Kirchhoff's First Law

Spec mapping: OCR H556 Module 4.3 — DC circuits (Kirchhoff's first law: $\Sigma I_{\text{in}} = \Sigma I_{\text{out}}$ at any junction in a steady-state circuit; consequence of conservation of electric charge; application to series and parallel branches). Refer to the official OCR H556 specification document for exact wording.

We've defined the quantities of circuit analysis — charge, current, drift velocity. Now we need our first law. Kirchhoff's first law (often called the junction rule or current law) is the statement that electric charge is conserved at any node of a circuit. It is the simpler of the two Kirchhoff laws (the second arrives in lesson 11), and together they let you analyse any DC circuit, no matter how baroque.

This lesson states the law precisely, derives it from charge conservation, applies it to worked junction problems, and develops the sign-convention discipline you will need for multi-branch circuit analysis later. We also explore the subtle aspects: why the law breaks down momentarily at switch-on, why it still applies in AC, and why a non-zero $\Sigma I$ at a node would build a field large enough to be unsustainable.

Statement of the Law

Kirchhoff's first law: The sum of the currents flowing into any junction in a circuit is equal to the sum of the currents flowing out of that junction.

In symbols (with currents into the junction taken as positive and out as negative):

$\sum_k I_k = 0$

Or equivalently:

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

A junction (or node) is any point in a circuit where three or more wires meet. At a junction the conductor branches into multiple paths, and the law tells you how the incoming current divides between (or recombines from) the outgoing branches.

The law applies at every junction in the circuit. If a circuit has three junctions, you can write Kirchhoff-1 three times — once per node — though one of the resulting equations will be linearly dependent on the others (a consequence of overall charge conservation).

Why It Must Be True — Conservation of Charge

Kirchhoff's first law is a direct consequence of one of the deepest principles in all of physics: the conservation of electric charge. Charge is neither created nor destroyed in any known process. In a steady-state circuit, this means that the rate at which charge enters a node must exactly equal the rate at which it leaves.

If $\Sigma I_{\text{in}} > \Sigma I_{\text{out}}$ , then charge would accumulate at the junction at a rate $dQ/dt = \Sigma I_{\text{in}} - \Sigma I_{\text{out}}$ . The accumulating charge would build an electric field that would push subsequent charges back out, very rapidly — in microseconds or less — restoring the balance. So even if a transient imbalance appears (e.g. when a switch is first closed), it self-corrects on a timescale far below the A-Level measurement scale.

Formally: at any node of volume $V$ enclosed by surface $\partial V$ ,

$\frac{dQ_{\text{node}}}{dt} = \Sigma I_{\text{in}} - \Sigma I_{\text{out}}$

In the steady state $dQ/dt = 0$ and Kirchhoff-1 follows. This statement is exact — at the level of fundamental physics, charge conservation is one of the few absolutely exact conservation laws (it follows from gauge symmetry in quantum field theory).

flowchart LR
  I1["I₁ = 3.0 A"] --> J((Junction))
  I2["I₂ = 2.0 A"] --> J
  J --> I3["I₃ = ?"]

In the diagram above, by Kirchhoff-1: $I_3 = I_1 + I_2 = 3.0 + 2.0 = 5.0$ A. Three amperes plus two amperes flowing in must equal five amperes flowing out, because charge has nowhere to accumulate.

A Labelled Junction Diagram

Below is the canonical OCR-style junction-rule diagram with all four signed currents shown.

Check: $I_1 + I_2 = 4.0 + 1.5 = 5.5$ A in; $I_3 + I_4 = 3.5 + 2.0 = 5.5$ A out. Kirchhoff-1 satisfied.

Sign Conventions

When writing the law in the algebraic form $\Sigma I = 0$ , you must adopt and stick to a sign convention. The standard choice:

Currents entering the junction: positive.
Currents leaving the junction: negative.

(The opposite convention is equally valid — "leaving positive, entering negative" — provided you use it consistently throughout a single problem.)

For example, if four wires meet at a junction carrying signed currents $+4$ A, $-1$ A, $-1$ A, $-2$ A (positive meaning into, negative meaning out), then:

$\sum I_k = +4 - 1 - 1 - 2 = 0 \checkmark$

If your sum comes out non-zero, something has been mis-signed — either you've drawn a current in the wrong direction or mis-applied your convention. The arithmetic check is itself a valuable diagnostic.

Common slip: changing convention mid-problem. If you start with "into-positive" at one node, keep that convention at every node throughout the same circuit. Switching mid-solution gives sign errors that are very hard to track down.

Decision Tree — Applying Kirchhoff-1

flowchart TD
  A["Identify all nodes (junctions) in the circuit"] --> B["At each node, list every connected wire and label its current"]
  B --> C["Choose a sign convention (e.g. in = +, out = -)"]
  C --> D["Write ΣI = 0 at each node"]
  D --> E{"Unknowns > independent equations?"}
  E -->|Yes| F["Add Kirchhoff-2 loop equations (lesson 11) until determined"]
  E -->|No| G["Solve the linear system"]
  G --> H["Check signs: negative answer = current flows opposite to assumed arrow"]

This decision-tree is the workflow for any DC circuit problem at A-Level. Kirchhoff-1 alone may suffice for simple parallel-resistor problems; combined with Kirchhoff-2 (loop rule, lesson 11) it solves any DC circuit.

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. Find the magnitude and direction of the current in the fourth wire.

Total current in: $\Sigma I_{\text{in}} = 2.0 + 1.5 = 3.5$ A.
Total current out (known): $0.8$ A.
Unknown current out: $3.5 - 0.8 = 2.7$ A.

So the fourth wire carries 2.7 A out of X. If, on solving, the answer had come out negative, that would tell us the assumed direction was wrong — the wire actually carries current into X with magnitude given by the absolute value.

Worked Example 2 — Parallel Resistors

Three resistors are connected in parallel between points A and B, with currents $0.40$ A, $0.25$ A and $0.15$ A flowing through them. Find the current from the battery into A.

Point A is a junction: one wire from the battery enters, three wires (to the parallel branches) leave. By Kirchhoff-1:

$I_{\text{battery}} = I_1 + I_2 + I_3 = 0.40 + 0.25 + 0.15 = 0.80 \text{ A}$

The same $0.80$ A must flow back out of B into the battery — 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

This is the origin of the well-worn rule "in parallel, currents add". It's not a separate principle; it's Kirchhoff-1 applied to the junction where the parallel branches diverge.

Worked Example 3 — Cascaded Junctions

A battery drives $1.2$ A into a junction X. The current splits at X into two branches: a top branch and a bottom branch. The top branch carries $0.7$ A. Find the current in the bottom branch.

At X (Kirchhoff-1): $I_{\text{bottom}} = 1.2 - 0.7 = 0.5$ A.

Further along, the bottom branch itself reaches a second junction Y, where it divides into two more wires carrying $0.2$ A and $I_y$ . Find $I_y$ .

At Y (Kirchhoff-1): $I_y = 0.5 - 0.2 = 0.3$ A.

The cascade — applying Kirchhoff-1 at each node in turn — is the workhorse for multi-junction circuit analysis. Each application reduces the unknowns by one until everything is determined.

Worked Example 4 — Negative Answer Tells You the Direction

A three-wire junction has assumed currents $I_1 = +5$ A (into), $I_2 = ?$ A (assumed into), $I_3 = +2$ A (out). Find $I_2$ .

Applying Kirchhoff-1 with "in = positive": $+5 + I_2 - 2 = 0 \Rightarrow I_2 = -3$ A.

The negative answer means our assumed direction was wrong. $I_2$ actually flows out of the junction at $3$ A.

This is one of the most powerful features of the algebraic Kirchhoff-1: you can guess the directions of unknown currents at the start, work consistently, and let the signs of the answers correct any wrong guesses.

Steady State vs Transient — A Subtle Point

Kirchhoff-1 applies in the steady state. During very brief transients — say the first nanosecond after a switch is closed — there can be a tiny imbalance at junctions as charge builds up on the surfaces of conductors to set up the necessary electric field for steady-state current flow. This transient lasts nanoseconds at most (RC times for stray capacitance) and is invisible at A-Level. For all the DC problems in OCR H556 you can assume Kirchhoff-1 holds exactly.

The law also applies to AC circuits, provided you interpret the current as the instantaneous value at a moment in time. The sum of instantaneous currents entering any junction is always zero in AC just as in DC. This is why ammeters in AC analysis still apply Kirchhoff-1 — the law is fundamentally a statement about charge, not specifically about DC.

The only situation in which Kirchhoff-1 visibly fails is at the plates of a charging capacitor (Module 6.1), where charge genuinely does accumulate on one plate. Maxwell resolved this by introducing the "displacement current" — a beautiful conceptual extension that you'll meet at undergraduate level — which restores charge balance by counting the time-varying electric flux through any closed surface, including one cutting between capacitor plates.

Worked Example 5 — Diagnostic Use of Kirchhoff-1

A student draws a circuit diagram with three wires meeting at a node. Their measurements give: wire 1 carries $5.0$ A in, wire 2 carries $1.5$ A out, wire 3 carries $2.0$ A out. The student claims this is complete. Apply Kirchhoff-1 to check.

$\Sigma I_{\text{in}} - \Sigma I_{\text{out}} = 5.0 - (1.5 + 2.0) = 5.0 - 3.5 = 1.5$ A unaccounted.

Kirchhoff-1 says the algebraic sum must be zero, so the student's diagram is missing a wire. There must be a fourth wire carrying $1.5$ A out. This is the diagnostic use of the junction rule: if your sum doesn't balance, you've mis-counted the wires (or one is genuinely unmeasured), and the physics tells you exactly how much current the missing branch must carry.

Worked Example 6 — Network with Three Junctions

A network has a $12$ V battery driving current into a junction P. At P the current splits into two branches: branch A carries $I_A$ to junction Q; branch B carries $I_B$ to junction R. From Q, two wires emerge with currents $2.0$ A and $1.5$ A. From R, two wires emerge with currents $0.8$ A and $1.2$ A. All four "leaving" currents return to the battery via further nodes.

Apply Kirchhoff-1 at each junction to determine $I_A$ , $I_B$ , and the total battery current.

Kirchhoff's First Law

Kirchhoff's First Law

Statement of the Law

Why It Must Be True — Conservation of Charge

A Labelled Junction Diagram

Sign Conventions

Decision Tree — Applying Kirchhoff-1

Worked Example 1 — Four-Wire Junction

Worked Example 2 — Parallel Resistors

Worked Example 3 — Cascaded Junctions

Worked Example 4 — Negative Answer Tells You the Direction

Steady State vs Transient — A Subtle Point

Worked Example 5 — Diagnostic Use of Kirchhoff-1

Worked Example 6 — Network with Three Junctions

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