Kirchhoff's Current and Voltage Laws

Ohm's law tells you what one resistor does when you know the voltage across it. That is enough for a torch. It is not enough for a radio, where a supply rail feeds a dozen branches at once, two of which push current back at each other. The syllabus names Kirchhoff's two laws explicitly, and they are the tool that turns a tangle of wires into arithmetic you can actually do. They are also, once you see them, almost insultingly simple: nothing accumulates at a junction, and nothing is created going round a loop.

Why one equation is not enough

Put three resistors in a single loop and there is only one current, so Ohm's law and the series formula finish the job. Now split the circuit so that current has a choice of paths, or add a second battery somewhere in the middle, and you have several unknown currents at once. One equation cannot solve for three unknowns. You need as many independent equations as you have unknowns, and Kirchhoff's laws are the machine for generating them — one equation per junction, one per loop.

Two definitions first, because the exam uses the words. A node (or junction) is any point where three or more wires meet. A loop (or mesh) is any path you can trace through the circuit that returns to where it started.

Kirchhoff's current law

At any junction in a circuit, the sum of the currents flowing in equals the sum of the currents flowing out. Equivalently, and more usefully once you are doing algebra: the algebraic sum of the currents at a node is zero, counting inward currents as positive and outward ones as negative.

The reason is conservation of charge. A junction is a point of wire. It has no capacity to store charge and no factory to make it. Whatever arrives in a given second must leave in the same second, or charge would be piling up at a point in a piece of copper — which does not happen. That is the whole argument, and it holds however many wires meet, whatever components are in them, and whether the currents are DC or AC.

Three wires meet, one brings in 3 A and another brings in 2 A: the third must carry away 5 A. There is no other possibility, and you never need to know a single resistance to say so.

Kirchhoff's voltage law

Around any closed loop, the algebraic sum of the EMFs and the potential drops is zero. Stated the way the ASOC paper states it: the sum of all the voltage drops in a closed circuit is equal to the applied voltage.

The reason is conservation of energy. Voltage is energy per coulomb. Walk one coulomb of charge all the way round a loop and you arrive back at the point you started from — the same point, so at the same potential. Every joule the source gave that coulomb must therefore have been handed over somewhere along the way, to the resistors, to the lamp, to the transistor. You cannot come back up the stairs having gained height.

Two immediate consequences the examiner likes:

• In a series circuit the drops must add up to the supply. Three resistors across 12 V dropping 5 V and 4 V leave the third dropping 12 − 5 − 4 = 3 V. You do not need the resistor values.
• The largest drop appears across the largest resistance. The same current flows through every part of a series circuit, so by V = IR the drop is proportional to the resistance. Where a resistor sits in the loop — nearest the positive terminal or furthest — makes no difference at all.

Signs: how to actually apply the laws

This is where beginners stall, so here is a procedure that always works. It has one rule you must believe: guess the direction of every unknown current, and do not worry about guessing wrong. If you guess backwards, the algebra returns a negative number, which is the arithmetic telling you politely that the current runs the other way. The magnitude will still be right.

1. Mark every branch with an arrow showing your assumed current direction, and give each one a name: I₁, I₂, I₃.
2. At each node, write the current-law equation: in = out.
3. Choose a direction to walk each loop — clockwise, always, so you never have to remember which way you went.
4. Walking the loop, add up the voltages with these signs:
   • Through a source from − to +, the potential rises: write +E. From + to −, it falls: write −E.
   • Through a resistor in the same direction as its assumed current, the potential falls: write −IR. Against the assumed current, it rises: +IR.
5. Set the total to zero. Solve.

Currents run downhill through resistors and uphill through sources: that one sentence carries the whole sign convention.

The potential divider

The most useful thing KVL gives you is a formula you will use for the rest of your life in electronics. Put two resistors in series across a supply and take the output from the junction between them.

The derivation is three lines. The two resistors form one loop, so by KVL the supply is shared between them, and by the series rule the current is the same in both: I = V in ÷ (R1 + R2). The output is the drop across R2, which is I × R2. Substitute:

In words: the supply divides in the ratio of the resistances.

Worked example 1 — the simplest divider

A 10 Ω and a 5 Ω resistor are in series across 15 V. What is the voltage across the 10 Ω resistor?

Total resistance 10 + 5 = 15 Ω, so I = 15 ÷ 15 = 1 A everywhere in the loop, and the drop across the 10 Ω is 1 × 10 = 10 V. By the formula directly: 15 × 10 ÷ 15 = 10 V. The remaining 5 V is across the 5 Ω, and 10 + 5 = 15 V as KVL requires.

Worked example 2 — a bias divider

A 12 V rail feeds R1 = 47 kΩ and R2 = 10 kΩ in series. What appears at the junction?

V out = 12 × 10 000 ÷ (47 000 + 10 000) = 12 × 10 ÷ 57 = 12 × 0.1754 = 2.11 V

Note that only the ratio matters. Changing both resistors to 470 kΩ and 100 kΩ gives the same 2.11 V while drawing a tenth of the current — but a divider that draws too little current is easily upset by whatever you connect to its output, because the load is then a third resistor in parallel with R2. The working rule is to let the divider pass roughly ten times the current the load will take.

The current divider

The current law gives the mirror image. Two resistors in parallel are connected to the same two nodes, so they see the same voltage; the current divides between them in inverse proportion to their resistances.

Note which resistance is on top: the other one. That is not a misprint. The easier path takes more current, so the branch with the smaller resistance gets the larger share. Which settles a favourite question: if branch A carries twice the current of branch B, then A must have half the resistance of B. It cannot be a voltage difference, because parallel branches always have the same voltage across them.

A two-loop network, solved twice

Worked example 3 — two sources, one load

A 9 V battery with a 10 Ω resistor in series, and a 6 V battery with a 10 Ω resistor in series, both feed a common 10 Ω load. Find all three currents. This is the arrangement above, and it is the situation whenever two supplies are paralleled onto one rail — a mains PSU and a backup battery, say.

Step 1 — the current law at node A. Both source branches bring current in, the load branch carries it away:

I₁ + I₂ = I₃

Step 2 — the voltage law round loop 1 (the 9 V source, R1, and the load R3):

9 = 10 I₁ + 10 I₃

Step 3 — the voltage law round loop 2 (the 6 V source, R2, and the same load):

6 = 10 I₂ + 10 I₃

Step 4 — solve. Subtract the third equation from the second: 3 = 10 I₁ − 10 I₂, so I₁ − I₂ = 0.3. From step 2, I₁ = 0.9 − I₃; from step 3, I₂ = 0.6 − I₃. Substituting both into step 1:

(0.9 − I₃) + (0.6 − I₃) = I₃

1.5 = 3 I₃, so I₃ = 0.5 A. Then I₁ = 0.9 − 0.5 = 0.4 A and I₂ = 0.6 − 0.5 = 0.1 A.

Step 5 — check both laws. Current law: 0.4 + 0.1 = 0.5 ✓. Voltage law round loop 1: (0.4 × 10) + (0.5 × 10) = 4 + 5 = 9 V ✓. Round loop 2: (0.1 × 10) + (0.5 × 10) = 1 + 5 = 6 V ✓. The load sits at 5 V above node B, which is why the 6 V branch contributes so little: it has only 1 V of headroom to push against, while the 9 V branch has 4 V.

Worked example 4 — one source, a branching load

A 12 V supply feeds R1 = 4 Ω in series with a parallel pair, R2 = 6 Ω and R3 = 12 Ω. Find every current and every voltage. Two loops again: supply-R1-R2, and supply-R1-R3.

1. Reduce the parallel pair. Product over sum: (6 × 12) ÷ (6 + 12) = 72 ÷ 18 = 4 Ω. Sanity check: 4 Ω is smaller than the smaller branch, 6 Ω. Good.
2. Total resistance. 4 + 4 = 8 Ω.
3. Supply current. I = 12 ÷ 8 = 1.5 A.
4. Divide the voltage. Across R1: 1.5 × 4 = 6 V. Across the pair: 1.5 × 4 = 6 V. KVL: 6 + 6 = 12 V ✓.
5. Divide the current. Both parallel branches have that same 6 V across them, so I₂ = 6 ÷ 6 = 1 A and I₃ = 6 ÷ 12 = 0.5 A. Or straight from the current divider: I₂ = 1.5 × 12 ÷ 18 = 1 A.
6. Check the current law at the branching node: 1 + 0.5 = 1.5 A ✓. And notice that the 6 Ω branch, having half the resistance of the 12 Ω branch, carries twice the current.