Solve Simple Circuit Problem: R1 = 2.00 Ω, R2 = 4.00 Ω, R3 = 8.00 Ω

  • Topic: Engineering 
  • Thread starter Thread starter Nexest
  • Start date Start date
  • Tags Tags
    Circuit
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
5 replies · 4K views
Nexest
Messages
5
Reaction score
0
Here is the problem:

The diagram below shows a circuit where; R1 = 2.00 Ω, R2 = 4.00 Ω, R3 = 8.00 Ω, V1 = 7.500 V, V2 = 10.00 V, and V3 = 8.00 V. (In solving the problems that follow, initially pick the current directions as shown. If the actual current turns out to be in the opposite direction, then your answer will be negative).

iy1wtc.jpg


Using the junction and loop rules I came up with these equations but apparently something is wrong, please help me:

I1 + I2 + I3 = 0

-V1 + I2R2 - V2 - I1R1 = 0

I3R3 - V3 + V2 - I2R2 + V1 = 0

EDIT:

I know how to solve this problem in theory with the equations but when I did I got the wrong answer so I'm checking if my original equations are right
 
Last edited:
Physics news on Phys.org
Firstly, this should probably be in the homework help forum.

Secondly, you haven't uploaded an image, or provided a link to somewhere else where this image is posted! (If you have an image, you should upload it to, say, tinypic.com, imageshack.us, or freeimagehosting.net, as opposed to uploading an image here, since it'll take some time for the mods to get around to approving the image / attachments).

EDIT: Can you report your own post to get your image approved faster? Or would the mods frown upon abuse of the abuse line?
 
I can't see the diagram you're talking about, but it looks like you have 1 KCL equation and 2 KVL equations written. You're probably looking for a third (independent) KVL equation to solve for the currents I1, I2 and I3.
 
Sorry, I wasn't aware of the image problem, is it working better now?
 
it shows up fine. you should be able to reduce that to two equations.
 
Yep, I can see it now.

Yeah just write two mesh equations and solve and that will get you the currents in all the branches of the circuit.

Hint: remember two mesh currents will be going through the center branch