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B      (*========================================================*)
B      (*                                                        *)
B      (*                 Elementary Electricity                 *)
B      (*                                                        *)
B      (*  Tutorial No. 1                                        *)
B      (*                                                        *)
A      (*                       OHM'S LAW                       *)
B      (*                                                        *)
B      (*                      Version 1.0                       *)
B      (*                                                        *)
A      (*  This program describes Ohm's Law which, written  in  *)
B      (*  equation form, is  I=V/R.   I  represents  electric   *)
B      (*  current,  V  represents potential difference,   and   *)
B      (*  R represents the opposition to the flow of electric   *)
B      (*  current.                                              *)
B      (*                                                        *)
B      (*  You will be expected to calculate the value of  any   *)
B      (*  one of the three variables, given the remaining two,  *)
B      (*  and enter your results into the computer.             *)
B      (*                                                        *)
B      (*  Written by T. J. Meyers              April 11, l985   *)
B      (*                                                        *)
B      (*========================================================*)
+                             OHM'S LAW     
EGeorg Ohm, a German schoolteacher, discovered in the 1820's that in  
Ea closed electrical circuit the current was directly proportional to 
Fthe potential difference and inversely proportional to the resistance.
1Expressed mathmatically, this relation is written
EI = V/R   where    I = current in                AMPERES             
E                   V = potential difference in   VOLTS               
E                   R = resistance in             OHMS                
EThis  "law"  applies chiefly to metals and direct-current circuits.  
DChanges in the temperature of a conductor, which occur when electric
Ecurrent flows through it, can limit the laws usefulness.             
DHowever, despite its limitations, Ohm's Law can be used in numerous 
Dsituations where a high degree of accuracy is not demanded, even in 
Dsimple alternating-current circuits.  The ability to manipulate the 
Dequation is a skill well-worth mastering. It only requires a bit of 
Esimple algebra.                                                      
ELet's see what you can do.                                           
ERewrite the equation I = V/R to solve for V.  Use upper-case letters.
V = IR
=Very good.  Let`s see what you can do with the next equation.
!Incorrect.  Give it one more try.
"Good.  Now let`s try the next one.
,Wrong again. You should review your algebra.
The right answer is 
V = IR
/Perhaps you can do better on the next equation.
<Which of the expressions below represents the correct answer
when you solve I = V/R for R.
2R = I/V    R = V/I    R = IV    R = none of these.
Correct.  Good thinking.
!Tst!  Tst!  Let`s try that again.
That`s nice.  You did it.
3You flamed out again.  I hope you have a parachute.
The correct answer is 
R = V/I
E                             SUMMARY                                 
DTo sum up, then, the three forms that Ohm's Law can take are:       
EI = V/R         V = IR         R = V/I                               
0You will note that V is ALWAYS in the numerator.
EYou will need these three equations later in the program so,         
Eif you are not sure of them, take time to write them down now.       
E                        (SO WRITE!)                                  
DTo use Ohm's Law intelligently you need to understand what the terms
Eof the equation mean.  The following paragraphs set forth a series   
Eof definitions in simplistic fashion.  For more detailed information 
you should consult a good text.
F                               CURRENT                                
FSimply, current can be defined as the rate of flow of charge. Charge, 
Fas you know, is of two kinds - positive and negative.  When  charges  
Fflow in fluids both kinds can be detected in motion.   In metals the  
Fflow of charge is associated with the drift of electrons.  Electrons  
Fcarry a negative charge.   The flow of charge in solid-state devices  
Fis more complex and beyond the scope of this program.                 
FIn any case, the rate of flow of charge is measured in a unit called  
Fthe AMPERE.  The ampere is named for Andre Ampere of France.          
FThe AMPERE is defined in terms of its magnetic effects.  Ideally, if  
Ftwo very long wires were placed parallel to each other in empty       
Fspace, precisely one meter apart, and if the same constant current    
Fwere present in each wire, and if each produced on the other a force  
Fof 2 x 10E-7 newtons for each meter of length, then the rate of flow  
Fof current in each wire is defined as 1 ampere.                       
FNeedless to say, The US National Bureau of Standards uses a more      
Fconvenient measuring technique.    Look it up and read about it.      
FREMEMBER:  the unit of current is the AMPERE.                         
)                     POTENTIAL DIFFERENCE
FWhen you lift an object from the ground to a bench-top you do work on 
Fit against the gravitational field.   If you were to push it sideways 
Foff the bench, the gravitational field would do work on it (about the 
Fsame amount you did lifting it) causing it to fall back to the ground.
FElectrical potential difference represents the capacity of an electric
Ffield to do work.  Electrical potential difference, like gravitational
Fpotential energy, can also be produced by work.   A simple example is 
Fthe charge produced by chemical action inside of a flashlight battery.
FThe unit of electrical potential difference is called the VOLT.       
FThe volt is named for Alessandro Volta of Italy.                      
FThe VOLT is a unit of WORK (not of force, as you may have thought).   
FTo be more precise:  if it takes 1 joule of energy to transfer        
F1 coulomb of charge from one point to another in an electric field,   
Fthe potential difference between the two points is 1 volt.            
FIf you find these terms unclear, look them up in your text book.      
FREMEMBER:  the unit of potential difference is the VOLT.              
G                               RESISTANCE                              
GWhen electrical energy acts to move charge from one place to another   
Git encounters opposition, more in some materials than in others. This  
Gopposition to the flow of charge is called RESISTANCE.                 
GThe electrical resistance of a conductor depends upon such factors as  
Gthe material from which it is made, its physical dimensions and its    
Gtemperature.  The longer the conductor, the greater the resistance; the
Glarger the cross-section of the conductor, the smaller the resistance. 
GIn most, but not all, materials, resistance increases with temperature.
GThe unit of electrical resistance is called the OHM.  It is named      
Gfor Georg Ohm of Germany.                                              
GThe OHM can be defined in terms of electric current and electric       
Gpotential difference as follows:                                       
GIf a potential difference of 1 volt produces a current of 1 ampere     
Gin a conductor, then the resistance of the conductor is said to be     
G1 ohm.                                                                 
GREMEMBER:  the unit of electrical resistance is the OHM.               
?                               PROBLEMS                        
?You cannot truly claim mastery of a procedure until you have   
?successfully demonstrated its application several times.       
?Therefore each of the following units will require you to solve
>problems using Ohm's Law.                                     
?It will be necessary to calculate answers to 3 decimal places, 
?then round off to two decimal places.  Use standard rounding   
?procedures.  Be precise.                                       
?Good solving!                                                  
#                            AMPERES
0Remember to round off your answer in the second 
0decimal place.                                  
Problem #1:    V = 
	     R = 
!               I  (in amperes) = 
!               Good.  You got it.
>               A mistake. Check your equation and recalculate.
!               I  (in amperes) = 
!               Correct this time.
               Wrong again.
%               The correct answer is 
Problem #2:    V = 
	     R = 
!               I  (in amperes) = 
/               Right.  Move to the next square.
/               Error.  Reconsider and try again
!               I  (in amperes) = 
2               Success.  Move to the next problem.
               Another error.
%               The correct answer is 
Problem #3:    V = 
	     R = 
!               I  (in amperes) = 
6               Very good.  Now let's try the next one.
4               Sorry about that.  Recheck and retry.
!               I  (in amperes) = 
&               OK this time.  Proceed.
/               Incorrect.  The right answer is 
Problem #4:    V = 
	     R = 
!               I  (in amperes) = 
0               Correct.  Go to the next problem.
-               Haste makes waste.  Try again.
!               I  (in amperes) = 
               You are correct.
-               Mercy!  The correct answer is 
Problem #5:    V = 
	     R = 
                I (in amperes) = 
               Right again!
'               Incorrect.  Recalculate.
                I (in amperes) = 
$              This is it.  Go ahead.
/               Incorrect.  The right answer is 
6                        VOLTS                         
6Remember to round off your answer in the second       
6decimal place.                                        
Problem #1:    I = 
    R = 
               V  (in volts) = 
-               That's it.  Take 1 giant step.
                Oops!  Try again.
               V  (in volts) = 
"               OK.  Take one step.
.               Not so.  The correct answer is 
Problem #2:    I = 
    R = 
               V  (in volts) = 
&               Excellent.  Move ahead.
/               A toe stub.  Catch your balance.
               V  (in volts) = 
!               Good.  Move ahead.
-               Sorry.  The correct answer is 
Problem #3:    I = 
   R = 
               V  (in volts) = 
.               A-OK.  Advance to next problem.
*               That's an OOPS!  Try again.
               V  (in volts) = 
)               Better.  Take 1 baby step.
.               A NO NO. The correct answer is 
Problem #4     V = 
    R = 
               V  (in volts) = 
*               Perfect.  Go to problem #5.
1               A slip of the pen.  Compute again.
               V  (in volts) = 
"               Good.  That's a GO.
-               Error!  The correct answer is 
Problem #5:    I  = 
   R = 
               V  (in volts) = 
"               That's a blast off.
)               Whoops.  That's a recheck.
               V  (in volts) = 
*               Solid.  Go to next section.
0               Not again! The correct answer is 
7                         OHMS                          
7Remember to round off your answer in the second        
decimal place.
Problem #1:      V = 
    I = 
                  R  (in ohms) = 
2                 A home run.  Let's try problem #2
(                 A wild pitch.  Do over.
                  R  (in ohms) = 
-                 A single.  Advance one base.
                 A strike out.
'                 The correct answer is 
Problem #2:      V = 
    I = 
                  R  (in ohms) = 
%                 A ringer.  Go to #3.
/                 A leaner.  Take another pitch.
                  R  (in ohms) = 
-                 A ringer.  Go to problem #3.
.                 Out of range.  The answer is 
Problem #3       V = 
    I = 
                  R  (in ohms) = 
-                 In the 10-ring.  Skip to #4.
+                 In the 7-ring.  Try again.
                  R  (in ohms) = 
1                 Bull's eye!  Go to next problem.
/                 Off the board.  The answer is 
Problem #4:      V = 
    I = 
                  R  (in ohms) = 
%                 Right on.  Go to #5.
0                 A near miss.  Take a fresh aim.
                  R  (in ohms) = 
-                 Right on.  Go to problem #5.
/                 Sorry.  The correct answer is 
Problem #5:      V = 
    I = 
                  R  (in ohms) = 
)                 A hit.  Congratulations.
&                 A miss.  Last chance.
                  R  (in ohms) = 
)                 A hit.  Congratulations.
1                 Too bad.  The correct answer is 
8                       CONCLUSION                       
7You have just completed the tutorial on Ohm's Law.  If 
8you were not satisfied with your performance on the     
8numeric portion simply run through it again.  You can   
8save yourself time by concentrating only on those areas 
8where you experienced difficulty.  Good luck!           
8On the other hand, you may have been outstandingly      
8successful.  In that case, CONGRATULATIONS!             
8                          END                           
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