marlon said:
Well, the only thing I was able to do is to apply the two Kirchoff's laws to the chain. In the first knod I have for the currents : I_0=I_1+I_2 Then the two knods in the middel (say I_3 is the current in the middle) = I_1=I_3+I_4 and below I_2+I_3=I_4
Then the second law , I follow the two closed subchains clockwisely
first chain : -I_1R-I_3R = I_2R
second subchain : -I_4R+I_5R+I_3R = 0
regards
marlon
You know, current can flow only across the chain if you connect a battery to it, between points A, B. See attached picture. The resultant resistance, RAB, is the voltage of the battery (E) divided by the current of the battery Io. RAB=E/Io. Don't forget that current flows out and back into the battery! Now yo have to calculate Io when you apply that voltage across A and B.
First take up current in all branches (as you have done already), but do not use more than necessary. Take the nodal law (Kirchhoff's first law) into account. See the attached picture. The arrows show the direction of current in each branch. The algebraic sum of the currents at each node should be zero. Three unknown currents are enough, you can combine the others from them. Now you use Kirchhoff's second law, for the loops. There are three independent of them, shown by the curved arrows (BACB, BCDB, ADCA).
Go round the loops following the arrows and calculate the electric potential at each node. The potential drops across a resistor by RI if you follow the direction of the current, and rises if you walk into the opposite direction. Imagine that you go uphill and downhill along a stream of water. The water flows downward, to lower potential. If you follow the flow, you go downward to lower potential. If you go in the opposite direction as the water flows, you rise to higher potential.
Assume that the potential is zero at your starting point. Start from here, and write the changes of potential as you go round a loop. When you rich back, the total change is zero, you get back to zero potential again. For example, walking round the loop BACB, you start from zero potential at B, then the potential rises up as you go to the + side of the battery, then it drops across the resistors, and then you get back to B, to zero potential again.
E - R*(Io-I1) - R*(Io-I2) = 0.
For the loop BCDB you get:
R*(Io-I2) + R*(I1-I2) -R*I2 = 0,
and for the loop ADCA:
-2R*I1-R*(I1-I2) + R*(Io-I1) = 0
Arrange the equations, solve for Io and calculate RAB = E/Io.
ehild
