# Spatial Matrix

#### T.Jackson

Joined Nov 22, 2011
328
Then you do this one ...

:: OHM's Law Quiz ::

Requirements:

• 6 resistors ...
• 5 specific Vrefs of: 1, 2.2, 3, 3.3, and 4.5VDC
• IT = 20mA
• Calculate value of R1 - R6
Rich (BB code):
              4.5V      3.3V      3V        2.2V      1V
|         |         |         |         |
5V o--/\/\/--+--/\/\/--+--/\/\/--+--/\/\/--+--/\/\/--+--/\/\/--o GND
R1        R2        R3        R4        R5        R6
Show all workings, natural voltages measurable from ground.

#### t_n_k

Joined Mar 6, 2009
5,455
Is there a trick?

R6=1/20mA=50Ω
R5=(2.2-1)/20mA=60Ω
R4=(3-2.2)/20mA=40Ω

and so on ....

R1=(5-4.5)/20mA=25Ω

I'm reluctant to open a zip file for fear of viruses.

#### T.Jackson

Joined Nov 22, 2011
328
That's wrong.

No need to fear of virus. I don't do that.

#### t_n_k

Joined Mar 6, 2009
5,455
Many thanks for the interesting challenge.

If you mean my answer to your question is wrong then you might be kind enough to help me correct my errors.

#### T.Jackson

Joined Nov 22, 2011
328
Many thanks for the interesting challenge.

If you mean my answer to your question is wrong then you might be kind enough to help me correct my errors.
Let me get a calculator and type it all out.

#### SgtWookie

Joined Jul 17, 2007
22,221
No, no calculators!

You must use a slide rule.

#### T.Jackson

Joined Nov 22, 2011
328
:: Problem Solution ::

Rich (BB code):
Step 1.  Calculate RT

RT = V / IT
= 5 / 0.02
= 250Ω

Step 2. Calculate R1  R5

Using Formula: R = (V - Vref) / V x RT

Therefore ...

R1 = V  Vref1
= 5 - 4.5
= 0.5 / V
= 0.1 x RT
= 25Ω

R2 = Vref1  Vref2
= 4.5 - 3.3
= 1.2 / V
= 0.34 x RT
= 60Ω

R3 = Vref2  Vref3
= 4.5 - 3.0
= 0.3 / V
= 0.06 x RT
= 15Ω

R4 = Vref3  Vref4
= 3.0 - 2.2
= 0.8 / V
= 0.16 x RT
= 40Ω

R5 = Vref4  Vref5
= 2.2 - 1.0
= 1.2 / V
= 0.24 x RT
= 60Ω

Step 3. Calculate Trimming R

R6 = RT   (R1 + R2 + R3 + R4 + R5)
= 250  (25 + 60 + 15 + 40 + 60)
= 50Ω

#### T.Jackson

Joined Nov 22, 2011
328
I think that's right. Been a while.

#### t_n_k

Joined Mar 6, 2009
5,455

I notice I got the same answers as you did despite being "wrong".

#### T.Jackson

Joined Nov 22, 2011
328

I notice I got the same answers as you did despite being "wrong".
Arh see when I just see numbers with no working I automatically assume it to be insanity and totally wrong.

You need to explain how you arrive at the answer.

#### t_n_k

Joined Mar 6, 2009
5,455
I understand. I'll keep that in mind. Cheers!

#### T.Jackson

Joined Nov 22, 2011
328
How would you do it if the Vref was required across each resistor instead?

#### Wendy

Joined Mar 24, 2008
22,546
Smell the smoke?

#### thatoneguy

Joined Feb 19, 2009
6,359
How would you do it if the Vref was required across each resistor instead?
Depends on how much current was needed at each voltage point.

i.e. set I=1mA and you have a lot higher resistor values, set I=20 A and you have very low resistor values (and a lot of heat).

#### thatoneguy

Joined Feb 19, 2009
6,359
Arh see when I just see numbers with no working I automatically assume it to be insanity and totally wrong.

You need to explain how you arrive at the answer.
He showed his work/formulas. The steps between could be done mentally instead of a calculator after making a few hundred voltage dividers for other circuits.

#### T.Jackson

Joined Nov 22, 2011
328
He showed his work/formulas. The steps between could be done mentally instead of a calculator after making a few hundred voltage dividers for other circuits.
Well he's extremely gifted if he can do all of that mentally. I sure as hell cannot.

#### T.Jackson

Joined Nov 22, 2011
328
I actually remember it all as a 'protocol' (set of step-by-step rules) -- that I was formally taught.

1. Decide on IT
2. Find RT
3. R = (the voltage that you want / the voltage that you have) x RT

I have memorized many things such as resistor colour codes in the form of sequential colours.

When you study, you write down on a piece of paper many times and you never forget.

Last edited:

#### steveb

Joined Jul 3, 2008
2,436
Well he's extremely gifted if he can do all of that mentally. I sure as hell cannot.
Actually, t_n_k is gifted at seeing easy solutions to complex problems. We've seen that here before. This problem might not be all that complex, but still a person might not see the simple approach right away. Often we take the hard way, out of habit.

In general, problems have easy solutions and difficult solutions, so it is important to not just dive straight into a problem and do it by the standard methods. Often a little thinking reveals a viewpoint that makes the whole problem seem trivial.

A computer program that was developed to solve very general resistor network problems, might not be able to find t_n_k's simplifying viewpoint on this problem, and it might write out a bunch of simultaneous equations and then solve them in a "brute-force" way. I'd be willing to bet that if students were given this problem on a test, some would dive straight in with simultaneous equations and waste 10 minutes that could be spent on the other problems. The calm guy (or girl) would take time to look carefully and then see the easy way.