Charlie:
I see what you are asking. I think Smart has shown one way to handle different numbers of resistors.
I would have used the basic current divider formula for two resistors in parallel. Then, to add more resistors, I would have used "equivalent circuits" to reduce the circuit to two parallel resistors. Then you are back to a circuit with only two resistors.
Basic Current divider:
IR1 = It * R2/(R1 + R2)
(notice you use R2 in the numerator to find current I1)
Resistance equals some constant (r) times the length L:
IR1 = It * (1/r*L2) /((1/r*L1 + 1/r*L2))
(the r's cancel out)
IR1 = It * (1/L2) /((1/L1 + 1/L2))
If R2 is actually several resistors, we can combine them into an Equivalent resistor, Req, using the formula for parallel resistors:
Req = 1/ ( 1/R2 + 1/R3 + 1/R4 + ...)
Then we substitute lengths:
1/r*Leq = 1/ (1/r*L2 + 1/r*L3....)
Again, the r's cancel out:
Leq = 1/L2 + 1/L3....
Now we just substitute Leq for L2 to solve for the current through R1. For each other current I2, I3, ..., we have to find the new Leq, and substitute it back into the equation with L2, L3, etc.
Steve
I see what you are asking. I think Smart has shown one way to handle different numbers of resistors.
I would have used the basic current divider formula for two resistors in parallel. Then, to add more resistors, I would have used "equivalent circuits" to reduce the circuit to two parallel resistors. Then you are back to a circuit with only two resistors.
Basic Current divider:
IR1 = It * R2/(R1 + R2)
(notice you use R2 in the numerator to find current I1)
Resistance equals some constant (r) times the length L:
IR1 = It * (1/r*L2) /((1/r*L1 + 1/r*L2))
(the r's cancel out)
IR1 = It * (1/L2) /((1/L1 + 1/L2))
If R2 is actually several resistors, we can combine them into an Equivalent resistor, Req, using the formula for parallel resistors:
Req = 1/ ( 1/R2 + 1/R3 + 1/R4 + ...)
Then we substitute lengths:
1/r*Leq = 1/ (1/r*L2 + 1/r*L3....)
Again, the r's cancel out:
Leq = 1/L2 + 1/L3....
Now we just substitute Leq for L2 to solve for the current through R1. For each other current I2, I3, ..., we have to find the new Leq, and substitute it back into the equation with L2, L3, etc.
Steve