Look up at the lights in your house. The kitchen light, the hall light, your bedroom lamp — each one has its own switch, and flicking one off leaves all the others glowing happily. Now compare that with an old string of fairy lights wired in a single loop, where one dud bulb kills the whole string. Same electricity, wildly different behaviour. The difference is how the parts are wired: your house lights are in parallel.

In a series circuit every component sits on one single loop, threaded in a line so the same current is forced through all of them. A parallel circuit is the opposite idea: the components sit on separate branches, each branch reaching straight across from one side of the supply to the other. The current now has a choice of paths — and that one change rewrites the rules for voltage, current and resistance.

Every branch gets the full supply voltage

Here is the first surprise. In a series loop the supply voltage is shared out between the components. In parallel it is not shared — every branch is connected directly across the two terminals of the supply, so every branch feels the whole supply potential difference.

If a V = 12\,\text{V} battery drives three lamps in parallel, each lamp has the full 12 V across it — not 4 V each. Write the branch pds as V_1, V_2, V_3 and the rule is simply

That is exactly why your house is wired this way: every socket and every light is a branch across the same mains supply, so every appliance receives the full mains voltage it was designed for, whether it is switched on alone or alongside twenty others.

The current splits — and the pieces add back up

The current, on the other hand, behaves the opposite way round. Charge streams out of the supply, reaches a junction where the branches meet, and splits — some charge going down each branch. Further on the branches meet again and the streams recombine. No charge is created or lost at a junction, so whatever flows in must flow out: the branch currents simply add up to the total current the supply provides.

Each branch decides its own current from its own resistance, using Ohm's law with the full supply voltage across it:

A low-resistance branch grabs a big current; a high-resistance branch takes a trickle — but the total leaving the supply is always the sum of all of them. (This "what goes in comes out" rule at a junction is really the law of conservation of charge.)

Play with a two-branch circuit

Below is a real parallel circuit: a 6 V supply at the bottom, and two branches reaching across it, each carrying a resistor. The pd across both branches is the full 6 V — that never changes. But dial each branch's resistance up or down and watch its current I = V/R respond, while the total current I_1 + I_2 climbing the left wire is always the two branch currents added together.

Try it: drop R₁ to 1 Ω and its current shoots up; push it to 6 Ω and it dwindles — yet the other branch never even notices. Then make both resistances small at once and watch the total current climb higher than either branch alone.

Worked example: two lamps in parallel

A 12\,\text{V} battery drives two lamps in parallel: R_1 = 4\,\Omega and R_2 = 6\,\Omega. Find each branch current, the total current, and the overall resistance.

Step 1 — the pd across each branch. Both branches sit straight across the battery, so each has the full 12 V: V_1 = V_2 = 12\,\text{V}.

Step 2 — each branch current, from Ohm's law.

Step 3 — add them for the total current.

Step 4 — the overall resistance. The supply pushes 5 A out at 12 V, so the whole circuit behaves like a single resistance

Look hard at that answer: 2.4 Ω is less than either branch — smaller even than the 4 Ω branch on its own. Adding a second path for the current made the circuit easier to push charge through, not harder. That is the headline result for parallel circuits, and the next card explains why.

More branches → less total resistance

It feels wrong at first: you add a resistor and the resistance goes down. But think about doors out of a crowded room. One door lets a certain flow of people through. Open a second door beside it and more people escape per second for the same push — the room has become easier to leave. Each extra branch is another door for the current.

So adding a parallel branch always lowers the total resistance, and the total is always below the smallest single branch — because that smallest branch is still there doing its job, and the new branch only adds even more current on top. For the exact value, the reciprocals add:

Check it against the worked example: \tfrac{1}{R_{\text{total}}} = \tfrac{1}{4} + \tfrac{1}{6} = \tfrac{3}{12} + \tfrac{2}{12} = \tfrac{5}{12}, so R_{\text{total}} = \tfrac{12}{5} = 2.4\,\Omega — exactly the value we got from V/I. Two ways round, same answer.

A neat special case: two identical resistors R in parallel give R_{\text{total}} = R/2. Three identical ones give R/3. Equal branches simply split the total resistance by how many there are.

Worked example: identical branches

Two identical 10\,\Omega lamps are connected in parallel across a 5\,\text{V} cell. What does each carry, and what is the total?

Each lamp has the full 5 V across it, so each carries I = \tfrac{5}{10} = 0.5\,\text{A}. There are two of them, so the supply delivers I_{\text{total}} = 0.5 + 0.5 = 1\,\text{A}. The overall resistance is R_{\text{total}} = \tfrac{5}{1} = 5\,\Omega — exactly half of one 10 Ω lamp, just as R/2 promised. Adding an identical second lamp doubled the current the battery has to supply, which is why a battery runs flat faster the more things you switch on in parallel.