Kirchhoff's laws — current and voltage rules — Physics

Before this

This leads to

A circuit that series and parallel cannot solve

Look at the circuit below. It has two batteries, a 12 V one on the left and an 8 V one on the right, and three resistors. Here is the question. How much current flows through each battery, and which way?

Two batteries and three resistors. The arrows are guesses for the directions of the three currents, I₁, I₂ and I₃. The two dots are junctions, where the current can split.

Try the method from the last lecture: "combine the innermost group first". It fails at once. The 2 Ω and the 6 Ω are not in series, because the current can split at the junction between them. They are not in parallel either, because there is a battery in the left branch. No two resistors here can be combined. The rules for series and parallel have nothing to say.

And there is a real puzzle in the answer. You might expect both batteries to push current up and out of their positive terminals. But the 12 V battery is stronger. It might push current backwards through the 8 V battery. Which one wins?

The last lecture promised that its two facts, "charge is conserved" and "energy is conserved", can solve circuits like this, where no groups can be combined. It also promised to give them their usual names. Here they are: Kirchhoff's laws. Gustav Kirchhoff wrote them down in 1845, while he was still a student at the University of Königsberg, as an exercise in his professor's seminar. (Source: Wikipedia, "Gustav Kirchhoff"; IEEE Circuits and Systems Magazine, 2023.) There is nothing new in the physics. What is new is a method that works for every circuit.

The first law: what goes into a junction comes out

A junction is a point where three or more wires meet, so that the current can split or join. In the last lecture you saw why charge cannot pile up anywhere in a steady circuit. If more charge arrived at a junction each second than left it, charge would build up there and push back on the charge arriving. So the flow in and the flow out must be equal.

Every second, 3.0 C of charge arrives at the junction, and 1.2 C + 1.8 C = 3.0 C leaves it. Nothing piles up.

Kirchhoff's first law (the junction rule)

At any junction, the total current flowing in equals the total current flowing out.

$Σ I_{in} = Σ I_{out}$ . This is conservation of charge.

The symbol $Σ$ (the Greek capital letter sigma) means "the sum of". Some books call this law KCL, for "Kirchhoff's current law".

In the opening circuit, the guessed currents $I_{1}$ and $I_{2}$ both arrive at the top junction, and $I_{3}$ leaves it. So the junction rule gives the first equation:

I_{1} + I_{2} = I_{3}

The bottom junction gives the same equation again, with in and out swapped. So it adds nothing new. In general, one junction in a circuit always repeats what the others have said.

The second law: round a loop, the potential comes back to where it started

A loop is any closed path round a circuit that ends where it began. The second fact from the last lecture was about energy. Every point in a circuit has a potential. A coulomb of charge that goes once round a loop ends where it started, at the same potential. So all the rises in potential round the loop must exactly cancel all the falls.

Where are the rises and the falls? A battery raises the potential of each coulomb that passes through it from its negative terminal to its positive terminal. The energy it gives each coulomb is its electromotive force, or e.m.f., with symbol $ε$ (the Greek letter epsilon). It is measured in volts, because 1 V = 1 J/C. Despite its name, the e.m.f. is not a force. It is energy per coulomb. Some books write it as $E$ . A resistor lowers the potential by $I R$ in the direction the current flows, because the charge transfers energy there.

Kirchhoff's second law (the loop rule)

Round any closed loop, the sum of the e.m.f.s equals the sum of the p.d.s across the resistors.

$Σ ε = Σ I R$ . Equivalently, the changes in potential round any loop add up to zero. This is conservation of energy.

Some books call this law KVL, for "Kirchhoff's voltage law". To use it, walk round the loop in one direction and keep track of the potential, like walking round a hill path and counting every step up and every step down. Four rules cover every step.

As you walk across…	The potential…	Because…
a battery, from − to +	rises by $ε$	the battery lifts each coulomb by $ε$
a battery, from + to −	falls by $ε$	you are going down the same step
a resistor, in the direction of your current arrow	falls by $I R$	current flows from high potential to low
a resistor, against your current arrow	rises by $I R$	you are walking uphill, against the flow

Do not learn the four rows as a list. Each one comes from the same picture: current flows downhill through a resistor, and a battery is a step up from − to +. If you can see the hill, you can rebuild the table.

That is it. That is the whole idea. The junction rule counts charge. The loop rule counts energy. Every circuit, however tangled, obeys both.

Solving the opening circuit

Now use the two laws on the circuit with two batteries. Step through the method below, one move at a time.

Step through — two batteries, three currents

Press Next to add one step. Read the reason before you go on.

Each step gives one line of the solution for the opening circuit and the reason it is allowed.

Guess the directions: I₁ up through the 12 V battery, I₂ up through the 8 V battery, I₃ down through the 6 Ω resistor. (Any guess works; a wrong guess just gives a negative answer.)
Junction rule at the top junction: I₁ + I₂ = I₃. (I₁ and I₂ arrive, I₃ leaves; charge cannot pile up.)
Loop rule, left loop: 12 = 2I₁ + 6I₃. (Walking with the arrows, the battery is a rise of 12 V and each resistor is a fall of I × R.)
Loop rule, right loop: 8 = 4I₂ + 6I₃. (The same walk on the right-hand side.)
Put I₃ = I₁ + I₂ into both: 8I₁ + 6I₂ = 12 and 6I₁ + 10I₂ = 8. (Two equations, two unknowns.)
Solve: I₂ = −2/11 A ≈ −0.18 A, then I₁ = 18/11 A ≈ 1.64 A. (Multiply by 3 and by 4, then subtract: 22I₂ = −4.)
I₃ = I₁ + I₂ = 16/11 A ≈ 1.45 A. (From the junction rule.)

I₂ is negative, so about 0.18 A flows down through the 8 V battery, into its positive terminal. The 12 V battery is pushing current backwards through the 8 V battery.

So the 12 V battery wins. About 0.18 A flows the "wrong" way through the 8 V battery, in at its positive terminal and out at its negative terminal. Charge passing through it backwards loses energy: the 8 V battery is taking in energy, not giving it out. If it is a rechargeable battery, it is being charged. This is exactly how a charger works: a higher e.m.f. drives current backwards through the battery being charged.

There is a quick check. Every branch connects the same two junctions, so every branch must show the same p.d. between them. Through the middle: $6 \times \frac{16}{11} = \frac{96}{11} \approx 8.73$ V. Through the left: $12 - 2 \times \frac{18}{11} = \frac{96}{11}$ V. Through the right: $8 - 4 \times (- \frac{2}{11}) = \frac{96}{11}$ V. All three agree. ✓

Notice that the exact fractions were kept until the end. Rounding $I_{2}$ to −0.18 A before using it makes the check come out about 0.04 V wrong. The physics is not wrong in that case. The rounding is.

Stop and think

A student gets $I_{2} = - 0.18$ A and says: "A current cannot be negative. I must have made a mistake. I will start again with I₂ pointing down." Is the student right?

Show the answer

No. There is no mistake. The minus sign is the answer telling you that your guessed direction was wrong. The current is 0.18 A, flowing opposite to the arrow you drew. Starting again with the arrow reversed would give +0.18 A and exactly the same physics. The one thing you must never do is change an arrow halfway through, after some equations have used it. Choose every direction once, at the start, and let the algebra correct you.

Change the second battery

The 8 V battery was being charged. What would it take for it to give out current instead? Try it below. Change the e.m.f. of the right-hand battery and watch all three currents.

Explore — the battery that is charged or discharged

The left battery is 12 V and the resistors are 2 Ω, 6 Ω and 4 Ω. Change the e.m.f. of the right-hand battery. Watch the current through it shrink, stop and reverse.

Try:

E.m.f. of the right-hand battery ε₂: 8 V

This table shows the three currents in the opening circuit for several values of the right-hand battery's e.m.f. ε₂. A positive I₂ means current flows up and out of the right-hand battery's positive terminal. A negative I₂ means it flows backwards, and the battery is being charged.

ε₂ (V)	I₁ (A)	I₂ (A)	I₃ (A)	Right-hand battery
0	2.73	−1.64	1.09	a plain wire: current flows down it
8	1.64	−0.18	1.45	being charged
9	1.50	0	1.50	no current at all
12	1.09	0.55	1.64	giving out current
16	0.55	1.27	1.82	giving out current

At exactly 9 V, no current flows through the right-hand battery. Below 9 V it is charged; above 9 V it gives out current.

You should have found a special value: 9 V. At 9 V, no current flows through the right-hand battery at all. Here is why. With the right-hand branch carrying no current, the 12 V battery drives current only through the 2 Ω and the 6 Ω, in series: $12 \div 8 = 1.5$ A. The p.d. across the 6 Ω is then $1.5 \times 6 = 9$ V. The right-hand branch sits across the same two junctions, so it has 9 V across it too. A 9 V battery there pushes up exactly as hard as the circuit pushes down, so nothing flows. Below 9 V, the circuit wins and charges it. Above 9 V, the battery wins and gives out current. At 0 V, the battery is just a wire, and the circuit becomes an ordinary series and parallel circuit. You can check that $I_{1} = 12 \div (2 + 2.4) \approx 2.73$ A.

Worked examples

Worked Example 1 — One loop

A 9.0 V battery is connected in series with resistors of 4.0 Ω, 4.0 Ω and 1.0 Ω. Find the current, using the loop rule.

Think first. There is only one loop and no junction, so there is one current, $I$ . Walk round with the current: up 9.0 V through the battery, then down $I R$ through each resistor.

9.0 = 4.0 I + 4.0 I + 1.0 I = 9.0 I

I = 1.0 A

Check. The series rule from the last lecture gives $R = 9.0$ Ω and $I = 9.0 \div 9.0 = 1.0$ A. ✓ For one loop, the loop rule is the series rule.

Worked Example 2 — Two batteries facing each other

A 12 V battery and a 3.0 V battery are in one loop, with their positive terminals facing each other, and resistors of 3.0 Ω and 6.0 Ω. Find the current and its direction.

Think first. Guess that the current flows out of the 12 V battery's positive terminal. Walk round in that direction. You go up 12 V through the 12 V battery. Through the 3.0 V battery you go from + to −, which is a fall of 3.0 V.

12 - 3.0 = 3.0 I + 6.0 I

I = \frac{9.0}{9.0} = 1.0 A

Check. $I$ is positive, so the guess was right: the 12 V battery wins. Opposing e.m.f.s subtract. If the 3.0 V battery were turned round, the e.m.f.s would add: $I = 15 \div 9.0 \approx 1.7$ A. ✓

Worked Example 3 — The parallel rule from Kirchhoff's laws

A 6.0 V battery is connected across two resistors in parallel, 3.0 Ω and 6.0 Ω. Use Kirchhoff's laws to find every current, and show that they agree with the parallel rule.

Think first. Call the battery current $I$ , and the branch currents $I_{a}$ (3.0 Ω) and $I_{b}$ (6.0 Ω). There are two small loops, each containing the battery and one resistor.

Loop through the 3.0 Ω: $6.0 = 3.0 I_{a}$ , so $I_{a} = 2.0$ A. Loop through the 6.0 Ω: $6.0 = 6.0 I_{b}$ , so $I_{b} = 1.0$ A. Junction rule: $I = I_{a} + I_{b} = 3.0$ A.

Check. The parallel rule gives $R = \frac{3.0 \times 6.0}{3.0 + 6.0} = 2.0$ Ω and $I = 6.0 \div 2.0 = 3.0$ A. ✓ The series and parallel rules are not separate laws. They are Kirchhoff's two laws, used on circuits simple enough to have shortcuts.

Three traps

Trap 1 — The wrong sign on a resistor

The fall $I R$ is "a fall" only when you walk across the resistor in the direction of your arrow for that current. When your loop crosses a resistor against its arrow, the potential rises by $I R$ . In the right loop of the opening circuit, walking round the other way, against the arrows, gives $- 8 = - 4 I_{2} - 6 I_{3}$ : the same equation, multiplied by −1. Choose a direction for the walk, and look at each arrow as you cross it.

Trap 2 — Changing an arrow halfway through

Once a current has an arrow, every equation must use that same arrow. If you draw $I_{2}$ upward for the junction rule and downward for a loop, the equations describe two different circuits, and the answer is meaningless. A negative answer is not a reason to change an arrow. It is the answer.

Trap 3 — Writing one loop too many

The opening circuit has a third loop: the outside path through both batteries, 2 Ω and 4 Ω. Its equation, $12 - 8 = 2 I_{1} - 4 I_{2}$ , is just the left loop minus the right loop. It adds no new information. You need as many independent equations as there are unknown currents: here, one junction and two loops for three currents.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

(a) Currents of 2.5 A and 0.8 A flow into a junction. One wire carries 1.9 A out. Find the current in the fourth wire. (b) At another junction, 3.0 A flows in, and 1.0 A and 2.5 A flow out. A fourth wire carries a current $I$ , drawn as flowing out. Find $I$ , and say what its sign means.

Hint 1
Total in = total out at every junction.

Hint 2
For (b), write $3.0 = 1.0 + 2.5 + I$ .

Full answer
(a) In: $2.5 + 0.8 = 3.3$ A. Out so far: 1.9 A. So the fourth wire carries $3.3 - 1.9 = 1.4$ A out. Check: $1.9 + 1.4 = 3.3$ . ✓ (b) $3.0 = 1.0 + 2.5 + I$ , so $I = - 0.5$ A. The minus sign means the guessed direction was wrong: 0.5 A flows into the junction along that wire. Check: in $3.0 + 0.5 = 3.5$ , out $1.0 + 2.5 = 3.5$ . ✓
A 9.0 V battery and a 1.5 V battery are in one loop, with their positive terminals facing each other, and resistors of 5.0 Ω and 2.5 Ω. (a) Find the current. (b) Find the p.d. across the 5.0 Ω resistor. (c) Find the current if the 1.5 V battery is turned round.

Hint 1
Facing batteries oppose each other, so their e.m.f.s subtract.

Hint 2
$9.0 - 1.5 = (5.0 + 2.5) I$ .

Full answer
(a) $7.5 = 7.5 I$ , so $I = 1.0$ A, out of the 9.0 V battery's positive terminal. (b) $V = I R = 1.0 \times 5.0 = 5.0$ V. (c) Now the e.m.f.s add: $10.5 = 7.5 I$ , so $I = 1.4$ A. Check (a): the potential rises 9.0 V, falls 1.5 V, 5.0 V and 2.5 V: $9.0 - 1.5 - 5.0 - 2.5 = 0$ . ✓
In the opening circuit, the 8 V battery is turned round, so that its positive terminal faces down. Keep the same arrows for $I_{1}$ , $I_{2}$ and $I_{3}$ . Find all three currents.

Hint 1
Only the right loop changes. Walking up the right branch with the $I_{2}$ arrow, you now cross the battery from + to −.

Hint 2
The right loop becomes $- 8 = 4 I_{2} + 6 I_{3}$ . With $I_{3} = I_{1} + I_{2}$ , the equations are $8 I_{1} + 6 I_{2} = 12$ and $6 I_{1} + 10 I_{2} = - 8$ .

Full answer
Multiply by 3 and by 4: $24 I_{1} + 18 I_{2} = 36$ and $24 I_{1} + 40 I_{2} = - 32$ . Subtract: $22 I_{2} = - 68$ , so $I_{2} = - \frac{34}{11} \approx - 3.09$ A. Then $8 I_{1} = 12 + 6 \times \frac{34}{11} = \frac{336}{11}$ , so $I_{1} = \frac{42}{11} \approx 3.82$ A, and $I_{3} = \frac{8}{11} \approx 0.73$ A. Now both batteries push current round the outer loop the same way, so the currents are much bigger. Check: p.d. across the 6 Ω is $6 \times \frac{8}{11} \approx 4.36$ V; through the left branch, $12 - 2 \times \frac{42}{11} \approx 4.36$ V. ✓
A stretch. A 12 V battery is connected across two paths in parallel. Path 1 has 100 Ω on top of 200 Ω, joined at point P. Path 2 has 150 Ω on top of 300 Ω, joined at point Q. A sensitive meter is connected between P and Q. (a) Show that no current flows through the meter. (b) The 300 Ω is replaced by 330 Ω. Which way does current now flow through the meter?

Hint 1
If no current flows through the meter, each path is a simple series pair. Find the potential at P and at Q, measured from the battery's negative terminal.

Hint 2
The bottom resistor of each path takes its share of the 12 V: $V_{P} = 12 \times \frac{200}{300}$ .

Full answer
(a) Path 1: $V_{P} = 12 \times \frac{200}{300} = 8.0$ V. Path 2: $V_{Q} = 12 \times \frac{300}{450} = 8.0$ V. P and Q are at the same potential, so there is no p.d. across the meter, and no current flows through it. The guess "no current" is consistent with both of Kirchhoff's laws, so it is the answer. This arrangement is called a Wheatstone bridge, and it is "balanced" because $\frac{100}{200} = \frac{150}{300}$ . (b) $V_{Q} = 12 \times \frac{330}{480} = 8.25$ V, now higher than $V_{P} = 8.0$ V. Current flows from high potential to low, so it flows through the meter from Q to P. Check: a tiny change in one resistor gives a clear 0.25 V difference, which is why the bridge is used to measure resistance very precisely. ✓

The bigger idea

Kirchhoff's laws turn any circuit into a set of equations. Two batteries and three resistors gave three equations. A phone's processor has billions of transistors, and its designers cannot build a test chip for every idea. So they let a computer write and solve the equations instead. In April 1973, Laurence Nagel and Donald Pederson of the University of California, Berkeley presented a program called SPICE. It builds the equations for a circuit from the junction rule at every junction, and then solves them. (Sources: L. W. Nagel and D. O. Pederson, SPICE, Berkeley memorandum UCB/ERL M382, April 1973; Wikipedia, "SPICE".) Programs that grew from it are still used to test chip designs before they are built.

There is something worth noticing here. The laws inside those programs are the two facts you have used since the first circuit lecture: charge does not pile up, and energy is not lost. The method grew from one student's seminar exercise into the way every modern circuit is designed. The physics underneath has not changed. The next lecture uses the loop rule on something inside every real battery: a small resistance of its own, which explains why a battery's p.d. falls when it delivers a large current.

Key concepts in this lesson

Kirchhoff's first law says that the total current into any junction equals the total current out, because charge is conserved.
Kirchhoff's second law says that round any closed loop the e.m.f.s add up to the p.d.s across the resistors, because energy is conserved.
The e.m.f. of a battery is the energy it gives each coulomb, in volts, and it is not a force.
A guessed current direction is never wrong in a way that matters, because a negative answer means the current flows the other way.
A circuit needs as many independent equations as unknown currents, and one junction and one loop always repeat what the others say.
The series and parallel rules are Kirchhoff's two laws applied to circuits simple enough to have shortcuts.