# Circuit analysis (thevenin equivalence)

#### dfx

1. Homework Statement

http://img509.imageshack.us/img509/5658/circuituz5.jpg [Broken]

Find the current through resistor X for V = 7V

2. Homework Equations

Thevenin equivalence?

3. The Attempt at a Solution

Right I was advised a way to do this by assuming a 1A current flows through the resistor and then working back and scaling accordingly, but I also want to try this using Thevenin Equivalence. So I've worked out:

Vth = 0.46V and Rth = 4.28 ohms

Working for Rth:

Rth is the resistance seen accross X with V shortcircuited. Thus, starting from right with 1 ohms (i.e. X) going to left

= [(((((((1||2) || 3) -- 2) || 7) -- 2 -- 1) || 4 -- 2) || 5) -- 2] , where || denotes parallel, -- denotes series. I think this is where I've messed up, giving me 4.28 ohms

Working for Vth:

Jus use voltage division from 7V all the way down to give 0.46V accross X

So the Thevenin equivalent is:

0.46V in series with 4.28 ohms and then the load (resistor X) connected is 1 ohm in series.

This doesn't give the correct answer though. The current is supposed to be 0.125A.

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#### SGT

First of all: to find the Thévénin equivalent you must open the circuit at the resistor X and to calculate Rth you should work from left to right using your notation:
Rth = (((((((2||5)--2)||4)--2--1)||7)--2)||3)--3

For the voltage division you must work from right to left.
Calling V2, V3, V4, V5 the voltages at the nodes from left to right, you have:
Vth = V5 = V4*3/(3+2)
V4 = V3*7/(2+7+1)

etc

#### berkeman

Mentor
There are two main ways to solve this circuit easily -- folding the resistors up and then back out, or using KCL.

To fold the resistors up, start at the right and use parallel and series combintaions to end up with just one resistor connected to the voltage source V1. I think this is what you were doing, but I'd do the series and parallel combinations as separate steps, instead of trying to combine them all into one equation. Once you have the current out of V1, then unfold the resistors back out, calculating the current division at each unfolding step.

The other way would be to write the KCL equation for each node, and solve the simultaneous equations for the voltage difference across the far right resistor. Have you learned about KCL in your class yet?

EDIT -- Dang, SGT types fast! Beat me to the punch again.

#### dfx

Thanks you guys.

SGT, Wouldn't Vth be V6 i.e. the "node" between the 2ohm and 1ohm (X) resistors on the complete right, because Vth would be the voltage accross X? I do realize there is no V6 at present, but for the Vth wouldn't you be introducing a V6 once you connect a "voltmeter" accross X? Kind of confused as you can see.
Also, why can't you "sum" the resistors from left to right? I did "open them out" from right to left, but find it much easier to sum from left to right because you can do it in smaller steps.

berkeman, thanks, I'll try using current divison too and see if it's any different.

I think I've combined the resistances wrong. I'd be really grateful if someone could give my working a quick check and see if they get the same answers. Thanks!

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#### SGT

Thanks you guys.

SGT, Wouldn't Vth be V6 i.e. the "node" between the 2ohm and 1ohm (X) resistors on the complete right, because Vth would be the voltage accross X? I do realize there is no V6 at present, but for the Vth wouldn't you be introducing a V6 once you connect a "voltmeter" accross X? Kind of confused as you can see.
When you open the circuit, there is no current through the 2 ohm resistor, hence no voltage drop, so Vth = V6 = V5.
Also, why can't you "sum" the resistors from left to right? I did "open them out" from right to left, but find it much easier to sum from left to right because you can do it in smaller steps.
You can combine the resistors from left to right if you find it is easier, but you must remember that when you cancel the voltage source, the two leftmost resistors (2 and 5 ohm) are ion parallel. Also, the rightmost resistor (2 ohm) is open, so it must be summed to the combination of the other resistors.
berkeman, thanks, I'll try using current divison too and see if it's any different.
I think I've combined the resistances wrong. I'd be really grateful if someone could give my working a quick check and see if they get the same answers. Thanks!
Read above and compare with what you did.

#### hgmjr

Post withdrawn...

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#### berkeman

Mentor
Here is a table of the voltage and currents associated with each of the resistors. I assigned the reference designators for convenience.

The resistor whose current is to be computed is R10.

hgmjr

<< Attachment with complete solution deleted by berkeman >>
hgmjr,

Welcome to the Physics Forums (PF). However, one of the very strict rules that we have here is that we do *not* post complete solutions to homework questions. Our job is to give hints and help guide the original poster (OP) to figure out the problem on their own. We certainly appreciate help in the homework forums, but that help needs to be tutorial in nature, not just providing complete solutions.

#### hgmjr

Sorry for the mis-step. That is a good policy. I will comply with that mandate in the future.

hgmjr

#### dfx

You can combine the resistors from left to right if you find it is easier
Sorry, I meant to say combine from right to left since that lets you do it in smaller steps, not the other way round as you suggested in your first post (left to right). Also thanks for pointing out the 2 is in parallel with 5. I think that was probably it. Will get back to you.