[SOLVED] Help with Thévenin's theorem

[ayamingesh]

I've managed to solve Rth (3R/5) but I can't seem to get Vth right.
I would appreciate help solving Vth by using voltage dividing.


Thanks in advance.

**broken link removed**
**broken link removed** (numbered resistors)

 

[Marvel_tronix]

I think u did it like Vo*R2/(R1+R2). It's wrong because R2 has in parallel (R3+R4). So, it should be Req=R2||(R3+R4); Vo*Req/(R1+Req).
Then the voltage that it comes out to be say, V1, should be used like: Vth=V1*R4/(R4+R3).
Or, simply use nodal analysis. Assign voltage V1 to 1st node from left and next node has already voltage=Vth.
Write 2 equations corresponding to the 2 nodes with voltages V1, Vth. Solve them using cramer's rule or using calc.
U will get them in a step.

 

[ayamingesh]

I think u did it like Vo*R2/(R1+R2). It's wrong because R2 has in parallel (R3+R4). So, it should be Req=R2||(R3+R4); Vo*Req/(R1+Req).
Then the voltage that it comes out to be say, V1, should be used like: Vth=V1*R4/(R4+R3).
Or, simply use nodal analysis. Assign voltage V1 to 1st node from left and next node has already voltage=Vth.
Write 2 equations corresponding to the 2 nodes with voltages V1, Vth. Solve them using cramer's rule or using calc.
U will get them in a step.

Thank you for your answer. I've managed to solve Vth by using nodal analysis, but I still get wrong answer using voltage dividing.. Dunno what I'm doing wrong.

 

[Marvel_tronix]

Thank you for your answer. I've managed to solve Vth by using nodal analysis, but I still get wrong answer using voltage dividing.. Dunno what I'm doing wrong.

Provide ur solution using voltage divider.....I will let u know what's wrong

 

[ayamingesh]

1: (R+R)//R = 2R/3
2: Vo*(2R/3)/(2R/3+R) = 6/15*Vo (twice as much as the answer)

 

[SolidRaven]

If it's your first few attempts at using Thevenin/Norton there's a simple trick to it: Use it multiple times. Replace each divider by its individual thevenin/norton representation. It isn't really the fastest or best method, but it does tend to work.

If you wish to solve it by voltage divider the easiest method should be something along the lines of this:
\[{V}_{1} = \frac{{R}_{2}//({R}_{3}+{R}_{4})}{{R}_{1}+{R}_{2}//({R}_{3}+{R}_{4})} {V}_{0}\]
\[{V}_{Th} = \frac{{R}_{4}}{{R}_{3}+{R}_{4}} {V}_{1}\]

 

[ayamingesh]

If it's your first few attempts at using Thevenin/Norton there's a simple trick to it: Use it multiple times. Replace each divider by its individual thevenin/norton representation. It isn't really the fastest or best method, but it does tend to work.

If you wish to solve it by voltage divider the easiest method should be something along the lines of this:
\[{V}_{1} = \frac{{R}_{2}//({R}_{3}+{R}_{4})}{{R}_{1}+{R}_{2}//({R}_{3}+{R}_{4})} {V}_{0}\]
\[{V}_{Th} = \frac{{R}_{4}}{{R}_{3}+{R}_{4}} {V}_{1}\]

Thank you sir!

 

[Marvel_tronix]

If you wish to solve it by voltage divider the easiest method should be something along the lines of this:
\[{V}_{1} = \frac{{R}_{2}//({R}_{3}+{R}_{4})}{{R}_{1}+{R}_{2}//({R}_{3}+{R}_{4})} {V}_{0}\]
\[{V}_{Th} = \frac{{R}_{4}}{{R}_{3}+{R}_{4}} {V}_{1}\]

Already, given up there:-o:-:roll:

it should be Req=R2||(R3+R4); Vo*Req/(R1+Req).
Then the voltage that it comes out to be say, V1, should be used like: Vth=V1*R4/(R4+R3).

Anyways, u got it.