# Current division into three branches

Please see the image below for the circuit. I am trying to find the currents through each branch in the circuit. However one part, which is essential in finding currents through all the branches, confuses me: when current flows from the source (I1), it will divide into the three branches containing R2, R3 and R5. But how can I calculate the currents through these branches?
I know the current division formula when it divides into two branches, but this one is a different case with three branches. Any guidance would be appreciated.

• Combine resistors (series & parallel) until you only have 2 branches. Then determine current through each branch, extract backwards. Apr 21, 2016 at 18:46

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.

• 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