Kirchhoff's Laws

When series and parallel run out

In the last lesson you took networks apart by eye. Two resistors end to end add. Two side by side combine as reciprocals. Keep collapsing pairs and eventually the whole box becomes one resistor, and Ohm's law finishes the job.

Now picture five resistors wired like a ladder with one rung: a source feeds node S, two resistors run from S down to two middle nodes A and B, two more run from A and B down to ground, and a fifth resistor bridges A straight across to B. Look for a series pair. There isn't one: every resistor has a third wire landing on one of its ends. Look for a parallel pair. There isn't one of those either. The network is perfectly ordinary and it cannot be simplified by inspection at all.

That is the situation Kirchhoff's two rules exist for. They never fail and they turn any circuit at all into simultaneous equations you can grind through. The cost is bookkeeping, and bookkeeping is where nearly all the marks get lost.

Charge does not pile up: the junction rule

Take any point where wires meet: call it a junction, or a node. Charge is conserved, and a wire junction is a tiny piece of metal with essentially no capacity to store charge. If more charge arrived per second than left, the node would build up charge without limit, and its potential would climb without limit. That doesn't happen in a steady circuit. So over any second, whatever flows in flows back out:

$\sum I_{\text{in}} = \sum I_{\text{out}}$

This is the junction rule, and it is nothing but conservation of charge in a steady state. If $2.50\ \mathrm{A}$ comes into a node on one wire and $0.80\ \mathrm{A}$ leaves on a second, the third wire carries $2.50 - 0.80 = 1.70\ \mathrm{A}$ away. Four wires, five wires, the same sentence applies.

A more compact version: pick a sign convention where current into the node counts as positive, and then $\sum I = 0$ at every node. You will use that form in a moment.

Potential is single-valued: the loop rule

Every point in a circuit has a potential, a number of volts, the way every point on a hillside has an altitude. Walk any closed path on a hillside and you end where you started, so the altitude changes must cancel. The same is true of potential:

$\sum_{\text{closed loop}} \Delta V = 0$

That is the loop rule. It is conservation of energy per unit charge: carry one coulomb all the way round and back to its starting point, and it must have the energy it started with.

Now the bookkeeping. You walk round the loop in a direction you choose, and every component you pass contributes one term.

A resistor carrying current $I$ has a potential drop $IR$ across it, with the current flowing from the high side to the low side. So if you walk along the current through a resistor you are walking downhill, and the term is $-IR$. If you walk against the current you are climbing, and the term is $+IR$.

A battery of emf $\mathcal{E}$ holds its positive plate $\mathcal{E}$ volts above its negative plate, no matter which way the current runs through it. Walk in at the minus plate and out at the plus plate and you have climbed, so the term is $+\mathcal{E}$. Walk in at the plus and out at the minus and the term is $-\mathcal{E}$. Notice that this has nothing to do with the current direction, only with which plate you meet first.

Those four cases are the whole of it. Write them down once, and every loop equation you ever write is an exercise in reading a diagram carefully.

Here is the rule in action on the simplest circuit that needs it: two batteries in one loop, pointing against each other, with total resistance $R$. Walk clockwise, taking $I$ clockwise too. You meet $\mathcal{E}_1$ minus-plate first, so $+\mathcal{E}_1$. You meet the resistor along the current, so $-IR$. You meet $\mathcal{E}_2$ plus-plate first, so $-\mathcal{E}_2$. Setting the sum to zero,

$\mathcal{E}_1 - IR - \mathcal{E}_2 = 0 \quad\Longrightarrow\quad I = \frac{\mathcal{E}_1 - \mathcal{E}_2}{R}$

Real batteries, and what a voltmeter reads

A battery is not a pure emf. It has some resistance inside it, called the internal resistance $r$, and the current has to fight through that too. What you can measure at the two terminals is the terminal voltage:

$V_{\text{term}} = \mathcal{E} - Ir$

when current $I$ is flowing out of the positive terminal. A $9.00\ \mathrm{V}$ battery with $r = 1.00\ \Omega$ delivering $1.50\ \mathrm{A}$ reads $9.00 - 1.50 \times 1.00 = 7.50\ \mathrm{V}$ at its posts. Push current backwards through it instead, as a charger does, and $I$ flips sign: forcing $5.00\ \mathrm{A}$ into a $12.0\ \mathrm{V}$, $0.400\ \Omega$ battery gives $12.0 + 5.00 \times 0.400 = 14.0\ \mathrm{V}$. That is why a car's electrics sit near $14\ \mathrm{V}$ with the engine running and near $12\ \mathrm{V}$ without.

For Kirchhoff purposes you treat the internal resistance as a plain resistor in series with an ideal emf, and it joins the loop equation like any other resistor.

Worked example 1. A $6.00\ \mathrm{V}$ cell of internal resistance $1.00\ \Omega$ and a $4.50\ \mathrm{V}$ cell of internal resistance $0.500\ \Omega$ are wired in one loop with a $3.00\ \Omega$ resistor, the cells opposing.

One loop, so one current. Call it $I$, clockwise, in the direction the bigger cell pushes. Walking clockwise you meet the $6.00\ \mathrm{V}$ cell minus-first ($+6.00$), its own internal resistance along the current ($-1.00I$), the $3.00\ \Omega$ resistor along the current ($-3.00I$), the $4.50\ \mathrm{V}$ cell plus-first ($-4.50$), and its internal resistance ($-0.500I$):

$6.00 - 1.00I - 3.00I - 4.50 - 0.500I = 0$

$1.50 = 4.50\,I \quad\Longrightarrow\quad I = 0.333\ \mathrm{A}$

The total resistance was just the sum of all three, and the net driving emf was the difference of the two. That pattern (net emf over total resistance) is what a single loop always collapses to.

Node voltages: one unknown per node, not one per branch

For circuits with several loops, you could assign a current to every branch and write a junction equation for each node and a loop equation for each loop. It works, and it is a lot of unknowns.

There is a cheaper route. Currents in a circuit are determined by potentials, so make the potentials your unknowns. Pick one node, call its potential zero, and call it ground. Every other node gets an unknown potential. Then the current from node X to node Y through a resistor $R$ is just $(V_X - V_Y)/R$ by Ohm's law, and you write the junction rule at each unknown node with every current expressed that way. Nodes held at a fixed potential by an ideal source are not unknowns at all: you already know them.

This is worth the effort the moment there are more branches than nodes, which is most interesting circuits.

Worked example 2. Two batteries in parallel across one load. Battery 1 is $12.00\ \mathrm{V}$ with $r_1 = 0.200\ \Omega$; battery 2 is $12.60\ \mathrm{V}$ with $r_2 = 0.300\ \Omega$. Both positive terminals meet at node A, both negative terminals at node B, and a $1.50\ \Omega$ load also joins A to B. What current flows in the load?

Set $V_B = 0$ and let $V_A = V$, the only unknown. Current out of battery 1 into node A is $(12.00 - V)/0.200$, because the ideal emf holds the far end of $r_1$ at $12.00\ \mathrm{V}$. Likewise battery 2 supplies $(12.60 - V)/0.300$. The load carries $V/1.50$ away. Junction rule at A:

$\frac{12.00 - V}{0.200} + \frac{12.60 - V}{0.300} = \frac{V}{1.50}$

Multiply everything by $1.50$ to clear the denominators: $7.50(12.00 - V) + 5.00(12.60 - V) = V$, so $90.0 - 7.50V + 63.0 - 5.00V = V$, giving $153.0 = 13.50V$ and $V = 11.333\ \mathrm{V}$. The load current is

$I_{\text{load}} = \frac{11.333}{1.50} = 7.56\ \mathrm{A}$

One equation, one unknown, no loops drawn at all. Check the branches if you like: battery 1 gives $(12.00-11.333)/0.200 = 3.33\ \mathrm{A}$ and battery 2 gives $(12.60-11.333)/0.300 = 4.22\ \mathrm{A}$, which add to $7.56\ \mathrm{A}$ as they must. If the numbers had been less kind, one of those could easily have come out negative (meaning the stronger battery is charging the weaker one) and that is a real, physical answer, not an error.

The bridge, at last

Back to the circuit from the opening. An ideal $6.00\ \mathrm{V}$ source runs from S (positive) to G (negative). From S, $100\ \Omega$ goes to A and $150\ \Omega$ goes to B. From A, $200\ \Omega$ goes to G; from B, $250\ \Omega$ goes to G. A galvanometer of resistance $50.0\ \Omega$ bridges A to B. What current does it carry?

Worked example 3. Ground G, so $V_G = 0$ and $V_S = 6.00\ \mathrm{V}$. Two unknowns left, $V_A$ and $V_B$.

Junction rule at A: in from S, out to G, out to B:

$\frac{6.00 - V_A}{100} = \frac{V_A}{200} + \frac{V_A - V_B}{50}$

Multiply through by $200$: $2(6.00 - V_A) = V_A + 4(V_A - V_B)$, so $12.0 - 2V_A = 5V_A - 4V_B$, which tidies to

$7V_A - 4V_B = 12.0$

Junction rule at B: in from S, in from A, out to G:

$\frac{6.00 - V_B}{150} + \frac{V_A - V_B}{50} = \frac{V_B}{250}$

Multiply through by $750$: $5(6.00 - V_B) + 15(V_A - V_B) = 3V_B$, so $30.0 + 15V_A = 23V_B$, and $V_B = (30.0 + 15V_A)/23$.

Substitute into the first equation and multiply by $23$: $161V_A - 4(30.0 + 15V_A) = 276$, so $161V_A - 120 - 60V_A = 276$ and $101V_A = 396$, giving $V_A = 3.921\ \mathrm{V}$. Then $V_B = (30.0 + 58.81)/23 = 3.861\ \mathrm{V}$. The galvanometer current is

$I_g = \frac{V_A - V_B}{50.0} = \frac{0.0594}{50.0} = 1.19\times10^{-3}\ \mathrm{A}$

It runs from A to B, because $V_A$ came out the higher. Two nodes, two equations, and the circuit that had no series or parallel pair anywhere in it is finished.

One thing that fell out for free: if $V_A$ had equalled $V_B$, no current would cross, whatever the galvanometer's resistance. That happens when the two dividers split the supply in the same ratio, which works out to $R_1R_4 = R_2R_3$ for the four outer arms. That is the balanced Wheatstone bridge, and it is how unknown resistances were measured for a century.

Symmetry, when the algebra would be ugly

Twelve $1\ \Omega$ resistors form the edges of a cube. Between opposite corners that is eight nodes and twelve branches, and nobody wants those equations.

Use the principle behind the balanced bridge instead: if two nodes are at the same potential, no current flows between them, so you may join them with a wire or cut the wire between them, and nothing changes.

Send current $I$ in at one corner. By the symmetry of the cube, the three edges leaving that corner are indistinguishable, so each takes $I/3$, and the three nodes they reach must all sit at the same potential. The three nodes adjacent to the far corner likewise all sit at one potential, and the far corner drains $I/3$ from each of them. The six middle edges connect the first trio to the second trio, and there are six of them sharing $I$, so each carries $I/6$. Walk a path from one corner to the other: $\tfrac{I}{3}(1) + \tfrac{I}{6}(1) + \tfrac{I}{3}(1) = \tfrac{5I}{6}$ volts. Divide by $I$:

$R = \tfrac{5}{6}\ \Omega \approx 0.833\ \Omega$

No simultaneous equations at all. Symmetry is not a different law: it is Kirchhoff's rules plus the observation that identical branches must carry identical currents.

Recognising when to reach for this

You are looking at a Kirchhoff problem when any one of these is true.

More than one source. Two batteries in different branches, or a battery buried inside one arm of a parallel pair, kills the series-parallel method outright. Any question about which battery is charging which is this.

A bridging element. A galvanometer, a voltmeter, or a resistor connected across the middle of two dividers. Trace the circuit and try to name a series pair or a parallel pair. If you cannot, stop trying and write node equations.

The question asks for one branch, not the total. "What current flows in the $6.00\ \Omega$ resistor" in a multi-branch network is a node-voltage question.

An unknown that isn't a current. A hidden resistor, an unknown emf, an ammeter reading you must work backwards from. Set the unknown as a symbol, write the same equations, and solve for it instead of for $I$.

Choosing a method is mostly counting. Fewer nodes than loops, use node voltages; fewer loops than nodes, assign a circulating current to each loop and write the loop rule for each.

Next you'll let the currents change with time. Put a capacitor into one of these branches and the junction and loop rules still hold at every instant, but one of the terms now depends on how much charge has already arrived, which turns an algebraic equation into an exponential. That is RC circuits and transients.