When Ohm’s law is not enough: solve any circuit with two simple rules

How to use this simulation

1. Start with the defaults: 12 V and 6 V batteries, with 4 Ω, 3 Ω and 6 Ω resistors. The three arrows show I₁ = 1.33 A, I₂ = -0.22 A and I₃ = 1.11 A.
2. Look at I₂. It is negative, so its arrow points down and its dots run the other way. The weaker battery is being pushed backwards by the stronger one.
3. Read the first row of the panel: the junction rule. The two currents entering the top junction are added and I₃ is taken away. The sum is zero for every slider position you try.
4. Read the three loop rows. In each, the battery voltage minus the voltage drops comes to zero. Raise E₂ slowly and watch I₂ shrink, reach zero and then change sign.
5. Set E₂ to zero. The right battery disappears and the circuit becomes one battery with two resistors in parallel. Compare the readings with the third worked example below.
6. Switch to electron flow to see the charges run opposite to the arrows. The arrows and the sums do not change, because they use the conventional current.

Why one law is not enough

Ohm’s law, V = IR, is perfect for one resistor. Series and parallel formulas stretch it to simple combinations. But look at the circuit in the simulation. The middle resistor R₃ is shared by two batteries. It is not in series with anything and not in parallel with anything, so no “equivalent resistance” trick can collapse it.

Gustav Kirchhoff, a German physicist, gave two rules in 1845 that work on any network, however tangled. They are not new physics. They are two things you already believe, written in a form you can calculate with: charge does not pile up, and energy is not created from nothing.

In exams these rules appear as “find the current in each branch” problems, usually with two cells and a few resistors. Once you can set up the equations, the rest is school algebra.

The two rules in plain words

The junction rule (also called the current law, KCL): at any junction, the total current flowing in equals the total current flowing out. A junction is a point where three or more wires meet.

The loop rule (also called the voltage law, KVL): if you walk once around any closed loop and add up every rise and drop in potential, the total is zero. You come back to the same point, so you must come back to the same potential.

The junction rule, with numbers

Take a junction where 5 A and 3 A flow in, and 2 A and some unknown current flow out. Coming in: 5 + 3 = 8 A. Going out must also be 8 A. Known outflows are 2 A and 1.5 A, which add to 3.5 A, so the missing current is 8 − 3.5 = 4.5 A.

In the simulation the top junction has I₁ and I₂ arriving and I₃ leaving. So the rule reads I₁ + I₂ = I₃, or I₁ + I₂ − I₃ = 0. With the defaults: 1.33 + (-0.22) − 1.11 = 0.

Notice you never needed to know which way the current really goes. You choose a direction for each current, write the rule, and the algebra tells you the truth.

The loop rule and its signs

Signs are where most marks are lost, so use one fixed habit. Pick a direction to walk around the loop. Then apply the table below to each component you cross.

Try it on a single loop first: a 12 V battery with 3 Ω and 5 Ω in series. The current is 12 ÷ (3 + 5) = 1.50 A. The drops are 1.50 × 3 = 4.50 V and 1.50 × 5 = 7.50 V. Walking round: +12 − 4.50 − 7.50 = 0. The battery’s rise is exactly used up by the resistors.

A method that always works

Here is the routine for any Kirchhoff problem. It takes a minute and it is the same every time.

• Label every branch current (I₁, I₂, I₃) and draw an arrow on each. Guess freely. A wrong guess costs nothing.
• Write the junction rule for one junction. For two junctions the second equation is the same as the first, so one is enough.
• Choose independent loops, enough to cover every branch at least once. Write the loop rule for each, using the sign table.
• You now have as many equations as unknown currents. Solve them by substitution or determinants.
• Read the signs of the answers. Positive means your guess was right. Negative means the real current runs against your arrow, and the size is still correct.
• Check with a loop you did not use. If its sum is not zero, find the slip.

The simulated circuit, solved by hand

The defaults are E₁ = 12 V, E₂ = 6 V, R₁ = 4 Ω, R₂ = 3 Ω and R₃ = 6 Ω. Currents I₁ and I₂ are assumed to go up their batteries and I₃ down the middle.

What a negative current means

I₂ came out as -0.22 A. That is not an error. It means the real current in the right branch flows down through the 6 V battery, from its positive plate to its negative plate, at 0.22 A. The 12 V battery is stronger and is pushing current backwards through the weaker one, which is exactly how a battery gets charged.

The power numbers say the same. Battery 1 delivers 16.00 W. Battery 2 delivers E₂ × I₂ = -1.33 W, and the minus sign means it absorbs 1.33 W instead of giving it. Together the batteries supply 14.67 W.

Where does that go? The resistors turn it into heat: 7.11 W in R₁, 0.15 W in R₂ and 7.41 W in R₃, a total of 14.67 W. Supply equals heat, which is the loop rule’s energy conservation seen from another side.

More worked examples

Example A (a symmetric circuit). Two equal 10 V batteries, outer resistors 5 Ω each, middle resistor 10 Ω. By symmetry I₁ = I₂, so I₃ = 2I₁. The loop rule gives 5 I₁ + 10 × 2I₁ = 10, so I₁ = I₂ = 0.40 A and I₃ = 0.80 A. The top junction sits at 8.00 V above the bottom one.

Example B (one battery only). Remove battery 2: E₁ = 12 V, R₁ = 4 Ω, R₂ = 3 Ω, R₃ = 6 Ω, and E₂ = 0. Now R₂ and R₃ are in parallel, giving 2.00 Ω. The total resistance is 4 + 2.00 = 6.00 Ω, so I₁ = 2.00 A. The junction voltage is 2.00 × 2.00 = 4.00 V, so I₃ = 0.67 A and I₂ = −1.33 A. Kirchhoff’s laws and the series-parallel method agree, and the minus sign just says the current in R₂ flows down.

A second route: the junction voltage

You can also solve this circuit by guessing the potential of the top junction. Let it be V. Then the current in each branch follows from Ohm’s law, and the junction rule gives one equation in V alone.

Here the conductances add up to 1/4 + 1/3 + 1/6 = 0.75 S, and the battery terms add to 12/4 + 6/3 = 5.00 A. So V = 5.00 ÷ 0.75 = 6.67 V, and I₃ = V ÷ 6 = 1.11 A. The same answer, with one equation instead of three. Use it when a circuit has many parallel branches and just two junctions.

Common mistakes

These four cost the most marks.

• Treating the arrow you drew as the truth. It is a guess; the sign of the answer corrects it.
• Using E and IR with the same sign when walking a loop. A battery gives a rise or a drop depending on direction, and so does a resistor.
• Writing the junction rule with three currents all entering. Unless one of them is negative, they cannot all flow in.
• Forgetting that a shared resistor carries a combined current. In the loop equations, R₃ multiplies I₃, not I₁ or I₂.

Exam corner

Marks are usually split like this: stating the rule or law, writing correct equations with a labelled diagram, solving, and giving direction and unit. A neat diagram with arrows is worth marks in itself.

Short questions ask for the statement of the junction rule and the loop rule, and the conservation law behind each (charge and energy). Numericals give two cells and three resistors. If a Wheatstone bridge is involved, remember it is a Kirchhoff problem too, and the next animation in this series explores it.

Revision list

Say these out loud before the exam.

• Junction rule: current in = current out, because charge is conserved.
• Loop rule: the potential changes around a closed loop add to zero, because energy is conserved.
• Across a resistor with the current: −IR. Against the current: +IR.
• Through a cell from − to +: +E. From + to −: −E.
• Unknown currents need as many independent equations as unknowns.
• A negative answer is a correct answer with the arrow drawn the wrong way.
• Check the result with one more loop.

Frequently asked questions