congratulations, Roberto!!!” Dave Conklin, on the IEEE

Award. Professional TI-89 programmer

“Ha! Some wise man finally discovered what so many

'meere students' have known for some time. It's so

good to see some recognition for you. You deserve it.”

Jay Myers, on the IEEE Award. EE major / University of

Texas at Arlington / Texas, USA

“Once again you did a great job and I'm sure I'm not the

only one who is sending you a ‘Thank you very much,

Roberto. Our EE-life is a lot easier with your

Symbulator’.” Kamil Malinski, on the release of v.3. EE

major / Technical University / Vienna, Austria

“Once again, good work! I thought Symbulator was

pretty much a finished and polished product and now

you pull new tricks out of your sleeve! ... Symbulator

has been a very valuable tool for me. By the way, I

recently passed my Professional Engineering exam.

Symbulator allowed me to check some of my answers. A

pretty hefty dual op-amp circuit was one I checked...”

Jim Wachtel, on the release of v.3. EE / Process Control

& Electrical Co. / St. Louis, Missouri, US



Circuit Analysis

with Symbulator

Volume 1: DC



Dedication

I dedicate both Symbulator and this book to the two persons that supported

me during the time it took to get them done. The first is my wife Monica,

who had the patience to live with a cyborg for all the years my mind was

focused on coding and documenting. My attention was elsewhere, but my

heart was always with you! The second is my father Tito, who invested timeand money in my education, and bought me every widget I fancied for

college, with a blind faith that one day something good would come out of it.

I’m happy to know that I made you proud, Dad, and I miss you! Finally, I

dedicate all my efforts to my daughter Sara: may you enjoy a happy, long life.

Acknowledgements

Many friends helped me tested the software and documentation: José Vega

(Panama), Tim Hutcheson (USA), Lars Frederiksen (Denmark), Joe Riel (USA),

Arne Harstad (Norway), Erwin Baert (Belgium), Charles Ware (USA), Doug

Burkett (USA), Reinhard Willinski (Germany), Kamil Malinski (Austria), Jake

Adams (USA), Daniele Martini (Italy), Rozgonyi Szabolcs (Hungary), Michael

Rans (UK), Alex Astashyn (Russia), Al Charpentier (USA), Nevin McChesney

(USA), Ivan Oro Yu (Panama), and Pepe Iborra (Spain). Three professors from

Universidad Tecnológica de Panamá allowed me to present Symbulator as

my undergraduate thesis: Medardo Logreira (chair), Eliane Boulet-Cabrera

and Marcela Paredes. To all of you, thanks!

Source of Problems

To make this book useful to actual students, problems were chosen from

widely used textbooks on circuit analysis, scanned and presented here as

they appear in the source. The textbooks are: Introductory Circuit Analysis

(11th Edition) by Robert L. Boylestad, Pearson - Prentice Hall, 2007;

Engineering Circuit Analysis (5th Edition) by William H. Hayt, Jr. and Jack E.

Kemmerly, McGraw-Hill, 1993; The Analysis and Design of Linear Circuits (5th

Edition) by Roland E. Thomas and Albert J. Rosa, Wiley, 2006; Circuit

Analysis: Theory and Practice (3rd Edition) by Allan H. Robbins and Wilhelm

C. Miller, Thomson – Delmar Learning, 2004; Elementary Linear Circuit

Analysis (2nd Edition) by Leonard S. Bobrow, Oxford, 1987; and

Fundamentals of Electric Circuits (2nd Edition) by Charles K. Alexander and

Matthew N.O. Sadiku. McGraw-Hill, 2004. The images from these books are

scanned and reproduced under the “fair use” principle.



Circuit Analysis with Symbulator

the premier symbolic simulator of linear networks for calculators

Volume 1: DC

Roberto J. Pérez-Franco

Facultad de Ingeniería Eléctrica (FIE)

Universidad Tecnológica de Panamá (UTP)

Engineering Systems Division (ESD)

Massachusetts Institute of Technology (MIT)



Table of Contents

Getting started ................................................................................................................. 9

Running a simulation ...................................................................................................... 10

Elements: e and r ........................................................................................................... 11

Voltage source and resistors .......................................................................................... 11

Illustrative examples ...................................................................................................... 13

Practice problems ........................................................................................................... 19

Element: j ....................................................................................................................... 29

Current source ................................................................................................................ 29

Illustrative examples ...................................................................................................... 30

Practice problems ........................................................................................................... 31

Tool: par ......................................................................................................................... 37

Dependent sources ........................................................................................................ 38

Element: s....................................................................................................................... 49

Short circuits................................................................................................................... 49



9

Getting started

What is Symbulator?

Symbulator (“Symbolic Circuit Simulator ”) is a 62KB TI-Basic program widely regarded

as the best simulator of linear circuit ever made for a calculator. It can perform directcurrent, alternating current, transient and frequency domain analyses of numerical

and symbolic linear circuits. It can find the Thévenin, Norton and two-port equivalent

of a circuit. It accepts elements such as resistors, inductors, mutual inductances,

capacitors, independent and dependent current and voltage sources, ideal operational

amplifiers, ideal transformers and six types of two-ports. It can also do Bode plots.

Where does it run?

The first five versions of Symbulator were made for the TI-89 calculator. Version 6 was

thoroughly tested in the TI-89 Titanium. Unfortulately, Symbulator does not run in the

TI-Nspire machines, since their programming platform is incompatible with the TI-89's.

Why would I use it?

Symbulator is extremely useful in solving a wide variety of circuits theory problems,

such as those taken by engineering students. In fact, a student with Symbulator and

the right knowledge on circuits theory can solve a lot of problems in circuit courses

much faster and neater than ever before. Its main advantage is that Symbulator allows

the student to focus in the conceptual understanding of circuit analysis, rather than

the mathematical technics used for their solution.

Symbulator is a fantastic tool when you need a symbolic approach to a small or

medium sized circuit composed of ideal, linear elements. You should use Symbulator

whenever you have to manage symbolic values or need symbolic results. When

compared to SPICE, PSpice or Electronic Workbench, this small software offers a

simpler way to define dependent sources, and includes special elements such as ideal

transformers and six different two-ports, usually not included in other simulators.

When is it most useful?

Symbulator is most useful with linear circuits with symbolic values, such as you find inundergraduate courses in circuit analysis. Students of Circuits I and II may benefit a lot

from this pocket-sized “expert system” for the numerical or symbolic solution of linear

circuits. It can help you with learning circuits theory, doing your homework or taking

that final exam. It is not, however, a replacement for your brain or an excuse to not

study your circuit theory classes: you must understand circuits theory to use it.

Symbulator is not really useful in two cases: it may be slow with large linear circuits

and is useless with circuits with non-linear elements (such as diodes and transistors.)



10

How much does it cost?

Nothing. It’s free. Has always been, will always be. (You are welcome!)

Who made this thing?

I did. I started writing it in April 1999, as a junior in Electrical Engineering at

Universidad Tecnológica de Panamá (UTP). Version 5 was done by January 2001.

Symbulator won 1st place in the 2000 IEEE Student Paper Contest - Latin America. It

got me a Distinguished Alum award from UTP in 2008; and was at least part of the

reason for the Outstanding Young Person award that the Panamanian Chapter of the

Junior Chamber International gave me in 2010. In June 2013 I have released version 6.

Get your calculator ready

You can download the latest version of Symbulator from this URL:

perez-franco.com/symbulator/s.zip

Symbulator relies on Lars Frederiksen’s DiffEq for Laplace transforms. Get it here:

perez-franco.com/symbulator/d.zip

Inside these two zips you will find the two files you need: s.tig and diff206.89g.

Transfer both files to your calculator, following the manufacturer’s instructions.

Both Symbulator and DiffEq run faster when they are archived. In order to archive

them, just execute the following two programs from the command line:

s\install() and dif\ install()

That is all! You are now all set. Have fun!

Running a simulationYou will learn how to simulate through the following examples. For now, let me just

say that we give Symbulator the circuit description as a string of elements separated

by semi-colons. The description can be stored in a variable or passed directly as an

argument. To run a DC simulation in Symbulator, we use an access program, or "gate",

called s\dc , which takes one argument: the circuit description in string form. Below

you will find plenty of examples of DC simulations.

Symbulator will run faster if you make MAIN your current folder and empty it before

each simulation. Symbulator will also run faster and more accurately if you use

integers and fractions in your element values instead of decimals. It is preferable to

simulate with exact numbers and then evaluate the answers using, e.g..

Ok, let’s get right into it!

Get the current

versions of

Symbulator and

DiffEq online

Install the

programs so they run faster

s\dc is the gate

to run a DC

simulation



11

Elements: e and r

Voltage source and resistors

Describing a resistor

In Symbulator, an ideal resistor is described as follows: first the name of the resistor,

which must start with the letter r, coma; second the name of the first node of the

resistor, another coma; third the name of the second node of the resistor, another

coma; and fourth the value of the resistor (in ohms). There is no coma at the end.

Example

An ideal resistor called r1, connected between nodes a and b with a resistance of 10Ω,

would be described thus: r1,a,b,10

Simulation answers

After the simulation in DC is complete, Symbulator stores a series of answers in the

calculator’s memory, labeled with easy to remember names for your convenience.

• The voltage of each of its nodes with reference to the ground is stored in a variable

called v and the name of the node. For example, for a node called 1, its voltage

with reference to ground is calculated and stored in a variable called v1.

• The voltage drop in the resistor , that is to say the voltage in the first node minus

the voltage in the second node, is stored in a variable called v and the name of the

resistor. For example, for a resistor called r5, the voltage drop is stored in vr5.

• The current through the resistor , flowing from the first node towards the second

node, is stored in a variable called i and the name of the resistor. For example, the

current flowing through a resistor called rx, from its first node towards its second

node, is stored in irx.

• The power consumed by the resistor is stored in a variable called p and the name

of the resistor. For example, for a resistor called r12, the power consumed is

stored in pr12.

What about conductances?

As an ideal circuit element, a conductance is not different from a resistor: they are the

same element, with the only difference that their values are given in different terms.

Resistance values are given in ohms and conductance values are given in siemens. A

conductance value can be converted into an ohm value by dividing 1 over the

conductance in siemens. Thus, for example, an 8S conductance between nodes a and b

could be described thus as a 1/8Ω resistor: r1,a,b,1/8

Examples with conductances will be presented later in this book, starting in page 30.

This is how

resistors are

described in

Symbulator.



12

Describing an ideal voltage source

In Symbulator, an ideal voltage source is described as follows: first the name of the

source, which must start with the letter e, a coma; second the name of the positive

node of the source, another coma; third the name of the negative node of the source,

another coma; and fourth the value of the source (in volts). No coma at the end.

Example

An ideal voltage source called e1, connected between nodes 3 and 0 , with a voltage of

12V (given as voltage of node 3 with regards to node 0), would be described thus:

e1,3,0,12

Simulation answers

After a DC simulation, for each voltage source in a circuit, Symbulator will store a

series of answers in the current folder of the calculator:

•

The voltage of each of its nodes with reference to the ground is stored in a variablecalled v and the name of the node. For example, for a node called 1, its voltage

with reference to ground is calculated and stored in a variable called v1.

• The voltage drop in the source, that is to say the voltage in the first node minus the

voltage in the second node, is stored in a variable called v and the name of the

source. For example, for a source called e5, the voltage drop is stored in ve5.

• The current through the source, flowing from the first node towards the second

node, is stored in a variable called i and the name of the source. For example, the

current flowing through a source called ex, from its first node towards its second

node, is stored in iex. The way the current direction is defined might seem

counterintuitive for voltage sources; it is such for consistency: the same directionis applied to all other two-node elements throughout Symbulator.

• The power consumed by the source is stored in a variable called p and the name of

the source. For example, for a source called e12, the power consumed is stored in

pe12; the power delivered is the negative of that, and can be found via -pe12

(where - is the negative sign, not the subtraction operator.) For sources, the choice

to store the consumed power instead of the delivered powered may seem odd; it

is such for consistency: for all elements, the power given is the consumed power.

• The equivalent resistance of the rest of the circuit, as seen by a source, is stored in

a variable called r and the name of the source. For example, the equivalent

resistance of a circuit as seen from a source called e2 is stored in variable re2.

What about dependent sources?

Symbulator makes no distinction between independent and dependent sources.

Because of this, it is just as easy to work with dependent or independent sources. The

only difference is that you, as a user, will input – for a dependent source – a value that

is a function of a current or a voltage, i.e. 8*vr1, or 0.3*ir2, or 2.3*(va-vb), etc.

Examples with dependent sources will be presented later, starting in page 38.

This is how

voltage sources

are described in

Symbulator.



13

Illustrative examples

Solving a numerical circuit

Exemplum docet. Let’s dive right in and solve a numerical linear circuit in direct current

using the s\dc gate. When I say a circuit is numerical what I mean is that we know thenumerical value of every element in the circuit. I have chosen Example 5.7 from

Boylestad’s Introductory Circuit Analysis (11e). Moving forward, I will refer to that

textbook as B11. For your benefit, the problem statement and the circuit schematic

were scanned and are reproduced below exactly as they appear in the textbook.

B11’s Example 5.7

Step 1: Describe the circuit

The first step is to describe the circuit. Circuit description starts with naming the

nodes. You can call the nodes anything you want, be it a number or a letter, as long as

the name is unique. You must always have one node called 0 (zero); this is your ground

node and has a voltage of 0V. I labeled the nodes starting in the ground and moving

clockwise: 0, 1, 2 and 3. It helps me to pencil the names in the schematic itself.

After naming the nodes, I am ready to describe the circuit in Symbulator notation.

When I only have one voltage source, like here, I enjoy naming it with a single letter: e.I describe the voltage source as follows: e,1,0,36 since its name is e, its positive

node is called 1, its negative node is called 0, and its value is 36 volts between these

nodes.

Now I describe the resistors. The first resistor I described as follows: r1,1,2,1000

because I name it r1, its first node is called 1, its second node is called 2, and its value

is 1000 ohms. The second resistor we describe similarly: r2,2,3,3000 and likewise

for the third resistor: r3,3,0,2000

Problem from

Example 5.7 in

Boylestad’sIntroductory

Circuit Analysis

(11e). Since it is

scanned, this is

exactly how the

problem

appears in the

textbook.

First step is to

describe the

circuit. Start by

naming the

nodes.

Your ground

should always

be node 0.

A numerical

analysis is possible when

we have the

values of all the

elements in the

circuit.



14

So we have that our circuit has four elements. We pass along this description to

Symbulator as a string, e.g. we open with a quotation mark, then enter the

descriptions of each element, separated with semi-colons, and then close with a

quotation mark. We can store this string in a variable. I like to call this variable cir

(short for circuit,) but you can call it something else if you prefer. I type this in my

Symbulator-ready calculator:

"e,1,0,36;r1,1,2,1000;r2,2,3,3000;r3,3,0,2000"cir

Once I press enter, the circuit will be stored in that variable.

Step 2: Run the simulation

We now ask Symbulator to simulate this circuit in direct current (DC), by typing this:

s\dc(cir)

After you press enter, Symbulator will quickly simulate the circuit stored in cir.

Symbulator will let you know when it is Done. It took 12 seconds in my calculator to

get 16 answers, and store them in variables in the current folder: the voltage in eachof the three nodes (v1, v2 and v3), for each of the four elements in the circuit its

voltage drop (ve, vr1, vr2 and vr3), current (ie, ir1, ir2 and ir3) and power consumed

(pe, pr1, pr2 and pr3), and the equivalent resistance as seen by the source (re).

Some alternatives

We could also have given Symbulator the command to simulate and the circuit

description in a single line, using two possible methods. The first is connecting the two

steps above using a colon, as shown below:

"e,1,0,36;r1,1,2,1000;r2,2,3,3000;r3,3,0,2000"cir:s\dc(cir):

The second is giving the circuit description as an argument of the s\dc gate:

s\dc("e,1,0,36;r1,1,2,1000;r2,2,3,3000;r3,3,0,2000")

Any of the three approaches yields the same result, and I use them interchangeably.

Step 3: Get the answers

Answer to question (a). The equivalent resistance seen by the source e is stored in: re .

Evaluate this expression. The calculator returns the right answer: 6000 (e.g. 6KΩ.)

Answer to question (b). Current IS is the current flowing through the source from node0 to node 1; thus, we can obtain it by evaluating - ie . Since this is a series circuit, we

could also find it as the current through any of the resistors: i r1 , i r2 or i r3 . In either

case, using, the calculator provides the right answer: .006 (e.g. 6mA.)

Answer to question (c). The voltage drop in resistor R1 is found evaluating vr1: the

calculator returns 6 (e.g. 6 V.) For R2, evaluate vr 2 to obtain 18 V. And for R3, vr3

gives us 12 V. These are all the right answers.

Answer to question (d). We evaluate -pe and get the right answer: .216 (e.g. 216mW)

The circuit is

passed to

Symbulator as a

string, elementsseparated by

semicolons

There is more

than one way to

skin a cat…



15

Answer to question (e). The power consumed by the resistors are found evaluating

pr1 , pr2 , and pr3 , using. The calculator returns the right answers: .036 (e.g.

36mW), .108 (e.g. 108mW), and .072 (e.g. 72mW) respectively.

Answer to question (f). Let’s ask the calculator if the sum of the consumed power in

the resistors equals the power supplied by the source. Evaluate this equation:

pr1+pr2+pr3=-pe

The calculator returns true. This is the right answer, and concludes the solution.

An alternative

You can ask for all the answers above in a single line, by using an array, such as this:

re, ir1,vr1,vr2,vr3,-pe,pr1,pr2,pr3,pr1+pr2+pr3=-pe

Using we evaluate this array, and get the same answers as before, also as an array.

Actually, we could solve the whole problem using a single line of text, as follows:

"e,1,0,36;r1,1,2,1000;r2,2,3,3000;r3,3,0,2000"cir:s\dc(cir):

approx(re, ir1,vr1,vr2,vr3,-pe,pr1,pr2,pr3, pr1+pr2+pr3=-pe)

An easy way to read this is to identify the colons as delimiters between things that you

would have asked in separate lines. So we see we have first the circuit description

being stored in a variable, a colon; second the command to simulate that circuit in DC,

another colon; and finally an array of all the answers we would like to evaluate.

Notice that we use integers as values for the simulation to run faster, and then

approximate the answers in the array by using the calculator’s command approx .

By showing you this, I just want to make you aware of the possibilities. But please donot rush into trying to solve a whole problem using a single line. There are times when

you may not be sure about your circuit description or about which variables to

evaluate as answers. In these cases, it is easier to solve the problem step by step: first

describe the circuit, then run the simulation and finally find the answers.

Practice problems of this type are found starting in page 18.

A word on symbolic problems

I call a circuit symbolic if one or more of the element has an unknown value. It is the

ability to simulate symbolic circuits that gives Symbulator its name and sets it apart

from other programs. I distinguish between two types of symbolic simulations.

Purely symbolic problems

The first type is when the desired answers are also symbolic, e.g. given as algebraic

functions in terms of unknown values (what I call a purely symbolic problem.)

We can evaluate

multiple

answers using

an array

You can

approximate

answers using

approx



16

Numerical-from-symbolic problems

The second type is when numerical answers are expected from a symbolic circuit

(what I call a numerical-from-symbolicproblem.) This solution is possible when they

give us additional information about the circuit. Numerical-from-symbolic problems

can be solved in two ways in Symbulator:

•

If they are simple, e.g. if there’s just one or two unknown values in the circuit,and we are only asked for one or two numerical answers, we can simulate the

circuit symbolically and then solve for the numerical answers using solve()

• If they are not simple, e.g. if there are many unknown values in the circuit, or

we are asked for many numerical answers, it is easier and more time-efficient

to use the expert mode of Symbulator, whose gate is s\Expert(cir)

Let’s see an example of each approach to solving numerical-from-symbolic problems.

Solving a symbolic circuit with solve

Let’s do a numerical-from-symbolic problem by means of the solve command. We

will obtain numerical values from the simulation of a circuit with some unknown

element values. Later we will solve this same problem using the expert mode.

B11’s Example 5.6

This is a very nice symbolic problem: we should be able to solve it into numerical

results because, even though the problem hides two values from us (e.g. the value of

the source E and of the resistor R1), it gives us in exchange two answers (e.g. the

equivalent resistance RT and the current I3) that we can use to solve for the unknowns.

Since this circuit is structurally identical to B11’s Example 5.7, we will use the same

names for the nodes. The circuit description is identical except for the element’s

values. As values for the elements in the circuit, Symbulator will accept numbers,

variables or even algebraic expressions. For this example, I chose to use e for the value

of source e and r1 for the value of the r1 resistor1. Make sure that the variables are

empty, either by emptying the current folder or by deleting them from the calculator’s

1 Giving a symbolic element a value equal to its name can be a useful trick to quickly remember

whose value it is. However, due to the potential for conflicts, it should be avoided in the expert

mode simulation, which stores all solved unknowns into variables, as the third example shows.

Symbulator has

an “expert”

mode for faster

and better

results

Unknown values

can be entered

as a variable.

Just make sure

the variable is

empty!



17

memory, thus: DelVar e,r1 . Below is how I described this circuit. I pass along this

description to Symbulator as a string, and store it in a variable.

"e,1,0,e;r1,1,2,r1;r2,2,3,4000;r3,3,0,6000"cir

Ask Symbulator to simulate this circuit in direct current (DC), by typing this:

s\dc(cir)

A moment later (16 seconds in my calculator2), Symbulator is Done, and we are ready

to answer the questions. Using the symbolic answers provided by Symbulator and the

known answers given by the problem, we can write two equations. We can then solve

these two equations for the two unknowns that interest us, using the calculator’s

solve command.3 Let’s explore what we have here.

The problem says that I3 is 6mA. In Symbulator, I3 is called ir3 (e.g. the current through

resistor r3.). If you evaluate ir3, you will see it produces an algebraic expression in

terms of the two unknowns, e and r1. This is what we call a symbolic answer. The

problem also says that RT is 12KΩ. As we saw in the previous problem, the equivalent

resistance as seen by the source e is given by re, which when evaluated gives another

algebraic expression in terms of r1. We can then write two new equations, re=12000

and ir3=6/1000, and solve them for the two unknowns we want: e and r1.

solve(re=12000 and ir3=6/1000,e,r1 )

An instant later we get the answers: e = 72 (volts) and r1 = 2000 (ohms). These are

the right answers. Not many other circuit simulators allow this flexibility.

Examples to practice this type of problems are found starting in page 23.

Solving a symbolic circuit with Expert Symbulator’s true strength is seen in numerical-from-symbolic problems like the one

we solved above, but using its “expert” mode of simulation, which cracks these

problems open even faster and can give you fully numerical answers to problems like

this with an equal number of unknown values and of answers provided by the

problem. Learning to use the expert mode pays off handsomely in terms of additional

power and speed. Let’s solve the same circuit from B11’s Example 5.6, this time using

the expert mode.

B11’s Example 5.6 – Redux We will use the same circuit description as before, with a single change: we will use rx

for the value of resistor R1, instead of the r1 value we used before. Like this:

2 If you want to know how long it took Symbulator to run a simulation, you have to make sure

your calculator’s clock is on, by typing ClockOn . As long as the clock is on, you can see how

long a simulation took by going to the Program IO screen (press F5 in the main menu.)

3 If you are not familiar with the solve command, I refer you to the calculator’s manual.



18

"e,1,0,e;r1,1,2,rx;r2,2,3,4000;r3,3,0,6000"cir

The reason for this is that r1 is a reserved variable in the calculator. The expert mode

tries to store all the values of the solved unknowns into variables of the same name.

Trying to store a value into r1 would result in an error.

Run the expert mode by typing this in the calculator:

s\Expert(cir)

When prompted, select DC and press Enter. When prompted, leave the float settings

as they are and press Enter4. Now you will see a prompt listing four “first level

variables”, namely v1,ie,v2,v3, and four “first level equations.” A quick examination

of the equations will show that they contain two additional variables that are not

listed in the previous field: the value of the source, e, and the value of resistor R1, rx.

So, add them to the “1st level var’s” list. Type this before the closing bracket:

,e,rx

so that it now reads v1,ie,v2,v3,e,rx.

Now we have six variables and four equations. You may recall from your algebra class

that you need an equal number of equations and unknowns in order to solve the set.

We are two equations short. The statement of the problem gives us the information

we need to write the two additional equations5. Go to the end of the text in the “1st

level eq’s” prompt, and type this in, leaving an empty space before the and:

and re=12000 and ir 3=6/1000

Now we have six variables and six equations. Press Enter and wait just a few seconds.

When Symbulator says “Done”, go ahead and retrieve the answers:

rx

You get 2000, the right answer.

e

You get 72, the right answer.

The speed advantage of the expert mode is not necessarily evident in this simple

problem. It does give you an idea of what the expert mode is all about: you get to halt

the simulation in mid-air and give Symbulator extra information. Had this circuit been

larger, the benefit of the expert mode in computation time would be clear.

Examples to practice this type of problems are found starting in page 27.

4 Do not change these float settings unless you have a good reason for doing so.5 We know the value for RT and the value for IS in terms of the variables listed in the prompt:

because RT is given by re, which is a function of e and ie, and IS is given by ir3, which is a

function of ie.

Variables r1

through r99 are

reserved variable

in the calculator.

This means you

cannot store

things into them.

If you want to

get numerical

answers, you

must have as

many equations

as you have

unknowns



19

Practice problems

Numerical problems

The following are practice problems taken from several textbooks. They were chosen

because they apply only the concepts that you have learned so far. This will allow you

to practice these concepts and reinforce them before moving on to new ones.

The problem below comes from Figure 1-26 (a) in Hyat and Kemmerly’s Engineering

Circuit Analysis (5ed). Moving forward, we will refer to that textbook as HK5.

HK5’s Figure 1-26

We are asked to determine the current, voltage drop and power consumed in each

resistor, as well as the power delivered by each voltage source. We are also asked to

check that the powers in the circuit add up to zero. Here is my solution.

I named the nodes thus: the bottom node is named 0, the top nodes, from left to right,

are named 1, 2 and 3. My description of the circuit is given below, followed after a

colon by the DC simulation command.

"e1,1,0,120;r1,1,2,30;e2,2,3,30;r2,3,0,15"cir:s\dc(cir)

When the simulation is done, you can ask the calculator for the answers you need:

• Evaluating i r1 or i r2 gets the current in the resistors: 2A

• Evaluating vr1 gets the voltage drop in the 30Ω resistor: 60V

• Evaluating vr2 gets the voltage drop in the 15Ω resistor: 30V

• Evaluating pr1 gets the power consumed in the 30Ω resistor: 120W

• Evaluating pr2 gets the power consumed in the 15Ω resistor: 60W

• Evaluating -pe1 gets the power delivered by the 120V source: 240W

• Evaluating -pe2 gets the power delivered by the 30V source: -60W. This

means this source is actually consuming 60W.

• Evaluating pr1+pr2+pe1+pe2 gets the sum of powers: 0W. As expected.

Wasn’t that easy? We could also have asked for all the answers with one array:

ir1,vr1,vr2,pr1,pr2,-pe1,-pe2,pr1+pr2+pe1+pe2

B11’s Example 5.20

The practice problems will get progressively more complicated as we move on. This

will allow you to build up your symbulating skills with confidence.



20

I named the nodes clockwise starting from the ground: 0, 1, 2, 3 and 4. Below is my

circuit description, given as an argument of the command to do the DC simulation:

s\dc("e1,1,0,50;r1,1,2,4;e2,2,3,12.5;r2,3,4,7;r3,4,0,4")

When it’s done, ask for the answers we need. Evaluating i r1 gets the current I: 2.5A.

Evaluating vr 2 gets the voltage drop in the 7Ω resistor: 17.5W


Since the ground is the bottom node, I named it 0. I named the top node 1.

s\dc("e,1,0,24;r1,1,0,10;r2,1,0,220;r3,1,0,1200")

a) Evaluating re gets the total resistance: 9.49 Ω

b) Evaluating - ie gets us the source current: 2.53A

c) Evaluating i r1 gets I1: 2.4A, ir2 gets I2: 0.11A, and i r3 gets I3: 0.02A.

B11’s Example 7.2

Determine I4, IS and V2. My solution below. I named the top node 1, and the other 2:

s\dc("e,1,0,12;r1,1,2,6800;r2,2,0,18000;r3,2,0,2000;r4,1,0,8200")



21

Answers: v2 is 2.51 V, - ie (e.g. IS) is 2.86 mA and ir4 is 1.46 mA.

B11’s Example 7.7

For your benefit, I have labeled the node names I used. My solution:

s\dc("e1,0,1,6;e2,0,2,18;r1,1,a,5;r2,a,2,3;r3,1,b,6;r4,b,2,2")

Answers: vr1 is 7.5V, vr 3 is 9V. For Vba, vb-va is -1.5V. For IS, -ie2 is 3A.

B11’s Figure 7.32

Determine I6 and V6. My solution below:

s\dc(“e,1,0,240;r1,1,2,5;r2,2,0,6;r3,2,3,4;r4,3,0,6;r5,3,4,1;r6,4,0,2”)

Answers: i r6 is 10A, and vr 6 is 20V.


Calculate the indicated currents and voltages.

My solution below:



22

s\dc("r2,2,3,8000;r1,3,4,4000;r3,1,2,12000;r4,1,4,24000;r5,1,0,

12000;e,4,0,72;r6,4,5,12000;r7,5,0,9000;r8,5,6,3000;r9,0,6,6000")

Answers: i r5 is 3mA, -ie (e.g. IS) is 7.36mA, and vr7 is 19.6V.

B11’s Example 7.4

Determine the currents I1, I2, IA, IB and IC, and the voltage drop areas A, B and C.

My solution below:

s\dc(“e,1,0,16.8;r1,1,2,9;r2,1,2,6;r3,2,3,4;r4,3,0,6;r5,3,0,3;r6,2,0,3”)

Current I1 is found via i r1 = 1.2A, I2 via i r2 = 1.8A, IA, via -ie = 3A, IB via i r3 = 1A and IC

via i r6 = 2A. The voltage drop in area A is vr1 = 10.8V; in both B and C it is v2 = 6V.


My solution below:

s\dc("e,1,0,28;r1,1,0,1600;r2,1,0,20000;r3,1,0,56000")

a) Evaluating re gets the total resistance: 1.44KΩ

b) Evaluating i r1 gets 17.5mA, i r2 gets 1.4mA, and i r3 gets 0.5mA

c) Evaluating -pe1 gets the power: 543mW



23

B11’s Figure 7.40

Determine Vb and Vc.

My solution below:

s\dc("e,a,0,120;r1,a,b,10;r2,b,c,20;r3,c,0,30;rl1,a,0,20;rl2,b,0,20;rl3,c,0,20")

Answers: vb is 66.21V, and vc is 24.83V.


Determine the current through each resistor.

My solution below:

s\dc("e1,1,0,15;r1,1,a,4;e3,3,0,20;r3,3,a,10;e2,0,2,40;r2,a,2,5")

Through an array, and using the approx command, we ask for all the three answers:

approx(ir1, ir2, ir3)

The calculator returns 4.77,7.18,2.41 meaning IR1 =4.77A, IR2 =7.18A and IR3 =2.41A.

These are the correct answers. In the rest of the book, we will often use this method

of asking for several answers at the same time by means of an array.



24

B11’s Example 7.6

My solution below:

s\dc("e,1,0,24;r1,1,2,6;r2,1,2,6;r3,1,2,2;r4,2,0,8;r5,2,0,12")

Answer: - ie is IS=4A, i r2 is I2=.8A, i r4 is I4=2.4A, vr1 is V1=4.8V, and vr5 is V5=19.2V.


My solution below:

s\dc("e1,a,0,20;e2,a,b,5;e3,c,0,8;r1,a,c,10;r2,b,c,4;r3,b,0,5")

Answers: va=20, vb =15, vc=8, va-vc= 12, vb-vc=7, i r2=1.75, IS via -ie3=-2.95


Find the voltage drop in the 3Ω resistor.

My solution below:



25

s\dc("e8,1,0,8;r2,1,2,2;r4,2,0,4;r6,2,3,6;r3,3,0,3;

r10,3,4,10;e1,0,4,1."):approx(vr3)

A single line instruction finds, via vr3 , that V3Ω is 1.1V.


Find the current through the 10Ω resistor in the network shown below.

My solution below:

s\dc("e15,1,0,15;r10,1,2,10;r8,1,3,8;r5,3,2,5;

r3,3,0,3;r2,2,0,2"):approx(ir10)

The colon allows us to use a single line instruction to find i r10 = 1.22A.


Find the voltage drop in the 2Ω resistor.

My solution below:

s\dc("e,1,0,240;r1,1,2,3;r2,2,3,4;r3,3,4,1;r4,4,5,2;r5,3,5,6;r6,2,5,6;r7,5,0,9"):approx(vr4)

A single line instruction gives us the voltage drop in R4, the 2Ω resistor: 10.67V

Circuits with ‘hidden source’

Sometimes the schematics of circuits are presented in such a way that sources of

voltage are not shown explicitly, yet their voltage is provided. These are what I call

‘hidden source’ problems. Below I offer two examples of these types of problems.



26

Both are taken from the textbook Circuit Analysis: Theory and Practice (3ed) by Allan

H. Robbins and Wilhelm C. Miller, to which from this point on we will refer as RM3.

RM3’s Example 7-5 (Hidden source)

Find the indicated currents and voltages.

In my solution below, notice I introduced two sources, one for 12V and one for -6V:

"e1,1,0,12.;r1,1,b,10;r2,b,a,10;r3,a,2,50;r4,b,2,30;e2,2,0,-6"cir:

s\dc(cir):approx(ir1, ir2, ir4,va-vb)The answer, .6,.2,.4,-2., indicates I1=.6, I2=.2, I3=.4 and Vab=-2. This is correct.

RM3’s Figure 7-16 (Hidden source)

Find the total circuit resistance, and the indicated currents and voltages.

This is how I named the nodes: the top, 1; the bottom, 2, and a and b as in the figure.In my solution below, notice I introduced two sources, one for -10V and one for -6V:

s\dc("r1,1,b,4000;r2,1,a,3000;r3,b,a,2000;r4,b,a,3000;

r5,1,b,1000;rt,a,2,6000;e1,1,0,-2;e2,2,0,-10"):

approx((v1-v2)/ irt , irt , ir1, ir2,va-vb)

The answer we get indicates that the equivalent resistance, given by (v1-v2)/IT, is

7.2KΩ, and that IT=1.11mA, I1=.133mA, I2=..444mA and Vab=-0.8V. This is correct.



27

Symbolic problems with solve

HK5’s Figure 1-24a ( solve)

Determine i x and v x in the following circuit.

I named the nodes thus: bottom is 0, top left is 1, top right is 2. I named the elements

according to their value: this facilitates remembering who’s who in the circuit. I also

defined the resistors’ nodes in the direction of the current indicated in the diagram.

s\dc("e18,1,0,18;ra,1,0,ra;r6,1,0,6;r5,2,1,5;evx,2,0,vx")

Explore the answers. Since ir5 (which we know is 12A) is in terms of vx, we can find vx:

solve(ir5=12,vx)

We get that vx is 78V, which is corrrect. Evaluating i r6 we find that ix is 3A.

The fact that we can find numerical answers in this problem can be quite puzzling until

one realizes that ignoring the value of RA doesn’t matter: due to the circuit’s structure,

it is not needed it to answer the two questions we have been asked.

Symbolic problems with Expert

B11’s Example 6.19 (Expert)

This problem, having three unknown element values and three known answers, is a

perfect candidate for the expert mode. Below is my circuit description.6

s\Expert("e1,1,0,e;r1,1,0,2000;r2,1,0,rr2;r3,1,0,rr3")

6 In the expert mode, we avoid using as values to be solved any variable between r1 and r99,

which are reserved variables. We also avoid using an element’s name as its symbolic value.



28

Select DC. Add the three unknowns to “1st level var’s”, thus: ,e,rr2,rr3 Add also

these three equations to “1st level eq’s”:

and ir1=8/1000 and ir2=10/1000 and ir3=2/1000

Press Enter. When Symbulator is Done, find IS and E by asking: approx(-ie1,e)

We get that IS is 0.2A and E is 16V.

By now you should be getting the idea of what the expert mode is all about. When

using the expert mode, you can ‘feed’ known answers to the calculator before it solves

the set of equations that are at the heart of the simulation process. This makes it

possible to solve these equations into numbers, as opposed to symbols. That way it is

easier – and faster – for both you and the machine to solve the circuit problem.

B11’s Example 7.12 (Expert)

Determine R1, R2 and R3 for the voltage divider supply. Can 2W resistors be used?

This problem is another perfect fit for the expert mode, because: (a) the target is to

obtain numerical values, and (b) we have N unknown element values, and in turn we

are given N numerical answers (e.g. currents, voltages.) In this case, we have six

symbolic (e.g. unknown) values and six numeric answers. Here is how I solved it.

s\Expert("e,a,c,72;r1,a,b,rr1;r2,b,0,rr2;

r3,0,c,rr3;rl1,a,0,rrl1;rl2,b,0,rrl2")

Choose DC. Add these six variables to the list of first level variables:

,e,rr1,rr2,rr3,rrl1,rrl2

Add these six equations to the list of first level equations:

and va-vc=72 and ir l1=20/1000 and vrl1=60 and

irl2=10/1000 and vrl2=20 and -ie=50/1000 and vc=-12

Enter. When Done, evaluate these variables to find the answers: rr1 = 1.33KΩ, rr2 =

1KΩ, rr3 = 240Ω. These are correct. Since all powers consumed in the resistors prr1 ,

prr2 and prr3 are smaller than 2W, it is possible to use 2W resistors in the design.



29

Element: j

Current source

Describing an ideal current source

In Symbulator, an ideal current source is described as follows: first the name of the

source, which must start with the letter j, and a coma, second the name of the first

node, another coma, third the name of the second node, another coma, and fourth

the value of the source (in amperes). No coma at the end. The value should be given in

terms of the current flowing through the source from the first node towards the second

node. This means that the value of the source is how much current leaves the source

out of the second node, and also how much current enters the source’s first node.

Example

An ideal current source called j1, connected between nodes 3 and 0 , with a current of

12A (flowing through it from node 3 towards node 0), would be described thus:

j1,3,0,12

Simulation answers

For each current source, in a circuit, Symbulator will store a series of answers in the

current folder of the calculator:

• The voltage of each of its nodes with reference to the ground is stored in a variable

called v and the name of the node.

• The voltage drop in the source, that is to say the voltage in the first node minus the

voltage in the second node, is stored in a variable called v and the name of the

source. For example, for a source called j5, the voltage drop is stored in vj5.

• The current through the source, flowing from the first node towards the second

node, is stored in a variable called i and the name of the source. For example, the

current flowing through a source called jx, from its first node towards its second

node, is stored in ijx. This may seem redundant, but it is actually rather handy.

• The power consumed by the source is stored in a variable called p and the name of

the source. For example, for a source called j12, the power consumed is stored in

pj12; the power delivered is the negative of that, and can be found via -pj12.

• The equivalent resistance of the rest of the circuit, as seen by a source, is stored in

a variable called r and the name of the source. For example, the equivalent

resistance of a circuit as seen from a source called j2 is stored in variable rj2.

This is how

current sources

are described in

Symbulator.



30

Illustrative examples

A simple example: J and R

B11’s Example 8.1Given the circuit below, determine the current and voltage drop in R1.

s\dc("j,0,1,10/1000;r1,1,0,20000"):ir1,vr1

In the line above we are doing three things: the first is giving Symbulator the

description of the circuit, the second is asking it to run a DC simulation of that circuit,

and the third is asking for the values of two variables: ir1 and vr1. We are applying the

tricks we learned before, of concatenating commands for the calculator through

colons, and of asking for the value of multiple variables by using an array.

The calculator returns .01,200., meaning a 10mA current and a 200V voltage drop.

An example with conductances

The following example is taken from the textbook Elementary Linear Circuit Analysis

(2ed) by Leonard S. Bobrow. From this point forward, I will refer to this book as Bo2.

Bo2’s Example 2.2

Given the circuit below, determine the voltages in the nodes.

As I explained before, in Symbulator all conductances are simulated as resistors. So,

the 4 siemens conductance becomes a 1/4 resistor, and so on. Below my description:

"j10,1,0,2;r12,1,2,1;r20,2,0,1/4;r30,3,0,1/3;r13,1,3,1/2;

j32,3,2,3"cir:s\dc(cir):approx(v1,v2,v3)



31

The answer, -1.3,.34,-1.12, indicates v1=-1.3V, v2=.34V, v3=-1.12V. This is correct.

An example with E, J and R

B11’s Example 8.2

Determine the values of VS, I1 and I2.

In the line below, we concatenate three commands using colons. The first stores the

circuit’s definition in a variable. The second asks Symbulator to run a DC simulation of

the circuit described in that variable. The third asks the calculator to provide us the

values of three variables that – given the circuit description – answer the questions.

"j,0,1,7;e,1,0,12;r,1,0,4"cir:s\dc(cir):v1,ie, ir

The calculator returns 12,4,3, meaning VS is 12V, I1 is 4A and I2 is 3A. These are the

correct answers. We will continue to use the single line instruction as we move on.

Practice problems

B11’s Example 8.15Determine the current through each resistor. My solution below, direction in blue:

s\dc("j6,0,1,6;r2,1,0,2;r6,1,2,6;r8,0,2,8;j8,2,0,8"):

approx(ir2, ir6, ir8)

We get these answers: 1.25,4.75,3.25. So IR2 is 1.25A, IR6 is 4.75A, and IR8 is 3.25A.

Bo2’s Drill Exercise 2.2 (Conductances)

Given the circuit below, determine the voltages in the nodes.



32

Below my solution:

"r10,1,0,1/3;r12,1,2,1/2;r13,1,3,1/2;r23,2,3,1/6;r20,2,0,1/8;

j12,1,2,17;j03,0,3,2"cir:s\dc(cir):approx(v1,v2,v3)

The answer, -2.,1.,.5, indicates v1=-2V, v2=1V, v3=0.5V. This is correct.


Determine the voltage in each node and the current through each resistor.

My solution below:

s\dc("j1,0,1,4;r1,1,0,2;r3,1,2,12;r2,0,2,6;

j2,2,0,2"):v1,v2,ir1, ir2, ir3

We get the following answers: 6,-6,3,1,1. So V1=6V, V2=-6V, IR1=3A, and IR2=IR3=1A.

HK5’s Figure 1-24b (Expert)


My solution below:



33

"j6,0,1,6;r5,1,2,5;r2,2,0,2;r1,1,3,1;r3,0,4,3;

j10,2,3,10;r,3,4,rx"cir:s\Expert(cir): ir1,vr

Select DC. Add ,rx to the list of variables and add and ir2=4 to the list of equations.

Run the simulation. The answer, -8,80, means that IX is -8A and that VX is 80V.

HK5’s Example 2.2 (Conductances)

Determine the voltages in the nodes. My solution below:

"j01,0,1,-8; j30,3,0,-25;j21,2,1,-3;r12,1,2,1/3;r23,2,3,1/2;

r13,1,3,1/4;r20,2,0,1;r30,3,0,1/5"cir:s\dc(cir):v1,v2,v3

The answer, 1,2,3, is correct: v1=1V, v2=2V, v3=3V.

B11’s Example 8.5

Determine the current I2. My solution below:

"j1,1,0,4;r1,1,0,3;e2,1,2,5;r2,0,2,2"cir:s\dc(cir):approx(ir2)

This gives us a value for I2 of 3.4A. This is correct.

Bo2’s Example 2.5 (Conductances)



34

Determine the voltages in the nodes, and the current through the voltage source.

"j,0,1,3;e,3,2,3;r12,1,2,1/7;r20,2,0,1/3;r30,3,0,1/5;

r13,1,3,1/2"cir:s\dc(cir):approx(v1,v2,v3,-ie)

The answer, -.5,-1.5,1.5,11.5, is correct: v1=-.5, v2=-1.5, v3=1.5, i=11.5

B11’s Example 8.22Determine V1 and V2. My solution below:

j1,0,1,6;r1,1,0,4;e,1,2,12;r3,1,2,10;r2,2,0,2;j2,2,0,4."cir:

s\dc(cir):approx(v1,v2)

The answer, 10.67,-1.33, tells us that V1 is 10.67V and V2 is -1.33V.


Determine V1, I1 and I2. My solution below:

"e,2,0,24;r1,1,2,6;r2,1,0,12;j,0,1,1."cir:

s\dc(cir):approx(v1,ir1, ir2)

The answer, 20.,-.667,1.67, tells us that V1 is 20V, I1 is -.667A and I2 is 1.67V.




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Determine I2 and I3.

“e1,1,0,20;r1,1,2,6;r2,2,a,4;j,a,0,4;r3,a,3,2;e2,0,3,12.” cir:

s\dc(cir):approx(ir2, ir3)

Answer: 3.33,-.666. This is correct.


Determine V1, V2, , I1, I2 and I3. My solution is found below the circuit schematic:

"e,3,0,64;r1,3,1,8;r2,1,2,4;j,1,2,2;r3,2,0,10"cir:

s\dc(cir):approx(v1,v2,ir1, ir2, ir3)

Answer: 37.82,32.73,3.27,1.27,3.27. You should know how to read these by now, but

here it is just in case: V1 = 37.82V, V2 = 32.73V, I1 = 3.27A, I2 = 1.27A, I3 = 3.27A.

RM3’s Example 9-12 ( solve)

If R3 is to be replaced with R4 and I4, determine the value and direction of the source.

The idea is to keep the same voltage drop and current flow between nodes a and b. To

keep them equal, we must know what they are. So we simulate the first circuit (left):

"e,0,b,20;r1,0,a,16;r2,a,b,40;r3,a,b,60"cir:s\dc(cir):va-vb,ir3

We find that the voltage drop is 12V and the current is 0.2A. Now we simulate the

circuit replacing R3 with a resistor R4 of 240Ω and a source j with value I4. Run this:

"e,0,b,20;r1,0,a,16;r2,a,b,40;r4,a,b,240;j,a,b, i4"cir:s\dc(cir)

Notice that both the voltage drop (given by va-vb) and the current (given by i r4+ij)

are algebraic functions in terms of i4. Now you can find i4 solving by voltage drop:



36

solve(va-vb=12., i4) …or by current flow… solve(ir4+ij=0.2,i4)

The result is the same: i4=.15. The required current source is .15A from a to b.

RM3’s Example 8-13

Solve for the currents through R2 and R3 in the circuit shown. My solution is below:

"r1,a,0,10000;r2,1,0,5000;r3,b,a,6000;r4,0,2,16000;j,a,b,2/1000;

e1,1,b,10;e2,b,2,8"cir:s\dc(cir):approx(ir2, ir3)

The answer, .00154,.00111 is correct: IR2 = 1.54 mA and IR3 = 1.11 mA.

B11’s Example 6.3 (Hidden source)

Determine VS and I1.

The problem presents a seemingly ‘hanging’ node with 20V. To simulate this, imagine

a ‘hidden’ 20V voltage source connected to the node.

"j,0,1,6;r1,1,2,2;r2,1,2,1;e,2,0,20"cir:s\dc(cir):v1,ir1

We get the answer: 24,2, which is correct. VS is 24V and I1 is 2A.

HK5’s Figure 1-24c (Expert)




37

"e,1,0,60;r8,1,2,8;r10,2,0,10;r4,2,3,4;r2,3,0,2;

j ,0,3, ix"cir:s\Expert(cir): ix,v3

Select DC. Add , ix to the list of variables. Add and ir8=5 to the list of equations. Run

the simulation, and you will get: 1,8. This is correct: IX is 1A and that VX is 8V.

B11’s Example 6.22 (Hidden source)

Determine I1.

The circuit seems to be open but has a current. Imagine around it a ‘hidden’ currentsource. My solution below:

"jt ,0,1,12/1000;r1,1,0,1000;r2,1,0,10000;

r3,1,0,22000"cir:s\dc(cir):approx(ir1)

We get I1 = 10.48mA, which is correct.

B11’s Example 6.21 (Hidden source, Expert)

Determine IS, I1 and I3.

This ‘hidden source’ problem is perfect for Expert. My solution:

s\Expert("js,0,1,is;r1,1,0,6;r2,1,0,3;r3,1,0,1"):approx(is, ir1, ir3)

Select DC, and leave float settings. Add this variable to the list: , is Also add this to the

list of equations: and ir2=2e-3 Run the simulation. The answer, .009,.001,.006, is

correct, since the currents are as follows: IS is 9mA, I1 is 1mA and I3 is 6mA.

Tool: par Symbulator has a tool to help you find the equivalent value of two resistors connected

in parallel. It is called par, and is used as follows: let’s say you have a 5Ω resistor and a

7Ω resistor connected in parallel, and you want to find their equivalent value. Type:

The par tool finds

the equivalent

value of two

resistors in

parallel



38

s\par(5,7)

You get 35/12 as the equivalent resistance. For approximate values: s\par(5.,7.)

returns 2.92 as the equivalent. The tool works with symbolic values, too:

s\par(r1,r2) returns the well-known equation for equivalent values of resistors in

parallel. Whenever you do not need to know the current through or power consumed

in each individual resistor, you can reduce resistors in parallel using the par tool as

part of your circuit description in a simulation. For resistors in parallel the voltage drop

is the same, even after a reduction. Below we show some examples of this.

Examples

As a first example, here is a simulation in which it makes sense to use par. In B11’s

Example 7.4 (shown in page 22), it makes sense to reduce R4 and R5 to an equivalent

resistor, since we do not need to know any answer particular to them.

s\dc("e,1,0,16.8;r1,1,2,9;r2,1,2,6;

r3,2,3,4;re,3,0,s\par(6,3);r6,2,0,3")

…provides the right answers, e.g. IB is i r3 = 1A and voltage drop in area B is v2 = 6V.

The second example is of a simulation in which using par makes no sense. In B11’s

Example 8.3 (shown in page 36), it makes no sense to reduce R1 and R2 using par,

because you need to know the value of the current through R1.

As third example, here is a simulation where you can reduce part of the resistors. In

B11’s Example 6.22 (shown in page 36), we must leave R1 alone, because we need to

know the current through it, but we can reduce R2 and R3 to their equivalent using par,

since we need to know no value particular to them:

s\dc("jt ,0,1,12/1000;r1,1,0,1000;re,1,0,s\par(10000,22000)")

…provides the same answer we got before, e.g. I1 = 10.48mA.

Moving forward, we will use the par tool whenever we feel it is appropriate. A similar

reduction of resistors in series is possible through simple addition whenever we do not

need to know specific answers for each resistor, such as the voltage drop or power

consumed in each, or the voltage in the node between them. The current through

series resistors is the same, so even through an equivalent you can get the current.

Dependent sourcesOne of my favorite scenes in cinema comes from The Dark Knight : the Joker (played by

Heath Ledger) is rolling on the floor of a Gotham City prison, taking a bare-knuckle

beating from an ever-more-frustrated Batman. Master of the situation and laughing

hysterically, the Joker says: “You have nothing! Nothing to threaten me with!”

Even though the movie had not been made yet, I remember feeling something along

the same lines – although maybe less hysteric – back in 1999 when I realized that one

of the consequences of having used a 100% symbolic implementation for Symbulator



39

meant that I could make any element’s value dependent on any answer of the circuit. I

could simulate voltage or current sources that were dependent on any voltage, current

or combination thereof, with the same ease that I could simulate a 12V source.

Here is what you need to know for simulating dependent sources on Symbulator:

nothing. There is nothing special to it, nothing at all. Just write the value as a function

of the circuit’s answers, using the variables that by now you should know well, and run

the simulation like it’s nobody’s business. With Symbulator, instead of fearing them,

you will laugh in the face of dependent sources, thinking: “You have nothing!” Booyah!

Without any more introduction, let me show you examples, so you can learn by doing.

Numerical examples

AS2’s Example 2.7

Find io and vo in the circuit below.

My solution below. I named the top note o (the letter) and used the bottom node as

ground so that when I ask for vo I get the voltage in node o, matching the book’s

variable vo. Likewise I name the resistor ro, so that I can ask for i ro to get current io.

None of this special labeling is necessary, but it’s kind of cool when the variable

Symbulator gives you is called the same as the one the book is asking for. Chalk it up to

my nerdiness. Notice that 5iro is interpreted by the calculator as 5 times iro.

s\dc("j i ,0,o,3;ro,o,0,4;jd,0,o,.5iro"):iro,vo

The answer, 6.,24., is correct: io = 6 and vo = 24.

Bo2’s Example 1.9

Determine I1, I2 and v.

This is my solution. The simulation took 10 seconds.

"j i ,0,1,2;r1,1,0,3;jd,0,1,4v1;r2,1,0,5"cir:s\dc(cir): ir1,v1,ir2

The answer, -5/26,-15/26,-3/26, is correct: I1=-5/26, I2=-3/26 and v=-15/26.

For Symbulator,

dependent

sources are just

sources. They

require no

special notation.



40

AS2’s Practice Problem 2.7

Find vo and io in the circuit. My solution below:

s\dc("j i ,0,o,6;ro,o,0,2;jd,o,0,iro/4;r8,o,0,8"):vo,iro

The answer, 8,4, is correct: vo = 8 and io = 4.

HK5’s Example 1-3

Determine the power delivered by each source and consumed by both resistors.

s\dc("ei,1,0,120;r1,1,2,30;ed,2,3,2vra;

ra,0,3,15"):-pei,-ped,pr1+pra

The answer, 960,1920,2880, is right: the independent source delivers 960W, the

dependent source delivers 1920W, and the resistors consume 2880W together.

AS2’s Practice Problem P2.6

Find vx and vo in the circuit. My solution below:

s\dc("ei,x,1,35;rx,x,0,10;ed,0,2,2vx;ro,1,2,5"):vx,vro

The answer, 10,-5, is correct: vx =10 and vo =-5.

Bo2’s Example 1.10



41

Determine v1, v2 and i. My solution below. The simulation took 14 seconds.

"ei,1,0,2;r1,1,2,1/3;ed,3,2,4*ir1;r2,3,0,1/5"cir:

s\dc(cir): ir1,vr1,vr2

The answers, -15/26,-5/26,-3/26, is correct: v1=-5/26, v2=-3/26 and i=-15/26.

AS2’s Example 2.6Determine vo and i in the circuit below. My solution is shown under the schematic:

s\dc("e12,1,o,12;ri,1,2,4;ed,2,3,2vo;e4,0,3,4;ro,o,0,6"):vo,ir i

The answer, 48,-8, is correct: vo =48 and i = -8.

HK5’s Drill Problem 1.11

Find the power absorbed by each element in the circuit. My solution below:

s\dc("r1,x,0,30;ei,1,x,12;r2,1,2,8;r3,2,3,7;

ed,3,0,4vx"):approx(pr1,pei,pr2,pr3,ped)

The answer, .768,1.92,.2048,.1792,-3.072, is correct.

AS2’s Example 3.6

Determine the value of Io in the circuit. My solution below:



42

s\dc("ei,a,0,24;ro,a,b,10;r12,b,0,12;r4,b,c,4;

r24,a,c,24;ed,c,0,4iro"):approx(iro)

The answer, 1.5, is correct.

Bo2’s Drill Exercise 1.12

Determine i, v and id.

We are given an unnecessary piece of information: since there are no unknown values,

giving us the 4V drop in the 2Ω resistor is superfluous. Don’t let this confuse you.

My solution is shown below the schematic. My simulation took 18 seconds.

"ei,1,0,10;r1,1,2,1;r2,2,3,2;r3,2,0,3;r4,3,0,2;

ed,2,3,ir1/2"cir:s\dc(cir): ir1,vr3,ied

The answer, 4,6,1, is correct: i=4, v=6 and id=1.

AS2’s Example 3.2

Determine the voltages at the nodes.

My solution below:

s\dc("j i ,0,1,3;jd,3,0,2ir2;r2,1,2,2;r4a,1,3,4;

r8,2,3,8;r4b,2,0,4"):approx(v1,v2,v3)

The answer, 4.8,2.4,-2.4, is correct.

AS2’s Example 3.4

Find the node voltages in the circuit. My solution below:



43

s\dc("r2,1,0,2;e,1,2,20;j,0,2,10;r6,2,3,6;rx,1,4,3;r4,3,0,4;

ed,3,4,3vrx;r1,4,0,1"):approx(v1,v2,v3,v4)

The answer, 26.67,6.67,173.33,-46.67, is correct.


Determine the voltage in each node. Notice the resistors’ values are given in siemens.

s\dc("j,0,1,6;ei,3,1,6;ed,2,3,3v1;r5,1,0,1/5;r2,1,2,1/2;r3,2,0,1/3;r1,2,3,1;r4,3,0,1"):v1,v2,v3

The answer, -1,2,5, is correct.

Bo2’s Example 2.7

Determine the voltages in all nodes. My solution below the schematic:

s\dc("j,0,1,1;r3,0,1,3;r4,2,1,4;r1,2,0,1;r2,2,3,2;

r5,3,0,5;ei,3,4,1.5;ed,4,0,2vr4"):v1,v2,v3,v4



44

The answer, 1.5,-.5,-2.5,-4., is correct.

Bo2’s Example 2.6

Determine the voltages in all nodes. My solution below:

s\dc("e1,0,1,1;e2,3,4,.5;ed,3,2,3vr4;j,0,4,2;r4,1,2,1/4;r1,2,0,1;r8,3,0,1/8;r2,2,4,1/2"):v1,v2,v3,v4

The answer, -1,-2.,1.,.5, is correct.

HK5’s Drill Problem 1-12

Find iA, iB and iC. My solution below:

s\dc("j l ,x,0,5.6;ra,0,x,18;jb,0,x,.1vx;r9,0,x,9;

j r,0,x, 2"):approx( ira, ijb, ir9)

The answer, 3.,-5.4,6., is correct.

Numerical-from-symbolic examples




45

Determine i, v, is and vs.

With one unknown value and one known solution, this problem is a job for Expert.

"es,2,0,vs;jd,0,3,2ir1;r7,0,1,7;r1,3,1,1;r3,3,2,3;r4,1,2,4"cir:

s\Expert(cir): ir1,vjd,-ies,vs

Add ,vs to the unknowns, and add and vr4=4 to the equations. Run the simulation.

My solution, including all the typing required inside Expert, took less than one minute.

The answer, 2,-9,-3,3, is correct: i=2, v=-9, is=-3 and vs=3.


Determine i, v and vd.

Notice that, since all element values are known, the tip we are given by the book –

namely, that the voltage drop in the 6Ω resistor is 1.5V – is totally superfluous.

"e,1,0,12;r1,1,2,1;r4,2,0,4;r10,2,3,10;r6,3,0,6;r2,3,4,2;

j ,4,0,vr10/15"cir:s\dc(cir):approx(vr10,ir2,vj)

The answer, 7.5,.5,.5 is correct: i=.5, v=7.5 and vd=.5.

Symbolic examplesTR5’s Exercise 4.2

Find vO and iO in terms of iS.


"j i ,0,x, is;r1,x,0,1000;r2,x,o,2000;jd,0,o,vx/500;

ro,o,0,500"cir:s\dc(cir):vo,iro

The answer, 1000*is,2*is, is correct: v O=1000 i S and i O=2 i S.



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TR5’s Example 4.4

Find vO and the equivalent resistance RIN, in terms of vS, when R1 is 50, R2 is 1000, R3 is

100, R4 is 5000 and g is 100mA (e.g. 100/1000).


"e,1,0,vs;r1,1,2,50;r2,2,o,1000;r3,o,0,100;r4,o,0,5000;

j ,0,o,100/1000vr2"cir:s\dc(cir):approx(vo,re)

The answer, .904*vs,10951., is correct: v O=.904 v S and RIN=10.95KΩ.

TR5’s Figure 4-4

Find the voltage drop, current, and power consumed by the 500Ω resistor, and the

ratio of that power to that delivered by the independent source, all in terms of i S.

I have not labeled the nodes in the figure, so you can practice doing it. My solution:

s\dc("js,0,1,is;r50,1,0,50;rx,1,0,25;jd,o,0,48irx;

r3,o,0,300;ro,o,0,500"):iro,vo,pro,pro/(-pjs)

The answers we get are correct: iO=-12is, vO=-6000is, pO=72000is2, and pO/pS=4320.

TR5’s Example 4.1

Find v O.

I have not labeled the nodes in the figure, so you can practice. Remember not using rc,

since it is a reserved variable, with a value of 1, in the calculator. My solution:



47

s\dc("ei,1,0,vs;rs,1,2,rs;rx,2,0,rp;

ed,0,3,r*irx;rrc,3,o,rrc;rl ,o,0,rl"):vo

This is the answer we get.

It is correct, as can be seen by comparing it with the textbook’s answer.

I don’t know of any calculator-based program that was able to provide this kind of

purely symbolic answer to a circuit simulator back in 1999 when I made Symbulator.

As a matter of fact, even today – fourteen years after that - I do not know of any other.

Bo2’s Example 1.11 (FET amplifier)Determine v2.

In my solution I named the value of the source v1, to keep it similar to the book. This

required avoiding naming any node as 1: if there was a node 1, Symbulator would

store in v1 the voltage of the node, creating trouble. There is no problem with using

r1 as a value, since nothing will be stored in that r1 value.

"e,a,0,v1;r1,a,3,r1;rg,3,0,rg; j,2,0,gm*vrg;

rd,2,0,rd;rl,2,0,rl"cir:s\dc(cir):v2

The simulation took 25 seconds. The answer I obtained is shown to the left.


TR5’s Example 4.5

Find iB. My solution is shown below the circuit schematic.



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"e1,1,0,vcc;rb,1,b,rb;e2,e,b,vƒ;re,e,0,re;rrc,1,c,rc;

j ,c,e,β*irb"cir:s\dc(cir): irb

The simulation took 69 seconds. Compare my answer to the book’s answer.

TR5’s Exercise 4.3

Find vO, in terms of the value in the circuit. For resistors, use their conductance value.

This is my solution.

"ei,1,0,vs;ed,2,0,µ*(vrx);r1,1,2,1/g1;r2,2,o,1/g2;

rx,1,o,1/gx;rl,o,0,1/gl"cir:s\dc(cir):vo

The simulation took a good 2 minutes to complete. We get this answer.




Element: s

Short circuits

Describing an ideal short circuit

Shorts are used mostly when we need to find out a current in a part of the circuit

where there is no element already. Otherwise, we would just define it as a single node.

In Symbulator, an ideal short circuit is described as follows: first the name of the short,

which must start with the letter s, and a coma, second the name of the first node,

another coma, third the name of the second node, another coma, and fourth – just for

the sake of providing an equal number of values in each element’s description, a zero.

Example

An ideal short circuit called s1, connected between nodes 3 and 0 (e.g. whose current

will be defined as flowing through it from node 3 towards node 0), would be described

thus: s1,3,0,0 The zero at the end is added because Symbulator requires symmetry.

Simulation answers

No power is consumed or voltage is dropped in a short. For each short in a circuit,

Symbulator stores only the current through it , flowing from the first node towards the

second, in a variable called i and the name of the short. I.e., the current flowing

through a short called sx, from its first node towards its second node, is stored in isx.

An example using short circuits

HK5’s Drill Problem 1-13

Find i1, i2, i3 and i4.

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

This is how short

circuits are

described in

Symbulator.

circuit analysis with symbulator - volume 1

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