According to the current division rule, if two resistors R1 and R2 are parallel then \$ I_{R1} = \frac{R2}{(R1 + R2)} \times I \$ . In this case which resistance value should I pick for the numerator to find \$ I_{R2} \$?
Electrical – Current division with more than two parallel resistor
currentcurrent-divider
Related Solutions
Does the voltage source also produce current, so that the current flowing to the parallel resistors is the sum of 40mA and the current produced by the voltage source?
If two components are in series, then the current through them is equal. That means the current through the voltage source is equal to the current through the current source.
Hint: I think the sensible approach is to simplify the circuit far enough to only have 2 branches, then calculate the current through them. Once you know the current, you can focus on the more simplified branch (but now knowing how much current flows through it as a whole), split it into two R's again and apply the familiar rule.
Full answer:
First of all you should know how one can simplify resistor circuits (reference).
Most people like to draw the circuit in a more suggestive way; grouping R's that are parallel and such.
simulate this circuit – Schematic created using CircuitLab
Group R1, R2, R4, R5 and R6
- begin with (R1 || R4) = R7; ie. 1/R7 = 1/R1 + 1/R4
- (R7 + R2) = R8
- R9 = (R8 || R5)
- R10 = (R9 + R6)
Calculate current; now you know the current over R10 and R3
Take R10 and calculate current between R8 and R5
Best Answer
Your 2 resistor formula won't work because there are 3 resistors in parallel. The current goes through all 3 resistors, so first you must calculate the total resistance. The formula for any number of resistors in parallel is 1/Rt = 1/R1 + 1/R2 + 1/R3 + ...
Once you have the total resistance you can multiply it by the total current to get the voltage drop. Once you have that, simply divide the voltage by each resistor's value to get the current flowing through it.