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Total resistance of a complex circuit when neither series or parallel rules can be applied.?
Someone is asking for my help, and it's been way to many years since I've been in school for this.
Here is the problem. I have 5 resistors in a network such that no two are either in series or parallel, so those rules won't work. I'm looking for the total resistance between points A and B.
Here is the circuit.
http://img.photobucket.com/albums/v74/puppy_raiser...
What I know.
If the 400 ohm resistor were an open R total would be 900 ohms and if a short, 600 ohms, so with 400 ohms (or any other value), the answer must be between 600 and 900.
What I think I remember.
It seems to me, you have to assume a voltage or current someplace, then using ohms law calculate other values, finally coming up with 3 equations and 3 unknowns. Solving for the 3 unknowns with basic algebra skills gives a quick path to the answer, but coming up with the 3 equations or even what the 3 unknowns are escapes me. Any help out there (with how you did it) would be much appreciated.
Thanks yipyap, but no.
If you combine the top two resistors, where are you going to connect the 400 ohm resistor?
Oops... sorry, maybe I didn't read far enough.
No, their explanation is for connecting the two nodes with a short, making two parallel circuits in series. Not the case in this problem. No two resistors can be replaced with a single resistor and have an equivalent circuit.
Thanks porteng
Excellent.
While that's not how I remember doing it, it works,
and beggers can't be choosers.
Thanks much!
3 Answers
- 1 decade agoFavorite Answer
Check out this link on page 73. It shows the derivation for the resistances and from that you can calculate the total resistance between the two terminals.
- Anonymous1 decade ago
Well the top two are in series and the bottom two are in series and both pairs are in paralell with each other.
I think this is the answer you're looking for:
http://www.mathpages.com/home/kmath385.htm
Yes i didn't read far enough myself! Tricky Q!
Tempted to get my meter out and work this out backwards!