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.
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.