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Thevenin's theorem

by ranju
Tags: theorem, thevenin
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ranju
#1
Jul16-14, 06:20 AM
P: 85
In the given circuit having a voltage source of 12V and current source of 6 A..we have to find thevenin circuit to left of terminals a &b..
Now if we'll find the thvenin resistance by shorting across 12 V battery and open circuiting across the current source.. as I solved it shud be.. 4//6//6 so the thevenin's resitance shud be 12/7 ohms..but this does'nt go with the given options..!!
Can you plzz point out ehre I am going wrong..??
the options given are12V & 16 ohm
12V & 12 ohm
20V&4 ohm
12V & 3 ohm
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ranju
#2
Jul16-14, 06:22 AM
P: 85
the another resistance in the pic is of 6 ohm..
psparky
#3
Jul16-14, 07:30 AM
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If you short the voltage and open the amp source....

You have 6 ohms in SERIES with 6 ohms which gives you 12 ohms. Better review your parallel and series theory. Picture sliding the two 6 ohm resistors where the shorted voltage source is. You would just simply add them in series. Since you opened the amp source, just totally remove that part for thevenin resistance. Parallel implies voltage across is the same...or two resistors share a "top" node and "bottom" node. Or you could say that there are two paths for the current to go thru. None of these cases apply...therefore series. Learning how to re-draw circuits will be critical for your development.

You then have 12 ohms in parallel with 4 ohms....The TOTAL Theveinin resistance is 3 ohms.

Using superposition again, The voltage source then contributes 3 volts (voltage division)and amp source contributes 9 volts (current division then V=IR) for a total of 12 volts.

12V & 3 ohm.

Boom.

ranju
#4
Jul16-14, 08:56 AM
P: 85
Thevenin's theorem

if you are saying that here 6ohm is parallel to other 6 ohm..
now what would u say about this circuit... what according to you is the thevenin's resistance??
ranju
#5
Jul16-14, 08:58 AM
P: 85
this is the given circuit...
psparky
#6
Jul16-14, 09:00 AM
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Quote Quote by ranju View Post
if you are saying that here 6ohm is parallel to other 6 ohm..
now what would u say about this circuit... what according to you is the thevenin's resistance??
The 6 ohm is NOT in parallel with the other 6 ohm.

They are in SERIES.

Re-read what I posted above. It is factual information you need to absorb.
ranju
#7
Jul16-14, 09:02 AM
P: 85
this is the circuit somewhat similar to thew prvs one.. what according to u is the Rth..??
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ranju
#8
Jul16-14, 09:03 AM
P: 85
sorry that was typing mistake.. by mistake I wrote parallel in place of series.. u avoid it and plzz cxheck out the ckt I posted now..
psparky
#9
Jul16-14, 09:07 AM
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Quote Quote by ranju View Post
sorry that was typing mistake.. by mistake I wrote parallel in place of series.. u avoid it and plzz cxheck out the ckt I posted now..
Ok. What is the thev resistance to you....and why? Look at the 3 rules of parallel I posted above.
Look at it this way too....when the sources are shorted or opened....finding thev resistance is the same as putting an ohmeter on the two points of interest....what would the ohmeter read?

Then I will comment.
ranju
#10
Jul16-14, 09:19 AM
P: 85
ohkk..Now I think I'hv got it somewhat.... but I just wanted to clear one thing is ther any difference finding equivalent resisitance and thevenin's resitance... because in the 2nd ckt diagram I gave to u ..the thevenin resiistance is 1 where as the equivalent resistance is 4..!!!..then why the diff.??
psparky
#11
Jul16-14, 09:24 AM
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Quote Quote by ranju View Post
ohkk..Now I think I'hv got it somewhat.... but I just wanted to clear one thing is ther any difference finding equivalent resisitance and thevenin's resitance... because in the 2nd ckt diagram I gave to u ..the thevenin resiistance is 1 where as the equivalent resistance is 4..!!!..then why the diff.??
1 ohm is correct because they are in parallel.
Voltage across is the same.
Resistors are attached by top and bottom node.
And there is two paths for the current to go thru.

Yatzee.

I think Equivalent resistance would be taken from the view of your source/sources.

Thevenin resistance takes it view from the points of interest.....terminals A and B from your pics above!! If you look at points A and B....they are not really even part of the circuit. Just points were somebody is sticking an ohmeter.

In your second pic, if you take the resistance from the sources, they are clearly in series...therefore 4 ohms.
ranju
#12
Jul16-14, 09:26 AM
P: 85
now why so...!!! why the view point is chnging...ths thng is actly confusing finding equivalent resiist is easy bt the concept of thevenin's resiistance is making me confused..!!
how come they become in series now.. its clear that they are in parallel..!!
psparky
#13
Jul16-14, 09:28 AM
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Quote Quote by ranju View Post
now why so...!!! why the view point is chnging...ths thng is actly confusing finding equivalent resiist is easy bt the concept of thevenin's resiistance is making me confused..!!
Re-read my above post again....I added a few things.

It takes a little while to get good at Thevenin. Make sure you know it backwards and forwards or you will get hammered on your next exam.

Thevenin can also be used to solve larger circuits....you will see as you progress.

The ohmeter "sees" different things taken from different points. Again, your equivalent resistance is taken from the source.
ranju
#14
Jul16-14, 09:32 AM
P: 85
see the thng is m nt able to get how from view of sources the 2 ohm resisitance r in series.. ???
what if we find even the equivalent resistances from the view of thevenin's resisitance?
psparky
#15
Jul16-14, 09:38 AM
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Quote Quote by ranju View Post
see the thng is m nt able to get how from view of sources the 2 ohm resisitance r in series.. ???
what if we find even the equivalent resistances from the view of thevenin's resisitance?
Look at it from the 8 volt source. Short the other source....get rid of the points a and b.
Slide the two resistors over to the right where the 12 volt source is shorted.

Clearly, they are in series. Add together you get 4 ohms.

If you look at the equivalent resistance from the thevenin resistance points, then yes the two will equal eachother....or become 1 ohm.

Equivalent and thev resistance are basically identical.....its just thevenin just picks its points which is generally different than the source.

Dont overcomplicate thevein. It is basically V=IR.

Norton is almost identical...again, V=IR.

In your next lab, take a circuit that is done in a breadboard. Short or open the sources, then take an ohmeter and read from different points. You will get different readings.
ranju
#16
Jul16-14, 09:44 AM
P: 85
now even m done asking the same sort of q. again n again..I still nt able to distinguish between the 2.. u agree tht if see from view of thevenin resistance thn equiv. resistance will be 1..!! In exam how will I be able to knw thn.. ??
ranju
#17
Jul16-14, 09:46 AM
P: 85
welll I think it'll be over-dramatic now..sticking to the same thng fr so long..leave it then..
well thanxx a lot for helping out..!!!
psparky
#18
Jul16-14, 09:47 AM
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Quote Quote by ranju View Post
now even m done asking the same sort of q. again n again..I still nt able to distinguish between the 2.. u agree tht if see from view of thevenin resistance thn equiv. resistance will be 1..!! In exam how will I be able to knw thn.. ??
Nobody said electrical engineering was easy.

Now that you are slightly more educated, work a bunch of problems. Also, you are now smart enough to ask your professor in class.

Re-read what I wrote several times over. Sleep on it and read them again.

Good luck.


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