The basic Noob Question ....

[Stratus]

Am reading through "Starting Electronics" and have come across the formula for resistors in parallel and can seem to get why the author has written the formula the way he has ?? and plus the fact that resistors in series follow what seems to be logical path, I would have thought that the same logic could have been applied to parallel resistors but apparently not or am I just missing something even more basic?? loll
I think this image may help
(the resistor values used in this example are 10k and 1k5 (the penny wont drop for me) what has me confused is the 0.000076 result I can see where he go it by dividing 15,000,000 into 11,500 , why then divide it the other way and get 1304 Ω and call that the answer sorry this probable sounds stupid but I'm just cant see the answer??

 

[Dr_Doggy]

what you have posted is the PRODUCT over SUM rule, the division question is upside down in your example!

 

[Stratus]

Hi Doggy thanks for the reply i have added a copy of the pages from the book with this explanation on it cuz the example that i uploaded earlier was copied from these pages ??

And one othere question is the answer always going to be in ohm's ?
Thanks

"Edit never added file"

 

[Stratus]

Just one other question that has me a bit confused. The diagram in the attached file shows a circuit made up of a battery 10v and a number of resistors and the question is to find a value for I by finding the overall resistance but before I can do that I have to decide which ones are in series, parallel and individual which is where I'm a bit stuck as some could be but in more than one category. You can see from the diagram which ones I've chosen was wondering if my choice was correct and if so, if I follow the rules for finding resistance for the individual groups of resistors and sum up my answers and apply ohms law I should come up with the correct value for I right ??

 

[misterT]

There are couple of mistakes in your parallel and series grouping. You can't calculate R5 and R9 in series because there is the resistor R4 in parallel with R5.
I would say that there are no "individual" resistors in that schematic. Every resistor is in parallel or in series with another resistor or a group of resistors.

First calculate:

Parallel:
R2||R3
R4||R5
R6||R7||R8

series:
R11+R? (can't read the number of the resistor below R11)

After you calculate those values, it is a good idea to redraw the diagram by replacing the parallel and series groups with the calculated values.. and then find new parallel and series groups etc.

 

[Dr_Doggy]

only the middle blue one can be done first(out of order) this is because it is truly series,, the other 2 blue ones are actually series parallel networks which must be simplified first

resistance is always in ohms! but we like to use scientific notation, so K= x1000 and M= x 1000000, and milli and micro

2k4 means 2400ohm, its just a scientific slang, it could also be 2.4Kohm

 

[Stratus]

Thanks Guys for the help, its taken 3 attemps so far and still before reading doggy__'s last post I can see why ill have to go for attempt 4 me thinks, oh buy the way R9 is the resistor below R11 and its value is 5K.
An the anwser i got for overall resistance is 10k48 so if there is only 10 volts that would mean there is only 9.5419847328244274809160305343511e-4 (thats paste from the Calc).
So im thinking ill have to post the images of how i worked out the over all resistance
God there most be and easier way?? Oh the joys of learning!!

 

[Dr_Doggy]

(it is hard to see all your values, it could help to post each step in your work)

ok,,, for now use 1/rt = 1/r1 + 1/r2.... to calc your parallel resistors just to keep in a trend for now,

first you must note there are NO independant resistors there,,, as i go through the steps redraw your circuit each time to follow.//

our object here is to simplify, to do this we will first compress all parallel branches, as ron says do R1 and R2 first(parallel calc) dont forget ..... then add R4 and R5(parallel again) ......... then do R6,7,8 ........

NOW it is cleaner and we can add the series resistors,, so it s simple adding(for simplicity the answer you got for R4parallelR5 will be called R4,5 K?)..... so first we take R4,5 and R9 and add them(series), then we take R2,3 and add it with R6,7,8 AND with R1 aswell(series),, now adding R9 and R11 together will also simplify that branch.

lets pause here since you are wondering about adding R10 to that series part , and the reason we couldn't do that is cause R10 has a parallel branch with R9,11 but now since we have added R9,11 we can now take that answer and simplify the parallel with R10 , once you have that branch simplified THEN you can add it to the HUGE SERIES part we just did,,,

simplify...... at the end you will get a resistor(equivalent) in parallel with R12......!!!!!

 

[Barry Wilkins]

Resistors in parallel

The reason for the problem is because it is a reciprocal .

Barry Wilkins

 

[KeepItSimpleStupid]

1/Rt = 0.00076, so Rt = 1315.789474

 

[Jugurtha]

Like the poster above said (KeepItSimpleStupid):

1/R=xxx so R=1/xxx.

(I thought I quoted it, but I'm not familiar with fora functionning quotes)

And one othere question is the answer always going to be in ohm's ?

Yes.

The product is Ohm.Ohm and the sum is Ohm, so the result of Ohm.Ohm/Ohm is Ohm.


When you work with literal expressions, any error in units (like a squared Ohm in a resistor), or "odd" result would just JUMP at your face, which is not possible when you work with numbers and replace each device by its value from the beginning.


This is why you want to always apply the actual numerical values as the very last step of the calculation, you want to work with the literal expression, simplify it to the maximum, and only then do N.A (Numerical Application) and actually replace each resistance with its value. Trust me, you'll avoid sooo many mistakes by doing this.

All the best.