Maths / Electronics question...

[DRM]

Blimey, it's easier to connect them up, then put a multimeter across them!

 

[Homers Double]

Or M+1 is the correct number of minions required to complete the task for you in this case.

 

[si_c]

you're looking for R_t the total resistance.

The formula is 1/R_t = 1/R_1 + 1 / R_2 + 1/ R_3

If the resistances are 2, 3 and 4 then it's as follows.

1/R_t = 1 / 2 + 1 / 3 + 1/ 4
= 6 / 12 + 4 / 12 + 3 / 12
= 13 / 12
So now we need to rearrange the numbers so that R_t stands on it's own.

1 = R_t * ( 13 / 12)
1 / (13 / 12) = R_t

R_t = 12/13 ( or 0.92) Ohms.

In your case, with three 9 ohm resistors its:

1 / R_t = 1/9 + 1/9 + 1/9
1 / R_t = 3/9

1 = R_t * (3 / 9 )
1 / (3/9) = R_t
9 / 3 = R_t
R_t = 3 Ohms.

You've just got to rearrange the fractions to put R_t in the right place.

 

[si_c]

Just to add: In my example above I used 12 as the common denominator - it's just the smallest number that 3,4 and 2 can divide into easily and makes doing the calculation a bit simpler. You can do it with a calculator and you don't need to worry about it.

In fact:

1/ R_t = 0.5 + 0.33333333 + 0.25
1 / R_t = 1.083333333
R_t = 1 / 1.083333333
= ~0.92

 

[ColinJ]

I wanted to calculate a reciprocal yesterday but my phone's calculator app doesn't have a reciprocal button. Rather than doing 1 / the-very-long-number-already-in-the-app (which would have involved retyping it), I just raised it to the power of -1 (^ button then -1)

Example of 1/pi...

(On the Android calculator app, click the symbols above the AC button to see an expanded set of buttons.)

 

[Dogtrousers]

IMO A spreadsheet is by far the best solution for this. Three cells in the first column one above each other to input your resistances. Three cells in the next column next to them with the reciprocals of the inputs. One cell below with the sum of the reciprocals. One cell below that with the reciprocal of that sum.

You can see and understand all the intermediate values and how the calculation flows. You can experiment with different inputs to see the effect they have.

That or write down each calculation on paper, which is the same but more manual.

No arsing around rearranging fractions or finding lowest common denominator or any of that which, although interesting in its own right, has absolutely nothing to do with the problem in hand.

 

[Ming the Merciless]

I don't understand how it works and probably won't

Imagine a river, that starts as one channel, splits into three, then combines back into one channel. When in equilibrium; the total of water flowing in the initial channel is the same as the total flowing through the three channels, as through the final single channel. The sum of the currents through the three channels matches the current through the single channel. It is the same for resistors in parallel; they represent paths for the current.

Each parallel resistor has the same potential difference / voltage across it. We know that the current through any one resistor is V / R. We know the sum of the currents through each parallel resistor matches the total current.

Thus we know that

C = C1 + C2 + C3

V/R = V/R1 + V/R2 + V/R3.

We can eliminate common terms on both sides by dividing by them. Thus divide both sides by V gives.

1/R = 1/R1 + 1/R2 + 1/R3

The equation form you are given.

 

[T4tomo]

O as I'm explaining this I realise I have confused you by making an error in my haste!! correct version below

Example 2
if it was say 3. 4 & 7 ohms
1/3+1/4+1/7 - so need a number they are all factors of. 3*4*7=84, so
i.e 28/84 + 21/84 + 12/84 = 61/84 o r=84/61 or 1.377ohms.

I can follow up until you added 28/84, then to mee it looks like we are back to random numbers. I can see where the 84 comes from but not the 28 or the 21, or the 7,

I understand how 56 is 1/3 of 84 (with help from a calculator) just not how we get 56 in the first place.

What I really need is a step by step instruction to follow. As said above, from experience I probably won't understand the equation, but if I have a set of steps to follow then I can be confident of getting the correct number at the end.

If that means "Add the top row and multiply the bottom, then divide the bottom by the total of numbers on top" or whatever that's what I'll do, then I'll practice until the exam, and forget it a day later.

So we were then just rewriting the equation but using 84 (3*4*7) as the denominator rather than 3, 4 and 7.
so 1/3 is just 28/84 (we use 4*7=28 to multiply top and bottom (i.e the factors in the common denominator that isn't in that fraction(1/3)
I.e 1* 4*7 over 3* 4*7
1/4 becomes 21/84 - multiple both by 3*7=21
1/7 becomes 12/84 multiple both by 3*4=12
28+21+12=61

 

[presta]

for identical resistors in parallel, the total resistance can be calculated by dividing the resistance of one resistor by the number of resistors

And even when they aren't identical.
Without a calculator, or writing anything down, 6k in parallel with 2k is 1.5k.
(6k/4, because the total number of 6k resistors in parallel is four).

 

[Yellow Fang]

I'm currently trying to practice the maths for parallel resistance, and the "worked examples" are frankly quite entertaining. Could someone with a more mathematical brain explain?

The equation given is:

1 / r (total) = 1 / r1 + 1 / r2 + 1/ r3

However my experience with maths is that I probably won't understand it: I just need to learn to use the equation and the number on the other end is usually the right one. However, this doesn't make sense to me because what use is knowing "1 / r (total)"?

The problem set for me was:



1 / r (total) = 1 / r1 + 1 / r2 + 1/ r3

= 0,1111 + 0,1111 + 0,1111 = 0,3333 Ω

So 1/ r (total) / 1 is 0,3333 Ω

I wasn't sure what to do with this so I looked up the solution. It said:



This didn't help, so I asked AI an AI programme. This is what AI said:



Okay, not sure how that should help, but I got:

1/r total = 0,1666 + 0,1333 + 0,5 = 0,7993Ω

But the "example" seems to go off the deep end:



Where did the "common denominator" come from, and why is it a 6? To me it looks like it substitutes a stack of random numbers to get the right answer.

Can anyone explain how to consistently use the equation to get an accurate answer? I am aware this is a sign of discalculia; I don't understand how it works and probably won't, but I've found that given the formula, with care I can get the right answer, even if it is like pulling the lever on a fruit machine.

I just need to pass the exam, have the certificate, and can get on with my real job, so any assistance is welcome.

Why are you studying this at your time of life?

 

[Andy in Germany]

Why are you studying this at your time of life?

Fair comment.

I'm working as an occupational therapist, and I happen to be working in a bike workshop with people who have psychological issues. I've been asked to take the German bike mechanics exam in June 2027 to make me better qualified for the job, and this involves a theory exam.

Because Germans, bless them, love to shoehorn the maximum amount of maths into any given qualification, I have to be able to understand the formulas complete formulas so it looks like I understand them.

The formula for working out spoke lengths when building wheels is coming up and that's even worse...

 

[Andy in Germany]

Imagine a river, that starts as one channel, splits into three, then combines back into one channel. When in equilibrium; the total of water flowing in the initial channel is the same as the total flowing through the three channels, as through the final single channel. The sum of the currents through the three channels matches the current through the single channel. It is the same for resistors in parallel; they represent paths for the current.

Each parallel resistor has the same potential difference / voltage across it. We know that the current through any one resistor is V / R. We know the sum of the currents through each parallel resistor matches the total current.

Thus we know that

C = C1 + C2 + C3

V/R = V/R1 + V/R2 + V/R3.

We can eliminate common terms on both sides by dividing by them. Thus divide both sides by V gives.

1/R = 1/R1 + 1/R2 + 1/R3

The equation form you are given.

Thanks, Ming, the problem is the same as it always is: I can understand the theory, but as soon as it becomes mathematical I can't see the connection, so "logical" steps don't make sense to me. In your example above, I can follow until "We can eliminate..." then my non-mathematical brain can't make the leap.

This is a known symptom of dyscalculia; I display several symptoms, but we didn't realise this until I learned to be an occupational therapist and were looking at learning difficulties. It's often described as "Dyslexia with maths"because it's not an intelligence problem, but a learning difficulty. An example of this is that I was consistently good at spelling tests at school, but still can't memorise times tables or do anything more than the most basic mental arithmetic.

My solution to this is to work out the method to deal with a formula and follow it rigidly, practice obsessively before the exam and then forget it. I'm also extremely careful about writing the units of measurement at every stage, so I don't get mixed up. I generally don't know what the number at the other end actually means, or really how I got it, but from experience, it's usually the correct number, so I get the point, pass the exam, collect the certificate, and do what I want to do afterwards.

 

[Yellow Fang]

Fair comment.

I'm working as an occupational therapist, and I happen to be working in a bike workshop with people who have psychological issues. I've been asked to take the German bike mechanics exam in June 2027 to make me better qualified for the job, and this involves a theory exam.

Because Germans, bless them, love to shoehorn the maximum amount of maths into any given qualification, I have to be able to understand the formulas complete formulas so it looks like I understand them.

The formula for working out spoke lengths when building wheels is coming up and that's even worse...

Bicycles or motorbikes?

 

[Andy in Germany]

Bicycles or motorbikes?

Bicycles: I don't want to think about how much maths they'd try to cram into a motorcycle apprenticeship....