Kirchhoff's First Law

Having defined charge and current, we now need our first rule for analysing circuits. Kirchhoff's first law is the statement that electric charge is conserved — it cannot be created or destroyed — applied to the specific context of currents at a circuit junction.

It is one of the two Kirchhoff laws required by OCR H556 (Module 4.3.1, circuits), and the simpler of the two. Together with Kirchhoff's second law (Lesson 11) it allows you to analyse any DC circuit, no matter how complicated.

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Statement of the Law

Kirchhoff's First Law: The sum of the currents entering any junction in a circuit is equal to the sum of the currents leaving that junction.

Equivalently, the algebraic sum of all currents at a node is zero:

ΣI_in = ΣI_out or ΣI = 0

A "junction" (also called a "node") is simply any point in the circuit where three or more wires meet.

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Why It Must Be True — Charge Conservation

Kirchhoff's first law is a direct consequence of the conservation of electric charge. If more current flowed into a point than out of it, charge would pile up at that point. This would build an enormous electric field that would immediately repel further incoming charge — in practice it does not happen. In a steady-state DC circuit, the rate of charge entering a junction must exactly equal the rate of charge leaving.

flowchart LR
  I1[I1 = 3 A] --> J((Junction))
  I2[I2 = 2 A] --> J
  J --> I3[I3 = ? A]

In the diagram above, by Kirchhoff's first law:

• Current in = 3 + 2 = 5 A
• Current out = I₃
• Therefore I₃ = 5 A

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Sign Conventions

When writing the law as "ΣI = 0", you have to adopt a sign convention. The usual choice is:

• Currents entering the junction are positive.
• Currents leaving the junction are negative.

(Either convention is valid, so long as you are consistent throughout a problem.)

For example, if three wires meet at a junction carrying currents +4 A, −1 A and −3 A (with the signs indicating direction), then +4 − 1 − 3 = 0. Check.

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Worked Example 1 — Four-Wire Junction

Four wires meet at a point X. Currents of 2.0 A and 1.5 A flow into X. A current of 0.8 A flows out of X along a third wire. What is the magnitude and direction of the current in the fourth wire?

• ΣI_in = 2.0 + 1.5 = 3.5 A
• ΣI_out (known) = 0.8 A
• Unknown current out = 3.5 − 0.8 = 2.7 A flowing out of X

If the answer had come out negative, we would conclude that the wire actually carries current into X instead of out.

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Worked Example 2 — Parallel Resistors

Consider three resistors connected in parallel between points A and B:

• R₁ carries a current of 0.40 A
• R₂ carries a current of 0.25 A
• R₃ carries a current of 0.15 A

What current flows from the battery into A?

Because A is a junction where the battery wire splits into three parallel branches, Kirchhoff's first law gives:

• I_total = I₁ + I₂ + I₃ = 0.40 + 0.25 + 0.15 = 0.80 A

The same 0.80 A must flow out of B back into the battery, because the reverse junction at B recombines the three branches.

flowchart LR
  B[Battery] -->|0.80 A| A((A))
  A -->|0.40 A| R1[R1]
  A -->|0.25 A| R2[R2]
  A -->|0.15 A| R3[R3]
  R1 --> BN((B))
  R2 --> BN
  R3 --> BN
  BN -->|0.80 A| B