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Thevenin's theorem 
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#1
Jul1614, 06:20 AM

P: 34

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 


#2
Jul1614, 06:22 AM

P: 34

the another resistance in the pic is of 6 ohm..



#3
Jul1614, 07:30 AM

PF Gold
P: 726

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 redraw 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. 


#4
Jul1614, 08:56 AM

P: 34

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?? 


#5
Jul1614, 08:58 AM

P: 34

this is the given circuit...



#6
Jul1614, 09:00 AM

PF Gold
P: 726

They are in SERIES. Reread what I posted above. It is factual information you need to absorb. 


#7
Jul1614, 09:02 AM

P: 34

this is the circuit somewhat similar to thew prvs one.. what according to u is the Rth..??



#8
Jul1614, 09:03 AM

P: 34

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..



#9
Jul1614, 09:07 AM

PF Gold
P: 726

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. 


#10
Jul1614, 09:19 AM

P: 34

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.??



#11
Jul1614, 09:24 AM

PF Gold
P: 726

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. 


#12
Jul1614, 09:26 AM

P: 34

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..!! 


#13
Jul1614, 09:28 AM

PF Gold
P: 726

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. 


#14
Jul1614, 09:32 AM

P: 34

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? 


#15
Jul1614, 09:38 AM

PF Gold
P: 726

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. 


#16
Jul1614, 09:44 AM

P: 34

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.. ??



#17
Jul1614, 09:46 AM

P: 34

welll I think it'll be overdramatic now..sticking to the same thng fr so long..leave it then..
well thanxx a lot for helping out..!!! 


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