# How is this possible? (circuits)

## Homework Statement

[PLAIN]http://img441.imageshack.us/img441/7796/81673480.jpg [Broken]
given this circuit find the total resistance of the circuit

I = V/R

## The Attempt at a Solution

ok I think that there are two ways of solving this, 1 by finding the total resistance directly from the resistors because some are in series some are in parallel

the other way is to find the total current and then the Rtot will be V/Itot

So, I found the I total using the online wolframalpha calculator which is

solve[{36 - 3y - 6(y+z)=0,36-6x-3(x-z)=0,-6x-3z+3y=0},{x,y,z}]

so
x = 24/7 ~~ 3.428571 and y = 36/7 ~~ 5.142857 and z = -12/7 ~~ -1.714286

x is I1 which flows over R1 y is I2 which flows over R4 and z is I3 which flows over R3

the answers are correct since my book has the same answers for the currents

so Itot = I1 + I2 = 3.428571+5.142857 = 8.571428 A

hence Rtot = 36*Itot = 4.2 Ω (which is the correct answer)

using the method with parallel and series I said that R5 and R3 are in parallel, hence, R35 = 2 Ω

R35 is in series with R4 hence R345 = 2 + 3 = 5 Ω

R345 is in parallel with R12 = 9 Ω

SO Rtot = 45 / 14 = 3.21428571

I can't understand why im getting different results, can anyone help me? thanks

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phyzguy
When you combine two parallel resistors, the equations assumes that the resistors are connected in parallel, which means the two resistors are a parallel path between two circuit nodes. R3 and R5 are connected to the same node at one end but not at the other end, so they are not in parallel. No two resistors in this circuit are truly in parallel, so you have to set up a system of equations with the various currents as the unknowns, which is what Wolfram Alpha did. Your second analysis is incorrect.

When you combine two parallel resistors, the equations assumes that the resistors are connected in parallel, which means the two resistors are a parallel path between two circuit nodes. R3 and R5 are connected to the same node at one end but not at the other end, so they are not in parallel. No two resistors in this circuit are truly in parallel, so you have to set up a system of equations with the various currents as the unknowns, which is what Wolfram Alpha did. Your second analysis is incorrect.

thanks for the answer, but could we assume that they are in series?

let's just leave the exercise if my prof gave me this circuit and told me find the resistance of each resistor, how would I do it using parallel and series theorems? thanks in advance

phyzguy
I don't know of a way to analyze a network like this with only series and parallel equations. Maybe someone else does.

Only series and parallel equations won't to work, since none of them are truly in series or parallel. Kirchhoff's laws are the way to go, I'm affraid.

YOU SHOULD NOT AND CANNOT USE Paralled and Series Laws because there another resistance in the iddle ! The oonly way to solve this is Kirchoff laws..Folow the Junction law and you'll get it !

No, there's an easy way, almost by inspection. I'm writing it up in a second post.

1) split the top into two 36 Volt sources
2) remove the three ohm resistor
3) compute the voltages in the (now seperate) dividers. It works out to 12 and 24 volts.
4) make a Thevenin equivalent for the center node. This means you now have a 12 volt generator in series with a (3||6=2) ohm resistor on the left, and on the right you have a 24 volt generator with a 2 ohm resistor on the right. You may now connect the center 3 ohm resistor back into the circuit. You have a 3 resistor divider with a 12 volt source on one side and a 24 volt source on the other. The resistors are 2 3 and 2 ohms in series.
5) the voltage on the left of the 3 ohm resistor is 12+(24-12)/(2+3+2)*2 = 24/7 + 12.
6) the voltage on the right of the 3 ohm resistor is 24-(24-12)/(2+3+2)*2 = 24 - 24/7.
7) you now have the voltages at the two nodes of the 3 ohm resistor in the original circuit. work out the current through the bottom two legs of the original circuit, a 3 ohm on the left and a 6 ohm on the right. Add them up, call it I.
8) 36/I = the resistance of the original circuit.

By insepction. (Except I'm bad at arithmetic. Use a calculator while you insepct.)

Edit: It comes out 4.2. If someone tells me how to post pics, I'll draw schematics of the transformations and equivalent circuits.

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