Circuit analysis (thevenin equivalence)

In summary, the current through resistor X is 1A and the Thévénin equivalent is 0.46V in series with 4.28 ohms and then the load (resistor X) connected is 1 ohm in series.
  • #1
dfx
60
1

Homework Statement



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

Find the current through resistor X for V = 7V

Homework Equations



Thevenin equivalence?

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 across 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 across 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.

Any feedback/help/advice very much appreciated!
 
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  • #2
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
 
  • #3
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.
 
  • #4
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 across 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" across 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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  • #5
dfx said:
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 across 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" across 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.
 
  • #6
Post withdrawn...
 
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  • #7
hgmjr said:
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.
 
  • #8
Sorry for the mis-step. That is a good policy. I will comply with that mandate in the future.

hgmjr
 
  • #9
SGT said:
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 let's 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.
 

What is circuit analysis?

Circuit analysis is the process of studying and understanding the behavior of electrical circuits. It involves using mathematical tools and techniques to analyze the flow of current and voltage in a circuit and determine its overall performance.

What is Thevenin equivalence?

Thevenin equivalence is a technique used in circuit analysis to simplify complex circuits into a single voltage source and a single resistor. This simplified circuit has the same output voltage and current as the original circuit and can be used to analyze the behavior of the circuit more easily.

How is Thevenin equivalence calculated?

Thevenin equivalence is calculated by first finding the Thevenin voltage, which is the open-circuit voltage at the output terminals of the circuit. Then, the Thevenin resistance is found by short-circuiting all voltage sources and calculating the equivalent resistance at the output terminals. The Thevenin voltage and resistance can then be used to create the simplified Thevenin equivalent circuit.

What are the benefits of using Thevenin equivalence?

Thevenin equivalence helps simplify complex circuits and makes them easier to analyze and understand. It also allows for the replacement of a complex circuit with a simpler equivalent circuit, which can save time and resources in design and troubleshooting. Additionally, Thevenin equivalence is useful for determining the maximum power transfer in a circuit.

How is Thevenin equivalence used in practical applications?

Thevenin equivalence is commonly used in circuit design, analysis, and troubleshooting. It can also be used in the design of electrical power systems to determine the best configuration for maximum power transfer. Additionally, Thevenin equivalence is used in electronic devices such as amplifiers and sensors for accurate and efficient circuit analysis.

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