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Turbo Pascal Chain module | 1993-03-01 | 22.3 KB | 278 lines |
- To continue, press any key
- 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
- 7 Press spacebar to try again
- 7 To return to the menu
- 7 first, type (q,Q) to quit,
- 7 then, type (menu,MENU) at A>
- 9 Press (q/Q) to quit
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