Kirchhoff's Laws: Solving Parallel Resistors w/ Example

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The discussion focuses on solving for the current in parallel resistors using Kirchhoff's Laws. A circuit with three parallel resistors of 2, 3, and 4 ohms is analyzed, with a total current of 10 Amps after reconnection. The participant seeks a more intuitive method to determine the current in each resistor based on their resistance ratios. They successfully calculate the total resistance and voltage, then derive individual currents for each resistor. The conclusion emphasizes that combining resistors into simpler pairs can simplify calculations, although it requires understanding the inverse sum of resistances.
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Homework Statement


Say I have a circuit that splits into three parallel wires. They each have a resistors on them of 2, 3, and 4 ohms respectively. They reconnect and their final current is 10 Amps. Is there a more intuitive way to find the current on each wire based on the ratios of ohms, rather than finding the voltage and using that to find the individual currents?

I know if the there were two parallel wires with resistors one each, I could sum the two R's together and use that as a denominator and the resistor of one wire as the numerator and multiply this fraction with the final current to find the current in the opposite wire. I want to know if this would work on more than 2 parallel wires and how. If it can, could somebody please explain using the example above.


Homework Equations


The sum of the currents from each wire should equal the final current after they are reconnected.

Voltage is same across parallel resistors.

Sum of R (parallel): R-1= ((1/R1) + (1/R2)...)

V=IR

The Attempt at a Solution



Sum R = 0.923 ohms
V = (10A)(0.923ohms) = 9.23V
I1 = (9.23V)/(2ohms) = 4.615A
I2 = (9.23V)/(3ohms) = 3.077A
I3 = (9.23V)/(4ohms) = 2.308A
 
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Find the resistance R' of the parallel combination of 2 and 3 ohm.
Then find the current in 4 ohm using the formula
I' = I*R'/(R + R')
 
oh! Duh. I can combine the other two resistors into one and compare it as a two wire problem. Too bad it's not intuitive enough as to just add and multiply things, since I need to find the inverse sum of the two resistors first. Thnx.
 
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