The discussion revolves around calculating current distribution in a circuit involving resistors in series and parallel configurations. The original poster seeks assistance specifically with part B of the problem, having already calculated the equivalent resistance.
Some participants have provided insights into the current distribution based on resistance ratios, while others express confusion about how to mathematically represent these distributions. The discussion is ongoing, with various interpretations being explored.
There is mention of specific resistor values and the need to construct equations to solve for current distribution, indicating a focus on mathematical representation. The original poster also notes uncertainty in demonstrating the principles discussed.
rohanprabhu said:if the current through the [itex]5 \Omega[/itex] resistor is 2A, then the current through the assembly of the 3 resistors i.e. the assembly of the [itex]6 \Omega[/itex], [itex]3 \Omega[/itex] and [itex]2 \Omega[/itex] is also 2A as both these assemblies are in series.
So, for the assembly of three resistors, you have the incoming current as 2A. When the current splits in more than two arms, the current in each arm is inversely proportional to current. This follows from the formula V = IR, as all three arms are in parallel, they are across the same potential difference and I becomes inversely proportional to R.
So, if the arms have resistances in ratio 1:2:3, the current in each arm will split in the ratio 3:2:1 i.e. the arm with the lowest resistance will have the highest current through it.. however the ratio still holds.
In your example, the resistances are in the ratio 6:3:2. Hence the current will be in the ratio 2:3:6. How does this divide into 2A of current? Construct a linear equation and solve for it. You shall have your answer...
jcpwn2004 said:Thanks for the help, I don't really understand to show the science but basically the 6ohm resistor will get .33A, 3ohm will get .66A and the 2ohm will get 1A? How do I show that?