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Parallel Circuits

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More than one path

Components are in parallel when they are connected across the same two points, giving current more than one path.

Reading the schematic: straight lines are wires; the unequal parallel lines are a cell/battery symbol; a zigzag is a resistor; a circle marked V is a voltmeter; a circle marked A is an ammeter. Labels beside the symbols tell us their values.
12 V 6 Ω 3 Ω 2 A 4 A Total current = 6 A

A useful first picture

RT and IT

The small T means total. RT is total resistance of the entire parallel network, and IT is the total current supplied by the source.

Think of a road dividing into two lanes and joining again later. Traffic can split between the branches. The total traffic entering the junction equals the total traffic leaving it.

What is really happening?

Every branch connected across the same two nodes has the same voltage across it. The current in each branch depends on that branch's resistance. The source supplies the sum of the branch currents.

A little history

Kirchhoff's junction rule formalized this idea: current entering a junction must equal current leaving it. It is an expression of conservation of electric charge.

Worked example

12 V 6 Ω 3 Ω 2 A 4 A Iₜ = 6 A

Both branches have 12 V across them.

6 Ω branch: 12 ÷ 6 = 2 A.
3 Ω branch: 12 ÷ 3 = 4 A.

IT = 2 + 4 = 6 A

The equivalent resistance is 12 V ÷ 6 A = 2 Ω.

Try these yourself

Two branches are connected directly across a 12 V source. What voltage is across each branch? 12 V R1 R2
Parallel branches share the same two connection points, so each branch has the full 12 V across it.
A 12 V source supplies one 6 Ω branch and one 3 Ω branch. Which branch draws more current? 12 V 6 Ω 3 Ω
At the same voltage, lower resistance draws more current. The 6 Ω branch draws 2 A; the 3 Ω branch draws 4 A.
If two branch currents are 2 A and 4 A, what current does the source supply? source branch branch 2 A 4 A Iₜ = ?
Branch currents add at the junction: 2 A + 4 A = 6 A.

A little deeper

For resistors in parallel:

1/RT = 1/R1 + 1/R2 + ...

The total resistance is lower than the resistance of any individual branch because the parallel network provides additional paths for current.

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