Kirchhoff's Laws

A car with its engine running has two power sources on the same wires. One is the battery. The other is the alternator, a generator that the engine turns to make electricity. Both feed the lights, the radio and everything else on the car's 12 V system. Is the battery helping to run the lights, or is the alternator charging the battery? That depends on the two voltages.

The series and parallel rules from lesson 02 cannot answer this. With a source in each of two paths, no resistor is purely in series or purely in parallel with another. Kirchhoff's laws can, and they work for any circuit of sources and resistors.

Nodes, branches and loops

A node is a point where two or more parts connect (lesson 02). All the wire joining them counts as one node, and some books call it a junction. In the instrument below, node A is where R1, R2 and R3 meet. Ground is a node too.

A branch is a path between two nodes that does not split on the way, so one current flows through every part on it. This circuit has three branches from A to ground: V1 with R1, R2 on its own, and V2 with R3.

A loop is any path around a circuit that ends where it started. The instrument checks two: the left loop through V1, R1 and R2, and the right loop through V2, R3 and R2.

Which way does the current flow?

A battery has two terminals, the ends you connect wires to, marked + and −. Schematics draw current leaving a source by its + terminal and coming back in at −. This is the conventional current direction: the way a positive charge would move. In a copper wire the moving charges are electrons, which drift the other way, but the arithmetic comes out the same.

In a circuit like this one, you often cannot tell which way a current flows until you solve it. So you pick a direction for each current, draw an arrow and stick to it. That choice is your sign convention. If an answer comes out negative, the current flows against your arrow, and the size is still right.

The instrument writes its convention under its controls, and its arrows show the real direction. V1 and V2 are ideal sources, which hold their voltage whatever current flows. A real battery sags a little, as a later lesson shows.

The current law: what flows in flows out

Kirchhoff's current law (KCL) says that at any node, the current flowing in equals the current flowing out. Charge cannot pile up at a node or appear from nowhere.

A USB hub without its own power adapter is a node you can hold. The current in its cable from the computer equals the currents its ports supply, plus the little its own chip uses. In the instrument, the "KCL at A" readout adds I1 + I3 - I2. Change any control and it stays at 0 A.

The voltage law: around a loop, it adds to zero

Kirchhoff's voltage law (KVL) says that if you go around any loop and add up the voltages, the total is zero. The rises equal the drops. Each voltage gets a sign as you walk past it:

• Through a source from − to +, you gain its voltage: +V.
• Through a resistor the same way as its current, you lose its voltage drop: −I × R.
• Through a resistor against its current, you gain it: +I × R.

Lesson 01's LED circuit is a single loop. The USB port gives a 5 V rise, the LED takes about 2.0 V and the resistor takes 3.0 V: 5 V − 2.0 V − 3.0 V = 0.

The formula at the top of the instrument, ΣI = 0, ΣV = 0, is both laws in short: Σ (the Greek letter sigma) means "add up all the", each with its sign.

Worked example: the voltage at A

Call the voltage at A, measured from ground, V_A. Once you know it, Ohm's law gives every current. R1 has 9 V − V_A across it, R3 has 6 V − V_A, and R2 has V_A:

• I1 = (9 V − V_A) / 1 kΩ
• I3 = (6 V − V_A) / 1 kΩ
• I2 = V_A / 1 kΩ

KCL at A says I1 + I3 = I2. Every resistor is 1 kΩ, so multiply both sides by 1 kΩ and the resistance drops out:

(9 V − V_A) + (6 V − V_A) = V_A

That gives 15 V = 3 × V_A, so V_A = 5 V. Since 1 V across 1 kΩ drives 1 mA, the currents are I1 = 4 mA, I3 = 1 mA and I2 = 5 mA. The instrument shows the same: "in 5 mA = out 5 mA".

KVL checks the answer. The left loop goes up 9 V through V1, down 4 V across R1 (4 mA × 1 kΩ) and down 5 V across R2: 9 V - 4 V - 5 V, which is 0 V. The right loop gives 6 V - 1 V - 5 V, also 0 V.

When one source is forced to absorb energy

A source's power is its voltage times the current leaving its + terminal. If that is positive, the source delivers energy. If current is pushed into its

• terminal instead, the power is negative and the source absorbs energy. A battery that absorbs energy is being charged.

Type 3 into the V2 box. V_A drops to 4 V, and I3 reads −1 mA, "from A to V2": it flows against its arrow, into V2's + terminal. Power V2 shows −3 mW, "absorbing (being charged)". The right loop reads 3 V + 1 V - 4 V: the 1 V across R3 is a rise now, because the loop walks through R3 against the real current.

This is how a car charges its battery. With the engine running, the alternator holds the wiring at about 14 V. The battery on its own sits at about 12.6 V (typical figures for a 12 V car battery), so the alternator pushes current into the battery's + terminal.

Predict before you try it

Somewhere between 3 V and 6 V there is a value of V2 at which I3 is zero, so V2 neither delivers nor absorbs. With V1 at 9 V and all three resistors at 1 kΩ, work out that value before you try it.

If no current flows through R3, it has no voltage across it, so V_A equals V2. With R3 carrying nothing, V1, R1 and R2 are a plain voltage divider from lesson 03: V_A = 9 V × 1 kΩ / 2 kΩ = 4.5 V. So V2 must be 4.5 V. Type 4.5 into the V2 box: I3 reads 0 A, "no current". Above 4.5 V, V2 helps to feed R2. Below it, V1 charges V2.

You have used these laws already

The series rule in lesson 02 is KCL: a node with only two parts on it has one current in and the same current out. The divider in lesson 03 used both laws: KCL gave one current through R₁ and R₂, and KVL gave V_in = I × R₁ + I × R₂.