# Wheatstone Bridge Circuit #2

1. Dec 9, 2013

### DunceKirchhoff

http://i42.tinypic.com/10sfxa1.jpg

Im having trouble with the bridge circuit.

I see the current has been given 2A

So 6.6=3i1+2(i1-i2)
5i1-2i2=6.6

So applying KVL I get

0=4i2+5i1-3(2-i1)
8i2+5i1=6

So using simultaneous equations I get-

5i1-2i2=6.6
5i1+8i2=6

So 10i2=12.6

i2= 1.26 But thats the answer for i1!!!!

Im new to physics so Im learning from scratch....but Im willing to put in the hours because I really need this qualification. Was considering hiring a tutor but money is scarce for me just now.

Any help would be appreciated :)

2. Dec 9, 2013

### Staff: Mentor

What happened to i3 which joins with I1 to become I4?

You'll need to employ a few more KCL relationships (or use some other analysis methods if you've covered them, such as mesh or nodal analysis).

Using just KCL and KVL I'd suggest writing KCL for each node to begin with so that you have a set of relationships between all the currents. Then write KVL for three loops and pound away at the substitutions...