You must draw the 7 loops with positive direction of circulation. Call the loop-currents i1 . . i7. Now follow exactly what kirchhoff states.Well ok,but how are you taking the potentials of the nodes?
You follow the arrow for i2: You will cross 4 resistors looping around, with the value R. therefore i2 results in a voltage drop = -4*R*i2.
One of the resistors in the loop is crossed by the current i3 in opposite direction, which yield a voltage rise = R*i3.
One other resistors is crossed by the current i1 in opposite direction, which yield a voltage rise = R*i1.
The sum of all these voltage changes must be zero, thus:
-4*R*i2 + R*i3 + R*i1 = 0.
Setting R = 1Ω, this can be rewritten:
-4*i2 + i3 + i1 = 0
Be very careful with the signs. ( That's why you have to make a drawing, so that you know what you are doing ). Having calculated the currents, remember to divide their values by 6, as the real value for R is 6Ω ( not 1Ω ).
Of course it will give 0: You have not connected the battery! One more loop including the battery is needed.Then,the solution of these six equations is i1=i2=i3=i4=i5=i6=0 !!!
Of course it will give 0: You have not connected the battery! One more loop including the battery is needed.
You should write it as: i2 - 4*i4 + i6 = 0 because that's what Kirchhoff states: The sum of . . . . . . = 0. Only the voltage of a battery/powersupply, you will write on the right side. This is important because Kirchhoffs laws are intended for a computer, that can very systematically build the set of equation itself. Just "show" it at drawing/diagram, and it will do it.
You are the one to learn the computer to do it by programming. So do it systematically yourself, and you can learn the computer to do the same. Then it will solve 50 equations for you at the speed of light.