Find Thevenin Voltage & Current Across R2: Tricks & Tips Included

  • Thread starter Thread starter phyznut
  • Start date Start date
  • Tags Tags
    Thevenin Voltage
Click For Summary

Homework Help Overview

The discussion revolves around finding the Thevenin voltage (Vth) and current across resistor R2 in a circuit analysis context. Participants are exploring techniques related to Thevenin's theorem, specifically focusing on the voltage across R2 when it is removed from the circuit.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to determine Vth after calculating Rth and seeks guidance on the tricks involved in this process. Some participants suggest considering the circuit with I2 set to zero and question the implications of this assumption. Others inquire about the use of the voltage divider and how it applies to the modified circuit.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the circuit's behavior when R2 is removed. Some guidance has been offered regarding the voltage divider approach, but there is no explicit consensus on the best method to find Vth yet.

Contextual Notes

Participants are navigating the implications of removing R2 and the resulting changes in circuit configuration, which may affect their calculations. There is a noted uncertainty regarding the voltage across points A and B when I2 is considered to be zero.

phyznut
Messages
16
Reaction score
0
I'm trying to find the current and voltage across R2 using thenenin.
http://img209.imageshack.us/img209/4606/thv15ph.png
I've managed to find Rth, which is equal to (R1 + R4)||(R3 + R5). That comes out to about 2.9 ohms. I need help with finding the Vth (Vab) though.
http://img208.imageshack.us/img208/5002/thv28nc.png
Can someone please help me. What are the tricks involved here?
 
Last edited by a moderator:
Physics news on Phys.org
Your second picture IS the trick ... with I_2 = 0, what is V_ab ?
That is, with I_2 = 0 , all the current thru R_1 goes thru R_3 , R_4 , and R_5 .
 
lightgrav said:
Your second picture IS the trick ... with I_2 = 0, what is V_ab ?
That is, with I_2 = 0 , all the current thru R_1 goes thru R_3 , R_4 , and R_5 .

Ok then so with R2 removed, the circuit is now series. That makes the current the same for the remaining resistors.

So if I_2 is 0, then the voltage between that point is also zero? That doesn't seem right though. I still don't get it, please give me another hint.
 
Can no one help.

Would I have to use the voltage divider? How do i use it?
 
Last edited:
Yes you could use the voltage divider formula. With R2 removed, you can think of [tex]V_{AB}[/tex] as the voltage drops across [tex]R_3 and R_5[/tex]

Look at it in this way. When you remove [tex]R_2[/tex], the circuit now becomes a series circuit. Look at where point A and B are. If you where to take a voltmeter and put the probes across points A and B. Your voltmeter would be giving you to combined voltage of R3 and R5.
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
2K
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
3
Views
2K
Replies
15
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K