Thevenin Theorem help

Discussion in 'Homework Help' started by bellers, Jan 27, 2013.

1. bellers Thread Starter New Member

Jan 27, 2013
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I need help with finding Vth in this circuit. I know the main steps when solving using Thevenin's theorem but the two voltage sources has thrown me.
In the solution, Vth has been calculated to be 6.67v. I can't understand how this has been calculated. So any help would be appreciated. Hopefully the attachment opens up correctly!

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2. WBahn Moderator

Mar 31, 2012
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Well, what are the main steps that you say you know and how is this problem different so as to throw you?

Redraw the circuit with the 1Ω resistor removed. Can you find the voltage across this opening use any analysis techniques you have learned up to this point?

3. bellers Thread Starter New Member

Jan 27, 2013
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I would take the 1ohm resistor out of the circuit, find Vth. Then Find Rth. Draw a Thevenin equivalent circuit with Vth, Rth and the 1ohm resistor and then work out the current in that circuit. No I do not know how 6.67v is being calculated as Vth.

4. Jim Hunter New Member

Jan 27, 2013
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I am new at this, too, but following the description in the ebook, I removed the 1Ω resistor, then used Kirchoff's Voltage Law to find the current in the resulting circuit. Once the current was known, the voltage drop across the remaining resistors could be calculated. You don't actually need to calculate the Thevenin resistance to get the voltage.

5. bellers Thread Starter New Member

Jan 27, 2013
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Hi Jim, thanks for the reply. I have got there now, with a bit of help from a workmate!
I was forgetting to take the voltage across the 4ohm resistor away from the supply voltage to get the voltage drop...which in this circuit would also be Vth. (Numpty!!)
I have updated the pdf so you can see my working although I'm not very good at using that "one note" software.
Like you say, I understand their are easier ways to do this, especially when working with 2 sources, but my lecturer was wanting us to solve it using 3 different theorems...

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6. WBahn Moderator

Mar 31, 2012
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All of these things will make it less likely you will make a mistake, more likely that mistakes will be caught, and more likely that you will get substantial partial credit on the remaining mistakes that sneak by.

7. bellers Thread Starter New Member

Jan 27, 2013
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Can you point me in the direction of any reading material that would help with the latter of the 3? Or even something that covers network theorem in general, I sometimes find myself not understanding what is happening at certain points of the circuits.

8. WBahn Moderator

Mar 31, 2012
18,093
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It's much more basic than "network theorem" and such. What you want to be sure that you have a firm handle on is what is known as the "passive sign convention".

In general, you get to pick, even randomly if you want, the directions of the voltages across every component in your circuit. It's not important which polarity you choose, just that you choose a polarity and are then consistent with it. For passive devices (i.e., devices that don't supply power -- although this is not the only definition of "passive" that is arround, but it's good enough for us right now) like resistors, capacitors, and inductors, the current through the device is positive if it flows from the positive terminal of the device (per your earlier voltage polarity choice for that device), through the device, and then out the negative terminal. For sources, the reverse is true -- current flows out of the positive terminal into the circuit and returns to the negative terminal. Now, we are talking about "conventional current", which is the assumption that current is made up of positive charges that flow from a positive voltage to a negative voltage (unless something, like a generator or a chemical reaction, provides the necessary energy to make the opposite happen). That most physical currents are made up of negatively charged electrons is conveniently ignored and, for almost all purposes, the two are completely equivalent, but the bookkeeping is much simpler using conventional current.