Two simple rules that let you analyse any DC circuit, no matter how complicated. Kirchhoff's Current Law (KCL) deals with junctions; Kirchhoff's Voltage Law (KVL) deals with loops.
Branch currentsJunctions & loopsConservation laws
2Laws
6Sections
📐The two laws, in plain English
Kirchhoff's Current Law (KCL): at any junction (node), the total current flowing in equals the total current flowing out.
Kirchhoff's Voltage Law (KVL): around any closed loop, the sum of the EMFs equals the sum of the voltage drops.
🧮Quick check: junction or loop?Interactive
Choose a scenario above, then press Check to see which law applies.
📗Key vocabulary
Node / junctionA point where three or more wires (branches) meet.
BranchA single path between two nodes, containing one or more components.
Loop / meshAny closed path you can trace around a circuit, returning to your start.
Conventional currentCurrent is treated as flowing from + to − around the external circuit.
A node is where wires meet, a branch is the path directly between two nodes, and a loop is any closed path you trace around.
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Why two laws?
Every DC circuit obeys two conservation laws at once
KCLConservation of charge
Charge can't pile up or vanish at a junction, so whatever current arrives must leave. KCL is what lets you write equations at nodes.
KVLConservation of energy
Every joule of energy the source supplies is used up by the components in the loop, so EMFs and voltage drops must balance. KVL is what lets you write equations around loops.
💡Why this matters
Every method you've used so far — Thévenin, Norton, Mesh Analysis, Superposition — is really just Kirchhoff's Laws applied in a clever shortcut. Understanding KCL and KVL properly makes every other theorem make sense.
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Kirchhoff's Current Law (KCL)
The sum of currents into a junction equals the sum of currents out
📐The rule
At any node, the algebraic sum of all currents is zero — currents flowing in are positive, currents flowing out are negative (or vice versa, as long as you're consistent).
Σ I(in) = Σ I(out)
📗Key vocabulary
Branch currentThe current in one specific branch of a circuit, given its own label (I₁, I₂, I₃…).
Sign conventionChoose "into the node" as positive before you start, and stay consistent throughout.
✅Worked example — explicit substitution
GivenI₁ = 2 A flows into node A, I₂ = 1 A flows into node A, I₃ flows out.
KCL statesI₁ + I₂ = I₃
SubstituteI₃ = 2 + 1
AnswerI₃ = 3 A
🧮Junction current calculatorInteractive
Enter the two currents flowing into the junction to find the current flowing out.
💡Common mistake
Mixing up direction. Always mark every arrow on your diagram before writing the KCL equation — if a current turns out negative once solved, it's simply flowing the opposite way to your assumed arrow, and that's fine.
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Kirchhoff's Voltage Law (KVL)
Around any closed loop, EMFs equal voltage drops
📐The rule
Starting anywhere on a loop and walking all the way round back to your start, the EMFs you pass through must add up to the same total as the voltage drops across the resistors.
Σ E = Σ I·R
📗Key vocabulary
EMFElectromotive force — the voltage a source (like a cell) supplies to the loop.
Voltage dropThe voltage used up across a component, found from I × R for a resistor.
Walking the loopTracing a single closed path once, adding EMFs and drops as you meet them.
🚰Voltage symbols
A simple relationship can be made between a tank of water and a voltage supply. The higher the water tank is above the outlet, the greater the pressure of the water, as more energy is released. The higher the voltage, the greater the potential energy, as more electrons are released.
⛰️Real-world analogy: water on a mountain
Picture a mountain stream. The water sitting at the top of the mountain has height — that height is stored energy, ready to push the water downhill. That's your EMF, E: the "height" a source lifts electrons to before they set off round the circuit.
As the water flows downhill, it crashes over boulders in the streambed. Every boulder it fights past uses up some of that height. Those boulders are your resistors: every one the current pushes through uses up some of the voltage, exactly like R₁ and R₂ in the loop.
By the time the stream reaches the bottom of the mountain, it has no height left — every last bit was spent getting past the boulders on the way down. That's Kirchhoff's Voltage Law: all of the "height" supplied at the top (E) has been used up by the boulders it passed (I·R₁ + I·R₂), with nothing left over. Watch the gauge on the left drain to zero exactly as the drop reaches the bottom.
✅Worked example — walking the loop
GivenE = 12 V, I = 3 A, R₁ = 3 Ω, R₂ = 1 Ω
KVL statesE = I·R₁ + I·R₂
Substitute12 = (3 × 3) + (3 × 1) = 9 + 3
Check12 = 12 ✓ loop balances
🧮Loop equation checkerInteractive
Enter the EMF, the loop current, and the resistors it passes through to see whether the loop balances.
💡Common mistake
Forgetting to keep the same walking direction all the way around. Once you decide to go clockwise, stay clockwise for the whole loop — don't switch direction halfway through.
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Combining KCL and KVL
Three worked examples: two sources sharing a branch, then three
📐The strategy — the standard procedure for applying Kirchhoff's Laws
Note down the known voltages and resistances (label any unknowns V₁, V₂, R₁, R₂, etc.)
Assign a current to each branch, choosing a direction (clockwise or anticlockwise) for each.
Label every branch current clearly on the diagram (I₁, I₂, I₃, etc.)
Write a KCL equation for each node.
Write a KVL equation for each independent loop.
Solve the resulting simultaneous equations for the unknown currents.
If a current comes out negative once solved, that simply means the true current flows opposite to the direction you originally chose — the arrow was drawn the wrong way round, but the working is still valid.
🧮Circuit solver — change the sourcesInteractive
Resistors stay fixed at R₁ = 3 Ω, R₂ = 4 Ω, R₃ = 2 Ω (Example 1's circuit). Change E₁ and E₂ to see how the branch currents change.
Enter E₁ and E₂, then press Solve to calculate I₁, I₂ and I₃.
📗Key vocabulary
Simultaneous equationsTwo or more equations, solved together, usually by substitution or elimination.
Shared branchA branch that belongs to more than one loop — it carries the combined current from the other branches.
Worked Example 1 — two sources, one shared branch
Step 0 · The circuit
Two sources, one shared branch
E₁ = 12 V drives branch 1 (R₁ = 3 Ω), E₂ = 10 V drives branch 2 (R₂ = 4 Ω), and both branches meet at Node A before flowing on together through the shared branch R₃ = 2 Ω. This circuit has 3 branches, 2 nodes and 2 independent loops.
What we want to find:The current through the shared branch R₃
Step 1 · KCL at Node A
Currents in must equal current out
Both I₁ and I₂ flow into Node A, and only I₃ flows out through the shared branch. Kirchhoff's Current Law says these must balance.
At Node A:I₁ + I₂ = I₃
At the return node:I₃ = I₁ + I₂
Step 2 · Loop 1 KVL
Walk around Loop 1
Going round Loop 1: E₁ drives the current in, R₁ drops I₁×R₁, and the shared R₃ drops I₃×R₃ — because I₃ also flows through this loop's return path.
Loop 1 equation:12 = 3·I₁ + 2·I₃
Step 3 · Loop 2 KVL, then solve
Substitute I₃ and solve the simultaneous equations
Loop 2 gives a second equation the same way. Substituting I₃ = I₁ + I₂ into both turns them into two equations in I₁ and I₂ — solvable by elimination.
Loop 1
12 = 3I₁ + 2I₃ → 12 = 5I₁ + 2I₂
Loop 2
10 = 4I₂ + 2I₃ → 10 = 2I₁ + 6I₂
Eq1 × 3:15I₁ + 6I₂ = 36
Subtract Eq2:13I₁ = 26 → I₁ = 2 A
Substitute back:2I₂ = 12 − 10 → I₂ = 1 A
Step 4 · The shared branch
Watch I₁ and I₂ combine into I₃
Back-substituting into the KCL equation gives the current through the shared branch — the two contributions simply add together.
I₃ = I₁ + I₂ = 2 + 1= 3 A
Step 1 of 5
Worked Example 2 — a second 2-loop circuit
Step 0 · The circuit
A different pair of sources, same method
E₁ = 21 V drives branch 1 (R₁ = 3 Ω), E₂ = 19 V drives branch 2 (R₂ = 4 Ω), and both branches meet at Node B before flowing on together through the shared branch R₃ = 5 Ω. This circuit has 3 branches, 2 nodes and 2 independent loops.
What we want to find:The current through the shared branch R₃
Step 1 · KCL at Node B
Currents in must equal current out
Both I₁ and I₂ flow into Node B, and only I₃ flows out through the shared branch.
At Node B:I₁ + I₂ = I₃
At the return node:I₃ = I₁ + I₂
Step 2 · Loop 1 KVL
Walk around Loop 1
Going round Loop 1: E₁ drives the current in, R₁ drops I₁×R₁, and the shared R₃ drops I₃×R₃ — because I₃ also flows through this loop's return path.
Loop 1 equation:21 = 3·I₁ + 5·I₃
Step 3 · Loop 2 KVL, then solve
Substitute I₃ and solve the simultaneous equations
Loop 2 gives a second equation the same way. Substituting I₃ = I₁ + I₂ into both turns them into two equations in I₁ and I₂ — solvable by elimination.
Loop 1
21 = 3I₁ + 5I₃ → 21 = 8I₁ + 5I₂
Loop 2
19 = 4I₂ + 5I₃ → 19 = 5I₁ + 9I₂
Eq1 × 9:72I₁ + 45I₂ = 189
Eq2 × 5:25I₁ + 45I₂ = 95
Subtract:47I₁ = 94 → I₁ = 2 A
Substitute back:5I₂ = 5 → I₂ = 1 A
Step 4 · The shared branch
Watch I₁ and I₂ combine into I₃
Back-substituting into the KCL equation gives the current through the shared branch — the two contributions simply add together.
I₃ = I₁ + I₂ = 2 + 1= 3 A
Step 1 of 5
Worked Example 3 — three sources, one shared branch (3-loop circuit)
Step 0 · The circuit
Three sources sharing one return branch
E₁, E₂ and E₃ each drive their own branch (R₁, R₂, R₃), and all three meet at Node A before flowing on together through the shared branch R₄. This circuit has 4 branches, 2 nodes and 3 independent loops.
What we want to find:I₁, I₂, I₃ and the shared current I₄
Step 1 · KCL at Node A
Three currents in, one current out
I₁, I₂ and I₃ all flow into Node A, and only I₄ flows out through the shared branch.
KCL equation:I₁ + I₂ + I₃ = I₄
Step 2 · Loop 1 KVL
Walk around Loop 1 (E₁ and the shared branch)
The dashed outline traces Loop 1: out along the top rail, down through R₄, back along the bottom rail, and up through E₁ and R₁.
Loop 1 equation:14 = 2·I₁ + 1·I₄
Step 3 · Loop 2 KVL
Walk around Loop 2 (E₂ and the shared branch)
The same idea, this time with E₂'s branch — the highlighted region is a bit smaller since branch 2 sits closer to the shared branch.
Loop 2 equation:12 = 1·I₂ + 1·I₄
Step 4 · Loop 3 KVL, then solve
Substitute I₄ and solve three equations at once
Loop 3 gives a third equation the same way. Substituting I₄ = I₁ + I₂ + I₃ into all three turns them into three equations in I₁, I₂ and I₃ — solvable by elimination, one variable at a time.
Loop 1
14 = 2I₁ + I₄ → 14 = 3I₁ + I₂ + I₃
Loop 2
12 = I₂ + I₄ → 12 = I₁ + 2I₂ + I₃
Loop 3
13 = 5I₃ + I₄ → 13 = I₁ + I₂ + 6I₃
Eq1 − Eq2:2I₁ − I₂ = 2
Eq2 − Eq3:I₂ − 5I₃ = −1
Combine & solve:I₃ = 1 A, I₂ = 4 A, I₁ = 3 A
Step 5 · The shared branch
Watch I₁, I₂ and I₃ combine into I₄
Back-substituting into the KCL equation gives the current through the shared branch — all three contributions simply add together.
I₄ = I₁ + I₂ + I₃ = 3 + 4 + 1= 8 A
Step 1 of 6
💡Cross-check
Whether a circuit has two sources or three, the branch-current method never changes: label every branch, use KCL once at the shared node to cut down the unknowns, then use KVL around each loop. Example 3 just means one more equation to solve alongside the others.
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Practice questions
Type your answer and check it against the correct value
📐Before you start
Reminder: KCL — currents in = currents out at a node. KVL — EMFs = voltage drops around a loop.
🧮Scratchpad calculatorInteractive
Use this to check your working as you go.
📗Remember
KCLΣ I(in) = Σ I(out) at a node.
KVLΣ E = Σ I·R around a loop.
Question 1 · KCLBasic
At a junction, I₁ = 5 A and I₂ = 3 A flow in. What current I₃ flows out?
Question 2 · KCLBasic
At a junction, I₁ = 7 A flows in. I₂ = 4 A flows out on one branch and I₃ flows out on another. Find I₃.
Question 3 · KVLBasic
A loop has EMF E = 15 V. One resistor drops 6 V. What does the remaining resistor drop?
Question 4 · CombinedExtended
A two-source circuit has E₁ = 11 V, R₁ = 2 Ω, E₂ = 11 V, R₂ = 3 Ω, and a shared R₃ = 1 Ω. Find I₁, I₂ and I₃.
Drag the sliders to set the two currents flowing into a junction. Watch I₃ update live, then answer below.
Current out, I₃ = 7 A
Question 6 · KVL, interactiveExplore
A loop has EMF E = 18 V and a fixed resistor R₁ = 2 Ω carrying current I. Drag the current slider until the second resistor's drop matches the value shown, then read off I.
Drop across R₁ = I × 2 = 6 V
Remaining drop needed across R₂ = 18 − 6 = 12 V
If R₂ = 4 Ω, which value of I on the slider makes the loop balance exactly?
💡Tip
For Question 4, find I₁ in terms of I₂ from one loop equation, then substitute into the other — exactly the method used in the worked example. For Question 6, the loop balances when the drop across R₂ (I × 4) equals the "remaining drop needed" shown above.
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Knowledge check
Six quick questions to test your understanding
Kirchhoff's Laws — quick fire
Score: 0/6Question 1/6
📜 The story behind Kirchhoff's Laws
Every circuit theorem you've used in this suite — Thévenin, Norton, Mesh Analysis, Superposition — is really just Kirchhoff's two laws applied in a clever shortcut. Here's where they came from, why they work, and where they're used today.
🕰️In short: Gustav Kirchhoff published both laws in a single 1845 paper, while still a 21-year-old student, as an extension of the conservation principles already known for charge and energy. Nearly 180 years later, they remain the foundation every other circuit-analysis technique is built on.
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Where it came from
⚡ Built on the shoulders of conservation laws
1845
Gustav Kirchhoff's circuit lawsA 21-year-old Kirchhoff published both the Current Law and the Voltage Law in the same paper — extending the already-known principles of conservation of charge and conservation of energy to electrical networks.
1847
Applied to real networksKirchhoff went on to use his own laws to analyse the resistance and current distribution in networks of conductors — some of the earliest systematic circuit analysis on record.
Late 1800s
Maxwell's loop currentsJames Clerk Maxwell popularised describing circuits using circulating "loop currents" rather than tracking every branch current separately — the idea behind Mesh Analysis, itself just a shortcut built on Kirchhoff's Voltage Law.
Today
Still the foundationEvery modern circuit-simulation package (SPICE and its descendants) solves a circuit by writing out KCL and KVL equations for every node and loop, then solving them automatically.
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The logic — why it works
🔁 Two conservation laws, nothing more
Kirchhoff's Current Law is really just a restatement of conservation of charge: charge can't be created or destroyed, and it can't pile up at a junction, so whatever current arrives must leave. That's the entire justification for Σ I(in) = Σ I(out) — no circuit theory required beyond that.
Kirchhoff's Voltage Law is a restatement of conservation of energy. Every joule of energy a source supplies to a loop is used up by the components in that loop — none of it can vanish, and none can appear from nowhere. That's why the EMFs and voltage drops around any closed loop must sum to the same total.
Because both laws are just conservation principles, they hold true for any circuit whatsoever — simple or enormously complex, linear or not. That's exactly why every other technique in this suite (Thévenin, Norton, Mesh Analysis, Superposition) can be derived from them: they're not separate laws, just efficient shortcuts built on the same two foundations.
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Where it's actually used
🔧 Real applications
🖥️Circuit simulation softwareSPICE-style simulators build and solve KCL/KVL equations automatically for every circuit you draw, however complex.
⚡Power network analysisWorking out current flow and losses through interconnected loops in local and regional power distribution grids.
📱PCB and chip designVerifying that current and voltage budgets balance correctly across every trace and node on a circuit board or integrated chip.
🔋Battery and charging systemsModelling how current splits across multiple cells or charging branches sharing a common return path.
🎓The foundation for every other theoremThévenin, Norton, Mesh Analysis, and Superposition are all systematic shortcuts for solving the KCL/KVL equations faster — not alternatives to them.
📗Key words on this page
Node / junction
A point where three or more wires (branches) meet — where KCL applies.
Loop / mesh
Any closed path you can trace around a circuit, returning to your start — where KVL applies.
Branch current
The current flowing in one specific branch of a circuit, labelled I₁, I₂, I₃…
EMF
Electromotive force — the voltage a source supplies to a loop.
Conservation law
A principle stating a quantity (like charge or energy) can't be created or destroyed, only moved or transformed.
👁️ Visual aid: nearly 180 years separate Kirchhoff's original paper from today's circuit simulators — but the same two equations sit underneath every point on this line.
🔍Try it yourself: trace the family treeInteractive
Choose a theorem above, then press Reveal to see which Kirchhoff law it's really built on.
💡Worth remembering: because KCL and KVL are conservation laws rather than circuit-specific tricks, they never fail — even for circuits so complex that no shortcut theorem applies cleanly. When in doubt, or when a shortcut method feels shaky, falling back to first-principles KCL and KVL will always get you the right answer.
