skivail said:
Can you talk to your teacher about this problem?
You can certainly solve the problem "brute force"... you can do KVL around the different loops... label each resistor with a current... then get around 12 equations with 12 unknowns... but that seems like the "wrong" way to go about the problem.
It seems to me like mesh analysis is the way to solve this. After getting the current through R4, R5 and R6, you can get 5 equations in 5 unknowns using mesh. It is still a large problem for a high school quiz... Still have the feeling I'm not seeing something.
Have a look at that site I gave you a link to.
Well, here's how I'd do it using mesh analysis:
Draw a loop clockwise inside each of the 6 squares. Each loop represents a current. You already know the current in the square in the top row to the right(with R4,R5,R6). Label each current: I1 (going through R1),I2 (through R2),I3(going through R3),I4(going through R12),I5(going through R13).
Then using KVL... for loop 1:
-(I1)(R11) -V1 - (I1)(R1) + V2 - (I1-I4)(R10) + V7 =0
Loop 2:
-V2-(I2)(R2)-(I2-I3)R8 - (I2-I4)R9=0
Loop 3:
-(I3-I2)R8 - I3(R3) - V3 - (I3-I5)R7=0
Loop 4:
-V7 - (I4-I1)(R10) - (I4-I2)R9 + V5 - (I4)(R12) + V4 = 0
Loop 5:
-V5 + (I5-I3)(R7) - V6 - (I5)(R13) = 0
So you've got 5 equations with 5 unknows.
Once you solve for I1,I2,I3,I4,I5... you can get the current anywhere in the circuit. The current through R8 for example is I2-I3 downward. If I2-I3 is negative then the current is I3-I2 (positive value) upward.
Not easy.
Sorry I couldn't help more. Hope someone else sees something I missed.