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Thevenines, how to find Rth

by hussain bani
Tags: thevenines
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hussain bani
#1
Apr18-13, 10:26 AM
P: 9
I short the batteries for Rth, but after doing short i dont understand how to find Rth, can sum 1 help me finding out ?
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psparky
#2
Apr18-13, 12:46 PM
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You are simply putting an ohmeter across Rab....what does the ohmeter see?

I would put the 4 ohm and the 16 ohm in parrallel....then add the 2 ohm in series. Put that result in parallel with the 15 ohm resistor........then add the 2 ohm resistor to that result in series. Then put in parallel with the 8, then parallel with the other 8....then add that result to the 5 ohm in series.

If you are having a hard time seeing which resistors are in parallel....parallel implies "voltage across is the same. If you look at the 16 ohm resistor and put a voltage meter across it......then slid it down so it is across the 4 ohm resistor.....the voltage would be the same for both. A.K.A parallel.
hussain bani
#3
Apr19-13, 01:47 AM
P: 9
I short the batteries for Rth, but after doing short i dont understand how to find Rth, can someone help me finding out ?

NascentOxygen
#4
Apr19-13, 02:42 AM
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Thevenines, how to find Rth

Quote Quote by hussain bani View Post
I short the batteries for Rth, but after doing short i dont understand how to find Rth, can someone help me finding out ?
Hi hussain bani. You short the batteries only on paper. You don't do this to real batteries, shorting them can be dangerous or damaging, and won't give you the right answer, anyway, because the chemical reaction inside changes drastically when a battery is abused like this.

So, on paper, you short the battery and determine the current that flows through that short circuit wire. Together with the no-load voltage, you can then calculate Rth = Vo/c Is/c
smashbrohamme
#5
Apr21-13, 06:52 AM
P: 91
In a circuit with other components, you can just take the batteries out and put a jumper wire where the batteries were.

Mathematically they just want you to bypass the batteries so they no longer have a effect on the circuit. Of course this is only for finding Rthevenin.
NascentOxygen
#6
Apr21-13, 11:20 PM
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Quote Quote by smashbrohamme View Post
In a circuit with other components, you can just take the batteries out and put a jumper wire where the batteries were.
In general, no. That won't work where the circuit includes active or non-linear devices (e.g., semiconductors).
Kevin2341
#7
Apr22-13, 12:23 PM
P: 55
My procedure when it came to doing thevenin equivalents, especially when you get nasty diagrams like that, was to start by simplifying the circuit (I have a thing against diagonal lines in electric diagrams).

Reading your question, what you want to do is find the equivalent resistance. Use all the rules and tricks you have learned up to this point and attempt to turn that mess of resistors into a nice and simple equivalent resistor. You will be doing quite a bit of parallel resistor combinations from what I have seen.

Something else to note, homework problems for circuit analysis are usually carefully rigged so that the answer is a number, but that is not always the case. (In my analysis of it, I had to guess on a couple numbers, but my number was near 9, but as you will know, simply putting the answer on your homework is NOT satisfactory for a full grade.)
psparky
#8
Apr23-13, 02:07 PM
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All the methods stated above are correct. If you are shying away from simplifying resistors, perhaps you need to step back and master that concept first. Re-drawing the circuit in a more familar shape is always preferred as mentioned above.

If you are competent in combining resistors, it should take you no longer than a minute or two to find the Rth in my opinion.

And my point about "what would a ohmeter read" is a legitimate statement. If you wire those resistors in real life and put an ohmeter across Rab....it will be the same as the calculated Rth....or the combined resistance.

Thevenin and norton will actually help you solve more complicated circuits once you learn it.
hussain bani
#9
Apr24-13, 03:53 AM
P: 9
Thanks everyone for their kind explanation, i got it...:)


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