Voltage Division with Multiple Voltage Sources Seperated by Resistors

AI Thread Summary
To find the voltage across the 10-ohm resistor in a circuit with multiple voltage sources, the circuit can be simplified by replacing the sources with a single equivalent voltage source. The voltage division formula can then be applied, resulting in a calculated voltage of 6/5V across the 10-ohm resistor. The discussion highlights the importance of understanding circuit simplification techniques for solving complex problems. Additionally, there are suggestions for improving image sharing to enhance clarity in future discussions. Overall, mastering these concepts is crucial for success in electrical circuit analysis.
TheCarl
Messages
21
Reaction score
0

Homework Statement


Find the voltage across the 10 ohm resistor.
http://www.spacemonkeybrewing.com/images/circuit.jpg

Homework Equations



KVL - Sum of Voltages in a closed loop equal zero.
Ohms - V=IR
Voltage Divide - Va1 = (R1/(R1+R2...))Vs

The Attempt at a Solution



This is only the second week into the introduction of electrical circuits and I'm not sure where to start here. It doesn't look like I can use KVL, Ohms law or the voltage division equation to solve this problem. I would know how to solve it if it only had a single voltage source or voltage sources that aren't divided by resistors. Any assistance would be greatly appreciated.
 
Last edited by a moderator:
Physics news on Phys.org
Hi TheCarl, http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

You can replace the 3 voltage sources by 1. Try that and see how you go. :smile:
 
Last edited by a moderator:
So are you saying that basically the circuit above would be equivilant to this:
http://www.spacemonkeybrewing.com/images/circuit2.jpg

And therefore V = (10Ω/(15Ω+10Ω+20Ω+5Ω))6V = 6/5V ?P.S. Thanks for the quick reply.
 
Last edited by a moderator:
TheCarl said:
And therefore V = (10Ω/(15Ω+10Ω+20Ω+5Ω))6V = 6/5V
That's the idea.
 
Thank you so much!
 
For future reference: your pic archive site seems slow or unreliable, and the pic is some of the time failing to appear. I venture that some people will be reading this and wondering where's the circuit we are discussing? For next time, I suggest that you try out another site, or else attach your pic to your post.
 
Here are the two photos in order in case they don't show up in the above threads.
 

Attachments

  • circuit.jpg
    circuit.jpg
    11.8 KB · Views: 448
  • circuit2.jpg
    circuit2.jpg
    9.2 KB · Views: 413

Similar threads

Replies
3
Views
2K
Replies
6
Views
2K
Replies
7
Views
2K
Replies
6
Views
4K
Replies
6
Views
2K
Back
Top