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CD School House 10
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CD School House - Education and Games (10.0) - Wayzata Technology (1995).iso
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TEACHAID
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PDDEMO.CAI
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1994-12-17
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10KB
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406 lines
.rem
.rem -----------------PRESS CTRL-G (Go) TO RUN THIS PROGRAM---------
.rem
.screen 0
.col 11,0
.rem l.cyan on black
.cls
.col 2
.rem green
.cen(*THE POTENTIAL DIVIDER CIRCUIT*)
.col 11
.skip 5
This demo is part of a tutorial written
to explain potential divider theory
to electronics students (16 - 18 year olds).
It shows how to use .img files to create a circuit diagram,
and to make a resistor grow in size!
.col 9
Contact Martin Rice, Newbury College of Further Education,
Oxford Road, Newbury, Berkshire, RG13 1PQ, England
for details of other .cai tutorials.
.locate 23,1
.col 2
.pause(*Press any key to continue...*)
.screen 2
.cls
.get lh.img
.put 120,22
.put 120,79
.get rv.img
.put 126,23
.put 126,55
.get lv.img
.put 126,47
.locate 14,1
Let's do another, still with easy numbers, but this time the top resistor
is 10k, and the bottom one is 5k. The voltage at the top is now +12V.
The process will again be:
calculate the current flowing by taking into account the total resistance,
then compute the voltage drop across the bottom resistor.
.locate 5,19
10K
.locate 9,19
5K
.locate 2,15
+12V
.locate 12,16
0V
.locate 23,1
.pause(*Press any key to continue*)
.rem*****************************************************************
.locate 23,1
.blank 1
@BEGIN QUESTION nocount
.locate 14,1
.blank 9
.locate 14,1
@TRIES 2
What is the value of the current flowing through the circuit?
@READ (*Please enter your answer here: *)
@RIGHT (*0.8mA*) Exact nocase
Good.
@RIGHT (*.8mA*) Exact nocase
Good.
@WRONG
Remember to calculate the total resistance, then use Ohm's Law.
Give the answer in mA, eg 1.2mA
.pause(*press any key to have another go*)
@WRONG FINAL
.locate 16,1
.blank 7
.locate 16,1
(You've had two tries at the answer.)
The correct answer is 0.8mA: 12V/15K = 0.8mA
@END QUESTION
.get vd.img
.put 126,47
.locate 7,11
0.8mA
.locate 23,1
.pause(*Press any key to continue*)
.rem*************************************************************
.locate 23,1
.blank 1
@BEGIN QUESTION nocount
@TRIES 2
.locate 14,1
.blank 9
.locate 14,1
Now calculate the voltage at the junction of the two resistors
@READ(*Please enter your answer here: *)
@RIGHT (*4v*) Exact nocase
Very good.
.pause(*Press any key to continue*)
@RIGHT (*4.0V*) Exact nocase
Very good.
.pause(*Press any key to continue*)
@WRONG (*8*) DECIMAL
Close! Your answer (8V) is the volt drop across the top resistor,
but we're interested in what's left across the bottom one!
.pause(*press any key to have another go*)
@WRONG
Use V = I*R to get the volt drop across the bottom resistor.
.pause(*press any key for another try*)
@WRONG FINAL
.locate 16,1
.blank 7
.locate 16,1
(You've given two wrong answers)
The correct answer is 4.0V: 0.8mA * 5k = 4.0V
.pause(*Press any key to continue*)
@END QUESTION
.locate 7,19
+4.0V
.locate 5,25
8V
.locate 9,25
4V
.get up-arrow.img
.put 192,22
.put 192,51
.get down-arr.img
.put 192,43
.put 192,72
.locate 14,1
.blank 9
.locate 14,1
Notice that the bottom resistor is 1/3 of the total resistance,
and (wait for it...)
.wait 12
the voltage at the junction is 1/3 of the total voltage.
.wait 12
Also, the top resistor is twice as big as the bottom one,
.wait 12
and has twice as much volt drop as the bottom one.
.pause(*Press any key to continue*)
.rem**************************************************************
.cls
.screen 2
Look at it like this:
.get lh.img
.put 120,112
.get rv.img
.put 126,88
.get lv.img
.put 126,80
.get rv.img
.put 126,56
.get res-grow.img
.put 126,54
.play (*t200 mf o3 l8 c*)
.put 126,52
.play (*t200 mf o3 l8 c+*)
.put 126,50
.play (*t200 mf o3 l8 d*)
.put 126,48
.play (*t200 mf o3 l8 d+*)
.put 126,46
.play (*t200 mf o3 l8 e*)
.put 126,44
.play (*t200 mf o3 l8 f*)
.put 126,42
.play (*t200 mf o3 l8 f+*)
.put 126,40
.play (*t200 mf o3 l8 g*)
.put 126,38
.play (*t200 mf o3 l8 g+*)
.put 126,36
.play (*t200 mf o3 l8 a*)
.put 126,34
.play (*t200 mf o3 l8 a+*)
.put 126,32
.play (*t200 mf o3 l8 b*)
.get lh.img
.put 120,31
.play (*t200 mf o4 l8 c*)
.locate 3,15
+12V
.locate 13,13
5K
.locate 7,12
10K
.locate 7,21
8V
.locate 13,21
4V
.locate 16,17
0V
.get up-arrow.img
.put 162,31
.put 162,85
.get down-arr.img
.put 162,76
.put 162,105
.get lv.img
.put 162,39
.put 162,68
.put 162,60
.locate 11,12
+4V
.wait 10
.locate 5,30
The idea is really very simple:
.locate 7,30
1: The bigger the resistor,
.locate 8,30
the more volt-drop there is.
.wait 12
.locate 10,30
2: All the volt-drops have to add up
.locate 11,30
to equal the total supply.
.locate 13,30
.wait 12
3: The output voltage is just
.locate 14,30
the volt-drop across the bottom resistor.
.locate 16,30
(This assumes there's 0V
.locate 17,30
at the bottom of the divider chain.)
.locate 22,1
.wait 12
Make sure you have taken that all in before you
.locate 23,1
.pause(*continue (press any key to do so)*)
.locate 1,1
.blank 1
.locate 3,30
┌───────────────────────────────────────────────┐
.locate 4,30
│ │
.locate 5,30
│ │
.locate 6,30
│ │
.locate 7,30
│ │
.locate 8,30
│ │
.locate 9,30
│ │
.locate 10,30
│ │
.locate 11,30
│ │
.locate 12,30
│ │
.locate 13,30
│ │
.locate 14,30
│ │
.locate 15,30
│ │
.locate 16,30
│ │
.locate 17,30
└───────────────────────────────────────────────┘
.locate 20,1
.blank 4
.locate 4,32
So, to find the output of a potential divider
.locate 7,32
1: Calculate what proportion of the total
.locate 8,32
resistance the bottom resistor is...
.locate 10,32
(in this case it's 1/3rd)
.wait 20
.locate 13,32
2: This also tells you what proportion
.locate 14,32
of the input voltage will come out.
.locate 16,32
(1/3rd of 12V is 4V.)
.locate 23,1
.pause(*Press any key to continue*)
.locate 3,30
┌───────────────────────────────────────────────┐
.locate 4,30
│ │
.locate 5,30
│ │
.locate 6,30
│ │
.locate 7,30
│ │
.locate 8,30
│ │
.locate 9,30
│ │
.locate 10,30
│ │
.locate 11,30
│ │
.locate 12,30
│ │
.locate 13,30
│ │
.locate 14,30
│ │
.locate 15,30
│ │
.locate 16,30
│ │
.locate 17,30
└───────────────────────────────────────────────┘
.locate 23,1
.blank 1
.locate 5,32
If the resistor values are too difficult
.locate 6,32
for mental arithmetic,
.locate 8,32
use a calculator to get the proportion,
.locate 9,32
then multiply by the input voltage.
.locate 11,32
The proportion is, of course,
.locate 13,32
(bottom resistance) 5K
.locate 14,32
─────────────────── eg ─── = 1/3rd
.locate 15,32
(total resistance) 15K
.locate 23,1
.pause(*Press any key to continue*)
.locate 3,30
┌───────────────────────────────────────────────┐
.locate 4,30
│ │
.locate 5,30
│ │
.locate 6,30
│ │
.locate 7,30
│ │
.locate 8,30
│ │
.locate 9,30
│ │
.locate 10,30
│ │
.locate 11,30
│ │
.locate 12,30
│ │
.locate 13,30
│ │
.locate 14,30
│ │
.locate 15,30
│ │
.locate 16,30
│ │
.locate 17,30
└───────────────────────────────────────────────┘
.locate 23,1
.blank 1
.locate 5,32
To summarise,
.locate 11,36
output = input * ───────────────────
.locate 10,36
(bottom resistance)
.locate 12,36
(total resistance)
.wait 15
.play(*t200 ml o1 l4 g l8 ab o2 cdef l2 g*)
.locate 20,1
But remember to use common sense, and the idea of proportions
to check that your answer looks right.
.wait 5
.play (*mn a l4 o1 b o2 l2 c*)
.locate 23,1
That completes the explanation part of this tutorial.
.pause(*Press any key to return to the menu.*)
@menu