## Introduction: About OHM and His LAW

OHM's LAW - What it is. How it works.

A personal LEARNING AID for the interested and patient learner. Just read the following pages with care or call them up using the HELP function under program execution.

A) Learn the color code for resistors through practice.

B) Learn to calculate rsistance for SERIES, PARALLEL, SERIES-PARALLEL circuits. Circuits are auto-generated and displayed. Answers are verified and a score card is kept. Chose a level of difficulty: (E)asy, (D)ifficult, (E)xpert. A calculator and notepad is needed for the more advanced levels. Just a notepad will do for the (E)asy level.

IMPORTANT: Answering questions on impulse will lead to failure and frustration. READ the instructions that preceed the cursor with care and respond in kind before going on (pressing ENTER).

REMEMBER: GIGO - Garbage In, Garbage Out

## Step 1: A Summary of What Is Available

HELP : Start by reading ALL the pages (5) in the (H)elp section. The pages are presented below along with some expanatory note.

## Step 2: Color Code

This page explains the color code for resistors.

SELECT THIS EXERCISE FIRST and get comfortable with the code.

Because of variations in color on different displays, an abbreviation (e.g. 'Viol' for violet) of the intended color is dispalyed just ahead of the cursor whenever a color interpretation is called for. A ribbon of colors is diaplayed on the upper right part of tha screen at all times.

This is a fun exercise !

## Step 3: General Rules - Resistance Network

These are the general rules for finding the total resistans (Rt) of resistance network. Note: Rt=Total resistance, Rs=Series resistance (really the Rt for a SERIES network), Rp=Parallel resistance (really the Rt of a parallel network), Req=equivalent resistance.

Read this page several time if necessary or until you are comfortable with the terminology.

## Step 4: General Rules

These are the general rules for series and for parallel circuits.

Instead of WIRES carying ELECRICITY, imagine that you have PIPES carrying WATER. Water PRESSURE is analogus to VOLTAGE, RESISTANCE to the flow of water (rough surface for exmaple) is analogous to RESISTANCE to the flow of electricity, The AMOUNT of water flowing in/out the PIPE is analogous to the CURRENT (AMPERES) that is available in an electical circuit. See the drawing below.

IN SERIES::: For a series of pipes connected in a closed loop::: Select a point in the loop as the beginning. That point is also the end point::: There is only one path through the loop::: The loop is like the figure z-e-r-o.

SERIES pipes are linked end-to-end. If a constant PRESSURE is maintained on the water in the pipe, you can draw the same AMOUNT of water in a given time no mater where you tape (or punch) the pipe. AMOUNT=CURRENT=AMPERES: ***IN A SERIES CIRCUIT THE CURRENT IS THE SAME IN ALL PARTS ***.

It makes sence that the total resistance in the series of pipes is equal to the SUM of the resistance in the individual sections of pipe. Where else could the resistance come from? ***IN A SERIES CIRCUIT Rt=R1+R2+....Rx*** Total resistance equals the sum of all the individual resistances.

Starting from a point in the SERIES loop, the PRESSURE will drop in successive sections of pipe because of resistance to flow. The sum of all the drops in pressure across all the sections of pipe in the SERIES loop is the same as the drop in pressure at the starting piont: ***IN A SERIES CIRCUIT, THE VOLTAGE DROP IS EQUAL TO THE SUM OF THE VOLTAGE DROPES ACROSS EACH COMPONENT***

IN PARALLEL::: Using the water pipe analogy, think of pipes connected together in the form of the figure e-i-g-h-t::: From any given starting point there is more than one path or branch in which water can flow. The secret for dealing with parallel circuits is to isolate each BRANCH and use the SERIES rules described above to find an EQUIVALENT RESISTANCE for each BRANCH(Req1 and Req2 in this case). This in effect reduces the circuit to TWO RESISTANCES (Req1 and Req2) in SERIES. As with SERIES circuits, TOTAL RESISTANCE is the sum of the individual resistances - Rpt=Req1+Req2 in this case.

Study the example in Photo 5 with care. Once you understand this example you will be confident when solving similar problems.

## Step 5: Parallel Resistance - How To

Study the example in this photo until you understand it thoroughly. Remember what was said about BRANCHES in step 4.

TIP: A quick way to find the sum of two parallel resistances is to divide their product by their sum:

Rt=(R1xR2)/(R1+R2).

If R1=20 and R2=30, then Rt=20x30/20+30=600/50=12

It follows that if R1=R2, then Rt= half of one of them.

If R1=20 and R2=20, then Rt=20x20/20+20=400/40=10

::: In the example below, think of 't' as TOP, 'v' as VERTICAL, and 'b' as BOTTOM.

::: BRANCH 3 consists of Rt3, Rv3, and Rb3 in SERIES. Therefore Rs3=Rt3+Rv3+Rb3=30+40+50=120 by our SERIES Rule. Rs3 is now in PARALLEL with Rv2 and the EQUIVALENT resistance, Req1, is calculated as described under TIP, above:

Req1=(Rs3xRv2)/(Rs3+Rv2)= (120x30)/(120+30)=3600/150=24

Now BRANCH 3 has an equivalent resistance of 24

::: BRANCH 2 consists of Rt2, Req1 and Rb2 in SERIES. Therefore Rs2=Req1+Rt2+Rb2=24+40+20=84 by our SERIES Rule. Rs2 is now in PARALLEL with Rv1 and the EQUIVALENT resistance, Req2, is calculated as described in TIP, above:

Req2=(Rs2xRv1)/(Rs2+Rv1)=(84x40)/(84+40)=3360/124=27

Now BRANCH 2 has an EQUIVALENT resistance of 27

::: BRANCH 1 consists of Rt1, Req2 and Rb1 in SERIES. Therefore Req3, now Rt= Req2+Rt1+Rb1=27+10+50=87, the TOTAL RESISTANCE of the circuit. VOILA !

## Step 6: A Summary of the Rules

Below is a summary of the rules of OHM's LAW and more.

Think in terms of pipes and water instead of wires and electrons if you need to.

## Step 7: A Summary of What You Can Do

You can select from 3 different kinds of exercises:

A - Color coding

B - Measurements : select a TYPE and a DIFFICULTY LEVEL

C - Design a resistance circuit and perform the calculations.

Your work will be verified.

ADVICE: Entering random responses will yield

GIGO (Garbage In, Garbage Out)

Practice with the easier problems first. You will gain confidence and expertise, than go on to the harder problems.

RUN R.EXE and wait for the program to load and auto execute.

Ignore the address, etc., on the title page.

## Share

### azeemteaching made it!

## Recommendations

We have a **be nice** policy.

Please be positive and constructive.

## 26 Comments

I've always known OHMs law and Rs in series and parallel BUT did NOT know (or forgot) how to calculate Networked Rs - so spending 10mins on STEP 5 was very satisfying especially as working from the RHS calculating series R taken in parallel with next R loop to the left and so on NOW seems very LOGICAL - A much appreciated intuitive tutorial! NOT wishing to be critical BUT why make life hard and write R.exe as a CONSOLE app rather than a Windows based program? Your successful project was my reason for registering with Instructables.com SO thanx AKA1 - so thanx

When I took Basic Electronics in 9th grade, this was a quiz the teacher put on the board the first day. He offered an extra A grade to anyone that produced the right equation that solved it...

I remember the same problem given by one of my teachers. Needless to say I did not get an 'A'. From what I remember, the solution is as follows:

thats a solution from point A to the furthest point B. If the points were on the same face, so that one resistor is in series with A, what would the resistance bee then? What about to the opposite side of one face, on your cube in the problem it would be the corner where you have written Vs

Currents cannot be used because they do not follow a symetrical pathway, from A to the closest point *im calling this B* you would have 3 possible pathways, and from A to the opposite corner of a face *I am calling this C* you would have 2 possible pathways.

Here is a blowup of the summary.

Point B is a point from where measurements must be made and it is NOT a three resister array point.

From B Rvr and Rvg are in parallel and the resistance for BRANCH B (RatB)=1/2(Rvg)=1/2(1/3R)=1/6R=R/6

RatB, Rvq and Rvs form a series BRANCH and:

Rt=RatB+Rvq+Rvs=(R/6)+(R/3)+(R/3)= (5/6)R

Therefore Rt=5/6 Meg ohms.

Here we go again! Been retired too long! Here is a more detailed explanation of the 12 Resistors Cube.

For simplification the VERTICIES of the cube are color coded and involve all 12 resistors: Bu(blue), Rd(red),Bl(black), Gn(green). 'R' is used as R=1 Meg ohms throughtout. See the diagrams ::: NOTE that the POINT B is NOT one of the verticies with an array of resistors. It is the point from which we make our final measurement.

For any vertex, say Bu, there are three equal resistors in parallel and their combined resistance is: 1/Rbu=(1/R1)+(1/R2)+(1/R3)=(1/1)+(1/1)+(1/1)=3

and Rbu=R/3 (one-third one of the resistors)

Since all the vertices are the same, we conclude:

Rbu=Rgn=Rrd=Rbl=R/3

:::In the diagram showing equivalent resistors ::: We want to measure from POINT B::: We note that we can connect; RED to B, BLACK to B, GREEN to B, but there is no path to connect BLUE to B except through one or more colors. We conclude the EQUIVALENT circuit shown::: A SERIES/PARALLEL CIRCUIT between POINTS A and B, and........

we calculate the BRANCH resistance in the parallel branch (R at B, RatB) as follows: RatB=(Red)(Green)/(Red+Green)::: RatB=(R/3)(R/3)/(R/3+R/3) or simply RatB=(1/3)(1/3)/(1/3+1/3)= (1/9)/(2/3)=1/6 or R/6.

::: We now have BRANCH circuit consisting of RatB,BLUE,and BLACK in series and Rt=RatB+RBlue+RBlack. Simply put

Rt=(R/6)+(R/3)+(R/3)=5/6R= 5/6 Meg ohms, the resistance between POINTS A and B on the Resistors Cube.

250 kOhms???

Some guessed correctly and incorrectly. Others actually built the bridge and measured it! Of course, none in either group got an 'A' out of it.

Half a meg

Again; Without an equation, there would be no credit wether your answer were right or wrong… Perhaps more conventional 2 dimensional schematics of the circuit would help:

I'm to buzzed for equations lets say it's six three meg resistors in parallel so 3 meg divided by number of legs equals half a meg.

I have a question for you. What if you're trying to find the resistance from the starting point to the point closest to it? If you could get back to me ASAP, it would be greatly appreciated.

From what I am told from my OLD electronics teacher...They used to use "Bad Boys Rape Our Young Girls, But Violet Gives Willingly (for SILVER and GOLD!)" when he was growing up but that slowly phased out due to its offensiveness. I dont think many people have a replacement for it, thus a device like this is not used.

awsome

Except for your typo… It's 1/3 + 1/6 + 1/3 = 5/6V, therefore 5/6MΩ or 833.33KΩ :)

Dear Las Vegas:

You did'nt say if you got an 'A' for solving the RISISTORS CUBE PROBLEM when in the 9th grade.

I have presented my solution with supporting comments; Rt=(5/6)Meg ohms. I am interested in knowing of other approaches to a solution and would like to know yours. Would you share it with me? If you decline I will not be offended.

Some time ago I was given the 'solution' of a respected collegue as Rt=(1/4)(1/3R)=1/12 Meg ohms. You can guess how he arrived at that conclusion. Most people just give a number as a solution. I am more interested in the logic (method) used to arrive at a solution. Hope to hear from you.

Sincerely, richjd.

My solution followed the same logic as cpotoso's. Labeling the other corners A1, A2, A3, B1, B2 & B3. Then following the current and applying Ohm's Law to determine the total resistance. Yes, I among others, did receive the credit. I did not solve it though until learning Ohm's Law later in the course. I have to admit that I do like your logic for the solution as well. I wouldn't have approached it from that direction. Also, thanks for the links. I've never seen that problem elsewhere on the net. Your collegue's solution was flawed to begin with and I suspect wasn't thought through long.

Find out more about the RISISTORS CUBE problem. Visit

http://www.radioelectronicschool.net/files/downloads/resistor_cube_problem.pdf

http://www.physics.upenn.edu/courses/gladney/phys151/bonus1_solution.html

richjd

this looks very useful for classroom use, are you teaching a class with it currently?

Hi dan: I wrote the program in 1989 when I was teaching Basic Electricity and Digital Electronics courses. I retire in 1992 and took several teaching aid programs which I had written in the 80s out of moth balls. I made extensive use of this program and I guess it is still pertinent today. Glad you liked it. Sincerely, Richard