Two batteries in one circuit? Work out what each would do alone, then add the answers.
The problem this solves
Ohm's law needs one voltage and one resistance. Give it a circuit with two batteries pushing current through the same resistor and it has nothing to work with — there are two voltages, and they are fighting over the same wire.
Superposition is the way out, and the idea is one you already use without noticing.
The idea: two taps filling a bath
A bath with two taps. The hot tap alone fills it at 3 litres a minute. The cold tap alone fills it at 5 litres a minute.
Open both. How fast does it fill? 8 litres a minute. You added them, and nobody had to teach you a theorem to do it.
Circuits behave the same way. Work out what each source would do on its own, then add the results. One hard problem becomes two easy ones — and you already know how to solve the easy ones with series, parallel and Ohm's law.
💡 If one tap were draining the bath, you would still add — you would just add a negative number. Same in a circuit: if the two sources push current in opposite directions through a branch, one partial answer is negative and the sum becomes a subtraction.
The catch: how to switch a source off
To find what one source does alone, the others must be turned off. "Off" means something different for each kind, and getting it backwards is the commonest way to lose this question entirely.
🔌 What "off" looks like
Voltage source (battery)Becomes a plain wire. A dead battery still conducts — it is a piece of metal, not a hole in the circuit.
Current sourceBecomes a gap. A current source pushing nothing is a sealed pipe; nothing gets through.
Either, with internal resistanceThe resistance stays. Only the EMF or the current output is removed, never the resistance.
💡 If you remember one thing: a dead battery is a wire. The current source is its opposite, so the other follows.
The formal statement
Worth knowing in the exam's own words, now that you know what it means:
In any LINEAR network with two or more independent
sources, the current (or voltage) in any branch equals
the ALGEBRAIC SUM of the currents (or voltages)
produced by each source acting alone, with all other
sources replaced by their internal resistances.
"Linear" matters. It means doubling the push doubles the flow — true of resistors, not true of diodes or transistors. "Algebraic sum" is the phrase doing the work about signs: add them, keeping their directions.
The step everyone trips on: the current divider
Every worked example below needs this, and it looks backwards the first time.
Two resistors side by side — 3 Ω and 4 Ω — and 7 A arrives at the fork. How much goes each way?
Current is lazy: more of it takes the easier path. So the 3 Ω branch gets the bigger share. That is the whole idea, and everything else is bookkeeping.
Current into a branch =
the OTHER branch's resistance
I × ───────────────────────────────
both resistances added
Into the 3 Ω: 7 × 4/(3+4) = 4 A ← bigger share
Into the 4 Ω: 7 × 3/(3+4) = 3 A ← smaller share
Check: 4 + 3 = 7 ✓ nothing lost at the fork
💡 Why the other resistance on top? Because a large resistance in the other branch means that path is refusing to take current, which pushes more of it down yours. The number on top is measuring how much the other path resists.
💡 The check that never fails: the smaller resistor must end up with the bigger current. If yours does not, you have flipped the fraction. Two seconds to verify.
Worked example
A 12 V source with 2 Ω in series on the left. A 6 V source with 3 Ω in series on the right. A 4 Ω in the middle that both push through. Find the current in the 4 Ω.
Step 1 — only the 12 V source
Switch the 6 V off, so replace it with a wire. Now look at what the 12 V source faces: its own 2 Ω in series, and then — because the 6 V source is now a wire — the 3 Ω and the 4 Ω both bridge the same two points, so they are in parallel.
3 Ω parallel 4 Ω:
R_p = (3 × 4)/(3 + 4) = 12/7 = 1.714 Ω
(parallel is always below the smallest — 1.714 is under 3 ✓)
Total the source must push through:
R_t = 2 + 1.714 = 3.714 Ω
I = 12 / 3.714 = 3.231 A
That 3.231 A splits at the fork. We want the 4 Ω's
share, so the OTHER branch (3 Ω) goes on top:
I′ = 3.231 × 3/(3+4) = 1.385 A
Check: the 4 Ω is the bigger resistor, so it gets less
than half of 3.231 ✓
Step 2 — only the 6 V source
Same again, mirrored. Short the 12 V source, and now the 2 Ω is the one parallel with the 4 Ω.
Both partial currents run downward through the 4 Ω, so they reinforce:
I = 1.385 + 0.462 = 1.847 A
(carrying full precision: 1.8462 A — the two
differ only by rounding the parts first)
Had they run in opposite directions, you would subtract. This is why you mark each partial current's direction on your sketch before adding anything — a sign slip loses the whole answer.
Proof it works: solve it a different way
You need not trust the method. Solve the same circuit with nodal analysis, which uses none of it:
Let V be the voltage across the 4 Ω. All current
leaving that node sums to zero:
(V−12)/2 + (V−6)/3 + V/4 = 0
Multiply by 12:
6(V−12) + 4(V−6) + 3V = 0
13V = 96
V = 7.385 V
I = 7.385/4 = 1.846 A ✓ same answer
💡 Worth doing once, on a problem you could already solve. Two unrelated methods agreeing is what makes the theorem believable — and after that you can use it on problems you cannot check.
The one trap: power does not superpose
Power in the 4 Ω from the TOTAL current:
P = I²R = 1.846² × 4 = 13.63 W ✓
Adding the two partial powers instead:
1.385²×4 + 0.462²×4 = 7.67 + 0.85 = 8.53 W ✗
Wrong by nearly 40%.
The reason is that power depends on I squared, and squares do not add: 2² + 3² is 13, while (2+3)² is 25.
💡 So: superpose the currents, then square. Never superpose powers. Examiners set this deliberately and the wrong answer looks entirely reasonable.
Exam notes
📝 What earns marks
Draw each sub-circuitMarks are awarded per step, so a separate sketch for each source scores even if the final addition goes wrong.
State which source is offAnd how — shorted or opened. Examiners look for it explicitly.
Mark directionsOn every partial current, before adding.
Typical weight6–8 marks for a two-source problem.
💡 Superposition needs linear components. A question containing a diode or transistor is testing whether you noticed — the answer is that superposition does not apply.
Syllabus points
Statement & application (numerical)
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