# Circuit Help

Discussion in 'Homework Help' started by TitaniumX, Apr 5, 2013.

1. ### TitaniumX Thread Starter New Member

Apr 5, 2013
2
0
The problem statement
What is the current through the 3 Ohms resistor in the diagram below?

Attempt at a solution

I tried to simplify the circuit by adding the 5ohm and 1ohm resistors because they are in series, an the same thing with the 3ohm and 1ohm resistor. I don't know how to apply I=V/R and Kirchoff's Rule since this is just a very odd problem with not too many given info. This problem has two voltage source so it's kind of confusing. I think I may have to use Kirchoff's Rules but I'm not sure how to approach this to get the answer. Any help would be great.

2. ### MrChips Moderator

Oct 2, 2009
12,648
3,458
Label the current flowing through the 8, 5 and 3-ohm resistors as I1, I2 and I3.

Now use Kirchoff's current and voltage laws to solve for I1, I2 and I3.

You have three unknowns. Hence you need three equations.

3. ### TitaniumX Thread Starter New Member

Apr 5, 2013
2
0
I tried to do that step, and I got the answer 2 Amp.

When I try to run a simulation of it on PSplice, it gave me a difference answer.

But none of the answers match the possible answer choices on the practice problem packet.

Can someone please show me the steps and calculations? Thanks

4. ### MrChips Moderator

Oct 2, 2009
12,648
3,458
In Homework Help forum, we will not do your homework for you.

5. ### WBahn Moderator

Mar 31, 2012
18,089
4,917
You need to show your work. Don't just say, "I tried and couldn't get it, so please do it for me."

If you tried to do "that step", then show us what you did in trying to do that step so that we can see what you have right and what you have wrong. We are not mind readers.

6. ### daniel.tippman New Member

Oct 7, 2012
1
0
Using KVL. Set up two current loops, I_left & I_right, in each branch of the circuit. Add up the voltage drops(Current*Resistance) as you walk around each current loop back to your starting point(watch your signs on the sources). For the center path, the current is the difference between I_left and I_right in one equation and the difference between I_right and I_left in the second equation. This method will result in two equations two unknowns.