Resistor circuit help

galwaydude18

i am a first year student studying electronics and computer engineering and have to repeat a few exams. i have to repeat electrical science and need some help. i have a question that requires me to find the total resistance of a circuit that includes both parallel and series resistors. i have found the total resistance of the circuit but I NEED TO KNOW how to find the current flowing in each resistor as well as the p.d across each resistor and the power dissipated in the circuit. i would appreciate it if some one could tell me how to do this!

elivsvonchiaing

Haven't had to look at this stuff in seventeen years - but here goes anyhow...

If you've worked out the resistance of the circuit you've done this simply by reducing the circuit into one resistor. You now simply need to work backwards. Before you start you know the pd across this: V= IR (for a current source) or just V for a voltage source. Next you have two resistors (logically still consisting of many). These are either in series or parallel. Parallel the pd is the same. In series. R1, R2. Pd across R1 is R1/R2 x V. Simply continue to iterate downward starting with V that applies to the logical equivalent of the sub-circuit.

This is probably a bit terse and would be far better with a picture. If this doesn't make sense why not post the circuit? Or if your nervous make up a similar one.

galwaydude18

Hi here is the question and the ciruit.

elivsvonchiaing

OK. I've got 9 Ohms for the circuit. The 20 + 5 come to : 20x5/25 = 4. This gives us two new resistors in parallel: 4 +5 = 9 and 10 + 8 = 18. These become: 9x18/27 = 6. We then just add the 3 ohm giving 9. The the pd across the 3 ohm is 3/(6+3) x180 = 60V. The pd across the 20 +5 and 10 +8 is therefore 180 - 60 = 120V. So the pd across the 5 ohm is 5/25 x 120 =24V and the 20 Ohm is 20/25 x 120 = 96V. The pd across the 10 and the 8 is
8/18 x 120 = 53.3333V (across the 8) and 10/18 x 120 = 66.6666V (across the 10).

Apologies for the R1/R2 it should have been R1/(R1+R2) and forget about the current- you don't need to worry about it. A real trip down amnesia lane for me!!!

galwaydude18

evilsconchiaing
do you by any chance have any idea how to do these as well?

Regards,
John

elivsvonchiaing

Uh God. I once did - the first three anyhow - can't read the last... Sorry, but I'll have to dig into me old college notes if I can find them. Think Thevenin was to convert the voltage source to current source for equivalency - If no one else is of help I'll be up my parents attic at some point - maybe not until the weekend. Sorry dude Memory like Swiss cheese - probably 'cos of the alcohol - the x-box - the TV - and only getting broadband before all of the latter had their toll on me

galwaydude18

no prob!! thanks for your help!!!! any one else got any ideas on how to do these questions?

elivsvonchiaing

Disappointed to see you got no posts here. Think the problem could be that most people who look at these threads and if they don't see 1-2 posts they assume its been solved. If they see 10 posts they may get interested - a debate perhaps. I'm signing off now but for what it's worth just post a reply, we can do 10 posts easily tomorrow night

galwaydude18

can someone please help me with these questions??? PLEASE!!!!!!???????

elivsvonchiaing

Dude, I found this link:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/thevenin.html

From reading it it seems the Thevenin equivalent (my comments earlier were confusing this with something else) is just a voltage source with a resistor. I get the Thevenin voltage at 1.81818V. 1k/(4K + 6K) x 20.

I get the Thevenin resistance by removing the voltage source and simplifying the circuit. First the 4k and 6k resistors become on 10k which gives us two parallels one 10k one 1k. R = (1k x 10k) / (1k + 10k) = 9.09M Ohm.

As I haven't looked at this stuff in too long I have no confidence in it. I would urge you to get someone to confirm or point out where I've gone horribly wrong

There's details of Norton here too - but I'm not going to touch this for fear of getting it horribly wrong.

galwaydude18

thanks for your help! im not sure if your answer is correct but i hope to have found out by the weekend!!

elivsvonchiaing

I'm really not sure its correct either! Just posting this (gratuitous) reply in the hope that the thread length get someone who knows interested

galwaydude18

please can someone help me? please im desperate here!!!!!!!!!!!!!

gobby

galwaydude18 wrote:

please can someone help me? please im desperate here!!!!!!!!!!!!!

Hmmm... Might be able to help you in a week or so. I'm guessing it might be a little late though. Repeat exams?? I've got the same problem so thats why I cant help for a wee while. That and the fact that im not in the country so I cant get at my old notes.

galwaydude18

ya would be a bit late in a weeks time!!! thanks for being helpful!! do you know any one that might be able to help me here?????

elivsvonchiaing

God I wish I didn't have to tell you this dude - but I was up my parents attic over the weekend and they've chucked the notes. They did say some years ago can we just dump your junk - I had action-man and lego in mind when I said yes

Someone, please help this dude!!!