Parallel Circuits - Physics for Pilot Assessments

A parallel circuit is one in which the current has more than one route available, and almost every rule from the series lesson turns over when it does. By the end of this lesson you should be able to say what is shared and what divides when branches are connected side by side, add branch currents to get the total, calculate a combined resistance and explain why it always comes out below the smallest branch, work the calculation backwards, and say why one lamp failing leaves the rest of a household lit.

Branches that start and end at the same two points

Take two resistors and connect both of their left-hand ends to one point in a circuit and both of their right-hand ends to another. Those two shared points are called junctions or nodes, and the two resistors now form separate branches between them. The immediate consequence is about voltage. The potential difference across a component is the difference in energy per coulomb between its two ends; both branches have the same pair of ends; so both branches have the same voltage across them. And if those nodes are connected directly to the supply, that shared voltage is the full supply voltage — not a share of it.

Voltage across parallel branches

V_1 = V_2 = V_3 = V_\text{supply}

$V_1, V_2, V_3$: the potential difference across each branch, in volts (V)
$V_\text{supply}$: the supply voltage, in volts (V) — every branch across the supply sees all of it

This is the first reversal, and it is worth saying in the same breath as the series rule so that the pair is learned together: in series the voltage divides and the current does not, and in parallel the voltage does not divide and the current does. Everything else about parallel circuits follows from those two sentences.

The current divides

At the node where the branches meet, the charge arriving has a choice, and it takes both routes at once — some of it down one branch and the rest down the other. How much goes each way is decided by the branches themselves: each branch has the full supply voltage across it, so the current in it is that voltage divided by its own resistance, quite independently of what the other branches are doing. The branch with the lower resistance takes the larger current, which is the reverse of the series case where the larger resistance took the larger voltage. Nothing is lost or created at a junction, so the currents in the branches must add up to the current in the shared wire feeding them.

Branch currents add to the total

I_\text{total} = I_1 + I_2 + I_3 + \dots

$I_\text{total}$: the current in the shared wire, drawn from the supply, in amperes (A)
$I_1, I_2, I_3$: the current in each individual branch, in amperes (A)

So a parallel circuit can be worked out branch by branch, and that is usually the quickest route through a question: find each branch current from the shared voltage and that branch's resistance, then add them for the total. The independence is real, and it is the practical point of the arrangement — switching one branch off does not change the voltage across the others or the current in them.

Both branches are connected to the same two nodes, so both have the whole 12 V across them and both carry 2 A. The shared wires carry the sum, 4 A, and the pair together behaves like a single 3 Ω resistance — lower than either branch on its own.

Why the total resistance falls

Compare the figure above with the series figure from the previous lesson. Both have a 12 V supply; the series pair drew 2 A, and this parallel pair draws 4 A. More current for the same voltage means less opposition, so the combined resistance of parallel branches is lower than that of either branch alone — and the mechanism is not mysterious. Adding a branch does not put anything in the way of the current already flowing; it opens a second door. Two doors let more people through than one, however narrow the second one is, so every branch you add lowers the total resistance and raises the total current. This is the reversal that catches people out, because in series adding a resistor always raised the total.

Putting a number on it takes a little more work than adding. What adds is not the resistances but their reciprocals — one divided by each of them, a quantity that measures how well each branch conducts rather than how badly.

Total resistance of parallel branches

\frac{1}{R_\text{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots

$R_\text{total}$: the resistance of the combination, in ohms (Ω)
$R_1, R_2, R_3$: the resistance of each branch, in ohms (Ω)
$\tfrac{1}{R}$: one divided by that resistance — add these, then invert the answer to get R

The step everyone forgets is the last one: what that calculation gives you is one divided by the total, so you have to turn the answer upside down at the end to get the total itself. Two shortcuts are worth having, because they cover almost every case a paper will set. If the branches are identical, the total is one branch divided by the number of them — two 6 Ω branches give 3 Ω, and three 6 Ω branches give 2 Ω. And for exactly two unequal branches, the total is their product divided by their sum, which for 6 Ω and 3 Ω gives $\frac{18}{9} = 2$ Ω. Use whichever is quicker, and then apply the sanity check that catches nearly every arithmetic slip: the answer must be smaller than the smallest branch. If it is not, something has gone wrong.

Example 1 — two equal branches

A 12 V supply feeds two 6 Ω resistors in parallel. Find the voltage across each, the current in each, the total current, and the combined resistance.

Both branches are connected straight across the supply, so each has the full 12 V across it.
Each branch current is $\frac{\text{voltage}}{\text{resistance}} = \frac{12}{6} = 2$ amperes.
The branch currents add: $I_\text{total} = 2 + 2 = 4\ \text{A}$
For the combined resistance, use the whole circuit: $\text{resistance} = \frac{\text{voltage}}{\text{current}} = \frac{12}{4} = 3\ \Omega$
Check it against the shortcut for identical branches: one branch divided by the number of them is $\frac{6}{2} = 3$ ohms. And 3 Ω is indeed less than either 6 Ω branch, as it must be.

Example 2 — two unequal branches

The same 12 V supply now feeds a 6 Ω resistor in parallel with a 3 Ω resistor. Find both branch currents, the total current, and the combined resistance.

Each branch still has the full 12 V across it — unequal resistances change the currents, not the voltage.
Through the 6 Ω branch: $\frac{12}{6} = 2\ \text{A}$ Through the 3 Ω branch: $\frac{12}{3} = 4\ \text{A}$
The lower resistance took the larger current, which is what you should expect.
Total current: $I_\text{total} = 2 + 4 = 6\ \text{A}$
Combined resistance: $R_\text{total} = \frac{12}{6} = 2\ \Omega$
Check with product over sum: $\frac{6 \times 3}{6 + 3} = \frac{18}{9} = 2\ \Omega$ Note that 2 Ω is below the 3 Ω branch, the smaller of the two — a total of 9 Ω, which is what adding them would have given, is the classic wrong answer.

Example 3 — working backwards to a branch

A 12 V supply delivers a total of 5 A to two parallel branches. One branch is a 6 Ω resistor. What is the resistance of the other branch?

Find the current in the known branch first: it has the full 12 V across it, so $\frac{12}{6} = 2$ amperes.
The branch currents add to the total, so the other branch carries 5 − 2 = 3 amperes.
That branch also has 12 V across it, so its $\text{resistance} = \frac{\text{voltage}}{\text{current}} = \frac{12}{3} = 4$ ohms.
Check the whole circuit: the combined resistance should be $\frac{12}{5} = 2.4$ ohms, and product over sum gives $\frac{6 \times 4}{6 + 4} = \frac{24}{10} = 2.4$ ohms. The two agree, and 2.4 Ω is below the smaller branch.

More than two branches, and how the current splits

Nothing about the reasoning changes when a third branch is added, and following it through is the quickest way to see why the total resistance keeps falling. Put three 6 Ω branches across the same 12 V supply and each one still has the full 12 V across it, so each still carries $\frac{12}{6} = 2$ A; the total is now 2 + 2 + 2 = 6 A, and the combined resistance is $\frac{12}{6} = 2$ Ω. Compare the sequence: one branch alone is 6 Ω, two branches are 3 Ω, three are 2 Ω, and four would be 1.5 Ω. Every branch added opens another route, so the current climbs and the combined resistance falls — never towards some limit part way, but steadily towards zero as branches accumulate. That is also the practical warning built into the arrangement: the shared wires and the supply carry the sum of all the branch currents, so they have to be able to stand it, and connecting too many things across one supply overloads those shared parts rather than the individual branches.

It is worth naming the pattern in how the current divides, because it is the exact mirror of the voltage-sharing rule in a series circuit. There, the current was common and the voltages split in direct proportion to the resistances, so the larger resistance took more. Here, the voltage is common and the currents split in inverse proportion to the resistances, so the larger resistance takes less. With a 6 Ω branch beside a 3 Ω branch, the resistances are in the ratio two to one and the currents come out in the ratio one to two — 2 A and 4 A, as Example 2 found. You can use that as a shortcut, but the safer habit in a timed test is the one used in the examples: work out each branch current from the shared voltage and that branch's own resistance, and let the proportions look after themselves.

Why one failure does not take out the rest

If a lamp in a parallel branch fails, its branch is broken and carries no current — but the other branches still run between the same two nodes, still have the same voltage across them, and so still carry the same current as before. The rest stay lit, and they stay at exactly their previous brightness rather than getting brighter or dimmer. This is why the lights and sockets in a building are wired in parallel: each can be switched independently, each gets the full supply voltage rather than a share of it, and a failure is confined to the thing that failed. The one thing that does change is the total: with one branch gone the combined resistance rises and the current drawn from the supply falls, by exactly the current that branch used to take.

Why it matters in aviation

Aircraft loads hang in parallel across a busbar, and both parallel rules are being used on purpose. Because every branch spans the same two points, each service receives the full busbar voltage rather than a share that depends on what else happens to be switched on, so equipment can be built for one known voltage. And because the branches are independent, a fault in one leaves the voltage across the others unchanged, and their currents with it, which is the electrical basis of redundancy: losing one item does not dim the rest.

What the arrangement does not protect is the shared part of the path. The current in the feeder, and the load the generator is carrying, is the sum of all the branch currents, so every service switched on adds to a total that generation has to meet, which is why that total is displayed and monitored. It is also why load shedding works. Removing non-essential branches lowers the total current without altering the voltage the remaining branches see, so the essential services carry on exactly as before from a smaller supply.

What the test asks

The commonest trap is adding parallel resistances as though they were in series — two 6 Ω branches offered as 12 Ω. Two defences work every time: the total must be smaller than the smallest branch, and it must never exceed it however many branches there are. The second trap is expecting a parallel branch to receive only part of the supply voltage, which is the series rule imported into the wrong circuit; each branch across the supply gets all of it. Expect an unequal-branch item where the distractor gives the larger current to the larger resistance, and expect the failed-lamp item, where the wrong options have the surviving lamps changing brightness. And read the wording carefully when a question asks for the current: in a parallel circuit, the current in the shared wire and the current in a branch are different numbers.