Lessons · Electrical · series and parallel together
Mixed circuits: shrink the parallel part first
When a circuit has both, replace each parallel group with its single equivalent resistance, then add what is left in series.
Hone is a place to practise a career, one idea a day. This is one of its lessons, written out in full and free to read without an account.
In beta. This lesson was written for Hone and has not yet been checked by a licensed electrician. Practice material, not professional advice. What that means.
What it is for
A run to a junction box that splits into two loads, or a long cable in series with a bank of lamps, is a mixed circuit. The source only ever sees one total, and finding it in the wrong order gives a wrong number that looks fine.
How to think about it
Draw it. Circle every parallel group and replace it with one resistor. Now the drawing is a plain series loop: add, then Ohm's law once for the current. Work back outward for the volts across each part.
Worked example
6 Ω ∥ 12 Ω = (6 × 12) / (6 + 12) = 4 ΩThe parallel pair becomes one 4 Ω resistor.
R = 2 Ω + 4 Ω = 6 ΩIn series with the 2 Ω cable. The source sees 6 Ω.
I = 36 V / 6 Ω = 6 AThe current in the cable and into the pair.
V across the pair = 6 A × 4 Ω = 24 VBoth branches see 24 V, because they are in parallel.
I in the 6 Ω branch = 24 / 6 = 4 A; in the 12 Ω branch = 24 / 12 = 2 A4 + 2 = 6 A. The branch currents add back to the total.
Your turn
A 3 Ω cable in series with a pair that reduces to 9 Ω. Write the total.
R = 3 + = 12 Ω
Solve one, graded on the server
The trap
Adding a parallel branch's resistance straight into the series total. The 12 Ω branch does not add 12 Ω to the loop; the pair adds 4 Ω. Reduce first, then add.