You have met the tidy rules for series and parallel circuits: current the same all round a single loop, or the supply voltage the same across every branch. But real circuits are rarely so obliging. Two batteries pushing against each other; a bridge of five resistors that is neither purely series nor purely parallel; a network with loops inside loops. Reach for "the series rule" or "the parallel rule" and you are stuck — the circuit is both at once, and neither.

In 1845 a 21-year-old student named Gustav Kirchhoff wrote down two short laws that crack every circuit, however tangled. They are not new physics — they are just conservation of charge and conservation of energy, wearing electrical clothes. Master these two statements and you can, in principle, solve any circuit ever drawn: write one equation per junction, one per loop, and solve them together. That is the whole of circuit analysis in two lines. Let's meet them.

The first law: charge cannot pile up (the junction rule)

A junction (or node) is any point where three or more wires meet. Charge is never created and never destroyed, and it cannot heap up at a point in a wire — so every coulomb that arrives at a junction each second must leave it again that same second. In terms of current (charge per second), the total current flowing in equals the total flowing out:

Picture a river reaching a fork: the water arriving up-stream has to equal the water going down the two channels — not a drop appears from nowhere, not a drop vanishes. So if a current I_1 flows into a junction and splits into I_2 and I_3, then

This is exactly why, in a parallel circuit, the branch currents add up to the supply current: the parallel rule I_{\text{total}} = I_1 + I_2 + \dots is just Kirchhoff's first law applied at the branch point. The junction law is the more general statement — it holds at every node of every circuit, no matter how the rest is wired.

The second law: energy is conserved round a loop (the loop rule)

Now follow a single charge all the way round any closed loop and back to where it started. A source of EMF \varepsilon (a battery or cell) gives the charge energy; each resistor takes energy from it as a pd drop IR. When the charge returns to its starting point it must be back at the same potential — energy given must equal energy taken. So, going once round any loop, the EMFs sum to the IR drops:

It is like a walk that ends where it began: every metre you climb you must come back down, so the heights cancel to zero. Round a loop, potential returns to its start, so the pushes (\varepsilon) and the drops (IR) balance exactly.

This is where the series rule comes from. In a single series loop with supply \varepsilon and resistors dropping V_1, V_2, \dots, the loop law reads \varepsilon = V_1 + V_2 + \dots — precisely "the supply voltage is shared." Kirchhoff's second law is that same idea freed from the single loop, ready for a circuit with many loops at once.

Two laws, every circuit

Here is the payoff. Take a circuit with several branches and several loops. Label an unknown current in every branch, choosing a direction for each (a guess — a wrong guess just comes out negative, which is fine). Then:

• Write a junction equation (\sum I_{\text{in}} = \sum I_{\text{out}}) at enough junctions;
• Write a loop equation (\sum \varepsilon = \sum IR) round enough loops;
• Solve the equations together for the unknown currents.

Because charge and energy are always conserved, these equations are always true and always enough. There is no circuit the two laws cannot reach — which is why every circuit-simulation program on Earth is, at heart, a machine for writing down Kirchhoff's equations and solving them.

See both laws working together

Below is a circuit with a \varepsilon = 12\ \text{V} cell driving current up the left wire to a junction, where it splits into an upper branch (current I_2, resistor R_2) and a lower branch (current I_3, resistor R_3), before recombining and returning to the cell. Drag the two branch currents.

Watch the first law: the current I_1 in the main wire is always I_2 + I_3 — set the branches and the trunk follows. And watch the second law: each branch sits in a loop with the cell, so its resistance auto-adjusts to R = \varepsilon / I, and every loop's drops sum back to the full 12\ \text{V} EMF. Two conservation laws, one picture.

Worked example 1: currents at a junction

Four wires meet at a junction. Currents of 3\ \text{A} and 2\ \text{A} flow in; a current of 4\ \text{A} flows out along a third wire. What does the fourth wire carry, and which way?

Apply the first law. Total in must equal total out. In total, 3 + 2 = 5\ \text{A} arrives. One exit already removes 4\ \text{A}, so the last wire must carry the rest:

The fourth wire carries 1\ \text{A} out of the junction. Had the sum come out negative, that would simply mean our guessed direction was backwards — the current would actually flow the other way.

Worked example 2: voltages round a loop

A single loop contains a 9\ \text{V} cell and three resistors in series carrying the same current. Two of them are found to drop 4\ \text{V} and 2\ \text{V}. What is the pd across the third?

Apply the second law. Going once round the loop, the EMF must equal the sum of the IR drops:

The energy the cell hands out (9\ \text{V} per coulomb) is spent exactly by the three resistors (4 + 2 + 3 = 9\ \text{V}) — none left over, none conjured from nothing. That is conservation of energy in one line.

Worked example 3: a two-loop network

Now the real prize — a circuit that is neither series nor parallel. A 2\ \Omega resistor carries the main current I_1 out of a 6\ \text{V} cell to a junction, where it splits into two parallel 6\ \Omega resistors carrying I_2 and I_3. Find all three currents.

Step 1 — the junction law. At the split, current in equals current out:

Step 2 — a loop law for each branch. Going round the loop through the cell, the 2\ \Omega resistor and the upper 6\ \Omega branch:

By symmetry the two 6\ \Omega branches carry equal current, so I_2 = I_3. Then I_1 = I_2 + I_3 = 2 I_2.

Step 3 — solve simultaneously. Substitute I_1 = 2 I_2 into the loop equation:

So I_3 = 0.6\ \text{A} and I_1 = 1.2\ \text{A}. Check: the two 6\ \Omega resistors in parallel make 3\ \Omega, in series with the 2\ \Omega gives 5\ \Omega, and I_1 = 6/5 = 1.2\ \text{A}. \checkmark Two little laws untangled a circuit the simple rules could not touch.