I have a kirchhoff type problem. What is the current that goes through all those resistors? My attempt at solving: R1=1ohm, R2=2ohm, R3=2ohm(middle one), R4=6ohm;
i2=i4+i3
Three loops can be made:
8v-1(i1)+4v-2(i2)-2(i3)-4v=0
4v+2(i3)-6(i4)=0
8v-1(i1)+4v-2(i2)-6(i4)=0
I made a 4X5 matrix to solve but I cant get the right answer. Im sure there is an easier way and im just not seeing it. Ill attach a pic once i post

Question

I have a kirchhoff type problem. What is the current that goes through all those resistors? My attempt at solving: R1=1ohm, R2=2ohm, R3=2ohm(middle one), R4=6ohm; 
i2=i4+i3
Three loops can be made:
8v-1(i1)+4v-2(i2)-2(i3)-4v=0
4v+2(i3)-6(i4)=0
8v-1(i1)+4v-2(i2)-6(i4)=0
I made a 4X5 matrix to solve but I cant get the right answer. Im sure there is an easier way and im just not seeing it. Ill attach a pic once i post

anonymous · Answer

attachment

roadjester · Answer

ok yeah you went overboard, you can only get 2 equations out of that circuit

anonymous · Answer

well you can make more loops but what equations are you thinking?

roadjester · Answer

ok you have two small loops, let's call the current through the left one $\large i_1$ and the one through the right one $\large i_2$

roadjester · Answer

so 
-8+$\large i_1$-4+2$\large i_1$+2($\large i_1-i_2)$+4=0;
-4+2($\large i_2-i_1$)+6$\large i_2$=0

roadjester · Answer

two equations, two unknows

roadjester · Answer

solve for $\large i_1$ and $\large i_2$

anonymous · Answer

yea i was doing loops and legs. loops only makes it easier, ill give it a try and see what happens

roadjester · Answer

all you need is two clockwise loops

farcher · Answer

You can sometimes solve yourself time and effort by simplifying the problem.  In your example replace the two resistors in the left hand arm with one and do likewise with the batteries. So 3 ohms and 12 volts.

anonymous · Answer

got it. thanks man

roadjester · Answer

np