Wendy: This is page 47 in the book. I've obviously got an aversion to math, calculator buttons and reading numbers off the screen. It's a good job this isn't all in megavolts or I'd probably not only fry circuits but also blow myself up and anyone else standing around me

Papabravo: That does give me peace of mind!


Update: Wendy, you also use it to calculate blood flow to your limbs??!

 

Tain't no problem. The idea is to have fun.

 

Well since it was my profession as well as my hobby we always just bought the resistors we needed. After I bought 5 of every conceivable value from 1 Ω to 1 MΩ and a few bags of 200 from Digi-Key it seems I just never ran through that stock and then SMT came along further decreasing the need for through hole components.

 

Thanks everyone. I seem to have a grasp of it... give me 10 minutes before I'm back with a confused face

 

Hello there,

As in post #14 we have for three resistors (simplified):
(R1*R2*R3)/(R2*R3+R1*R3+R1*R2)

and for four resistors:
(R1*R2*R3*R4)/(R2*R3*R4+R1*R3*R4+R1*R2*R4+R1*R2*R3)

which you can see has the general form of:
∏(r)/(∑ (∏(r,N-1)))

which in human language means:
"The product of all resistors divided by the sum of products of all combinations of all resistors taken N-1 at a time".

So the numerator is just the product of all resistors, no matter how many.
The denominator is the sum of products, where the products here are formed by taking every possible combination of resistors but only N-1 at a time, so if you have 4 resistors you form the products of all possible combinations of 3 resistors, summing the results of all these products.
You can start to see why we write this more commonly as:
1/(1/R1+1/R2+1/R3+...+1/Rn)

Another view is like this...

Start with the other formula for two resistors:
Rp=Ra*Rb/(Ra+Rb)

and calculate for R1 and R2 in parallel, then use that as one resistor in the above formula, we get:
(Ra*R1*R2)/(R1*R2+Ra*R2+Ra*R1)

and then replace Ra with R3 and get:
(R1*R2*R3)/(R2*R3+R1*R3+R1*R2)

which is the same formula as for three resistors in parallel. What this means is we can apply the original formula:
Rp=Ra*Rb/(Ra+Rb)

recursively when there is more than two resistors by FIRST calculating the parallel combination Rp of R1 and R2, then using that result to calculate the parallel combination of Rp and R3, and the result is the same as before: three resistors in parallel.

So that means we have at least THREE ways to calculate parallel resistors when there are more than two in parallel:
1. The direct reciprocal method.
2. The harder to formulate product over sums of products of N-1 combinations.
3. The two resistor formula applied recursively.

Take your pick
I have used #1 and #3 mostly in the past for real circuits, and #2 mostly for academic illustration.

 

The book then tells me that this will not work for calculating resistance for more than 2 resistors in parallel?

So (R1 x R2 x R3) / (R1 + R2 + R3) won't work?
Neither should (R1 x R2 x R3 x R4 x R5) / (R1 + R2 + R3 + R4 + R5) right?

You KNOW that these are both wrong just by considering the units. The first one has three resistances multiplied together in the numerator, so the units there are ohms-cubed. The denominator is the simple sum of three resistances, so it has units of ohms. When you divide the first by the second, you are left with ohms-squared, which is NOT the units of resistance that you need. Similarly, the units on the second is ohms raised to the fourth power.

It might help if you learn where these parallel resistance formulas come from to begin with. It sounds like your text either doesn't take the time to present it or that you passed over that part because you didn't see the point.

The key is to just apply Ohm's Law.

The relationship between the voltage and current in a resistor is:

\(
V \; = \; I \cdot R
\)

which is the same as

\(
I \; = \; \frac{V}{R}
\)

If we have N resistors in parallel, we know that they all have the same voltage across them (that is simply what it means for two components to be in parallel) and that the total current is the sum of all of the individual currents. This means that

\(
I_{tot} \; = \; I_1 \; + \; I_2 \; + \; I_3 \; + \; ...
\)

So now we replace the individual currents with the equation for those currents based on Ohm's Law, noting that the total current can be written, via Ohm's Law, in terms of the common voltage and the total (or equivalent) resistance:

\(
I_{tot} \; = \; \frac{V}{R_{eq}}
\)

So we get

\(
\frac{V}{R_{eq}} \; = \; \frac{V}{R_1} \; + \; \frac{V}{R_2} \; + \; \frac{V}{R_3} \; + \; ...
\)

We can divide every term on both sides by V, giving us:

\(
\frac{1}{R_{eq}} \; = \; \frac{1}{R_1} \; + \; \frac{1}{R_2} \; + \; \frac{1}{R_3} \; + \; ...
\)

This is good for any number of resistors and can be remembered easily as "the reciprocal of the equivalent resistance is equal to the sum of the reciprocals of the individual resistors."

If you just have two resistors, then you have

\(
\frac{1}{R_{eq}} \; = \; \frac{1}{R_1} \; + \; \frac{1}{R_2}
\.
\frac{1}{R_{eq}} \; = \; \frac{R_2}{R_1 R_2} \; + \; \frac{R_1}{R_1 R_2}
\.
\frac{1}{R_{eq}} \; = \; \frac{R_1 \; + \; R_2}{R_1 R_2}
\.
R_{eq} \; = \; \frac{R_1 R_2}{R_1 \; + \; R_2}
\)

The equations for 3 or 4 resistors can be gotten using a similar approach, but they aren't worth memorizing. You will work with two resistors in parallel a lot, so that equation you will almost certainly memorize whether you want to or not. For anything more, use the general formula for N resistors.

As for how to tell if a text is any good, that is often not that easy -- and what is good for me and my style of learning may be completely wrong for you. I have often found texts that looked promising but, after digging into them, have turned out to be very lacking for one reason or another. I have also found that it sometimes takes several texts to get overall good coverage of a given topic because one might to good at some aspects but be really bad on others.

 

Wow! how can it be so confusing? when in doubt, I've always broken it down to pairs of resistors in parallel, take any two, do the simple calculation, with the result obtained, do the same again for the next resistor (if calculating 3, or the result of another pair if 4 resistors etc) this saves so much confusion and always agrees with my accurate Ohm meter. (within reasonable tolerance depending on actual values of the resistors used) When I was working as a development engineer, excluding precision test equipment etc, any circuit that required better than 10% tolerance on any component, was considered a "bad" circuit for mass production, although there were of course exceptions which usually then included a pre-set for final adjustment. Nowadays components are generally more consistent.

 

@Dal1980

Did you notice how messy things look when you substitute the givens at the beginning of a problem?

Do the algebra first and substitute the givens at the very end. I suppose a problem like this is OK, but if the math consumes several pages and you've made an error it proves difficult to isolate a page or two when you're staring at numbers.

 

What about the difference of reciprocals?

This is a trivial example, but:

Say you need a 200 ohm resistor, and have the standard 220 ohm part.

1/200-1/220 =4.545e-4

= 1/0.0004545 = 2.2K

220 // 2.2K = 200 ohms.

 

What about the difference of reciprocals?

This is a trivial example, but:

Say you need a 200 ohm resistor, and have the standard 220 ohm part.

1/200-1/220 =4.545e-4

= 1/0.0004545 = 2.2K

220 // 2.2K = 200 ohms.

What about it?

 

Hola Dal,

Maybe you did not realise but the equivalent value of n resistors with the same value is just that value divided by n.

Also, when putting resistors in parallel, the only thing you can be always sure about is that the equivalent will be smaller than the smaller one.

 

This is the way Ive always found the easiest way to enter parralell resistor calculations on my Casio fx83gt plus calculator, cost £9.00 in sava centre!

 

To find the equivalent resistance of the following in parallel

1330 // 39000 // 17400, it takes me exactly 18 (counted) keystrokes using a Casio fx-570.

 

To find the equivalent resistance of the following in parallel

1330 // 39000 // 17400, it takes me exactly 18 (counted) keystrokes using a Casio fx-570.

I don't have a Casio, so I'd be interested in knowing what the 18 strokes were.

On an RPN HP it took 20 keystrokes. They are as follows:

1330(1/x)39000(1/x)(+)17400(1/x)(+)(1/x)

 

I don't have a Casio, so I'd be interested in knowing what the 18 strokes were.

On an RPN HP it took 20 keystrokes. They are as follows:

1330(1/x)39000(1/x)(+)17400(1/x)(+)(1/x)

The (your) complete correct sequence takes 21 keystrokes.

EDIT
Sorry I forgot you said RPN; it is 20
EDIT

 

The (your) complete correct sequence takes 21 keystrokes.

Not following.

On an RPN calculator it takes 20 keystrokes (unless you are counting the key that must be pressed to turn the calculator on).

1) 1
2) 3
3) 3
4) 0
5) 1/x
6) 3
7) 9
8) 0
9) 0
10) 0
11) 1/x
12) +
13) 1
14) 7
15) 4
16) 0
17) 0
18) 1/x
19) +
20) 1/x

What are the keystrokes on your Casio?

 

1 . 3 3 (x^-1) + 3 9 (x^-1) + 1 7 . 4 (x^-1) = (x^-1) =

Good night William.

 

I thought about doing it in kΩ (or leaving off the trailing zero of each value and then multiplying by 10 at the end) since I assumed you wanted an answer in ohms. If you allow for mentally moving the decimal point on the answer then I can leave of the trailing zero and have 17 keystrokes.

By letting the answer be in kΩ, the RPN method is just 16 keystrokes.

1.33(1/x)39(1/x)(+)17.4(1/x)(+)(1/x)

It would take another five, 1000(*), to get it back into ohms.

For most computations, RPN results in fewer keystrokes because the stack takes the place of parens (or, in many cases, intermediate application of the = key). But it has a brutal learning curve to get comfortable with it. Surprisingly, once you DO get comfortable with visualizing the stack in your head (the old RPN calculators only displayed the top of the stack, though you could scroll through it if you needed to), it becomes an extremely easy and efficient way to do most computations, even if they involve quite a few parens and/or operator precedence rules.

 

I don't have a Casio, so I'd be interested in knowing what the 18 strokes were.

On an RPN HP it took 20 keystrokes. They are as follows:

1330(1/x)39000(1/x)(+)17400(1/x)(+)(1/x)

13 on the Casio fx83

 

13 on the Casio fx83

So what were the 13? Seems a bit hard to imagine since 1.33, 39, and 17.4 account for 10 and I would think that there has to be at least one key stroke after each numeric value, so that's 13 right there.