Difficulty: Intermediate Prerequisites: Ohm's Law, series and parallel resistor combinations, basic circuit vocabulary (nodes, loops, branches).
Kirchhoff's rules let you solve circuits that cannot be reduced to simple series/parallel combinations. Most real circuits have multiple loops and multiple sources, which means you need a systematic way to write enough equations to find every unknown current and voltage. This topic sits at the heart of DC circuit analysis and underpins everything from power distribution to electronics design. You should already be comfortable applying Ohm's Law to single-loop circuits before tackling this material.
Kirchhoff's Junction Rule says currents in equal currents out at any node. Kirchhoff's Loop Rule says the sum of all voltage gains and drops around any closed loop is zero. Together, these two rules give you a system of linear equations you can solve for every unknown current in the circuit.
Kirchhoff's Junction Rule (KJR, Current Rule, KCL)
The algebraic sum of all currents entering a junction equals the sum of all currents leaving it. In simple terms, charge cannot pile up at a node, so whatever flows in must flow out.
Kirchhoff's Loop Rule (KVR, Voltage Rule, KVL)
The algebraic sum of all potential differences around any closed loop in a circuit is zero. Think of it as a round trip in elevation: if you walk in a complete loop, you end up at the same height you started.
Node (junction)
A point where three or more conductors meet. This is where you apply the junction rule.
Loop
Any closed path through a circuit. You choose a traversal direction (clockwise or anticlockwise) and sum voltage changes as you go.
Branch
A single path between two nodes carrying its own current. Each branch has one unknown current.
Sign convention for EMF traversal
When you traverse a battery from its negative terminal to its positive terminal, the voltage change is +V. Traversing from positive to negative gives -V.
Sign convention for resistor traversal
When you traverse a resistor in the direction of assumed current flow, the voltage change is -IR. Traversing against the assumed current gives +IR.
Identify all nodes, branches, and independent loops in the circuit.
Assign a current variable (I₁, I₂, I₃, ...) to each branch and draw an arrow for the assumed positive direction. If your answer comes out negative, the actual current flows opposite to your arrow.
Count unknowns. You need that many independent equations.
At each node, write: ΣI_in = ΣI_out.
For N nodes, you get N - 1 independent junction equations (the last node's equation is redundant).
Example from the source problem: at the top-left node, I₂ = I₁ + I₃.
Choose a traversal direction for each loop.
Walk around the loop. For each element, add the appropriate voltage change using the sign conventions above.
Set the total sum to zero.
Example, left loop of the source circuit: I₁R₁ - V₁ - I₃R₃ = 0.
Example, outer loop: I₂R₂ + I₃R₃ + V₁ + I₂R₆ - V₂ = 0.
Substitute the junction equation into the loop equations to reduce the number of unknowns.
Solve algebraically (or use matrix methods for larger circuits).
Once you have all currents, find any voltage drop with V = IR.
Given values:
V₁ = 18 V, V₂ = 12 V
R₁ = R₅ = 61 Ω, R₂ = R₆ = 141 Ω, R₃ = 50 Ω, R₄ = 135 Ω
R₄ and R₅ are in series, fed by V₂ alone in their branch. The current through that branch is I₄ = V₂ / (R₄ + R₅).
V₄ = I₄ · R₄ = V₂ · R₄ / (R₄ + R₅)
V₄ = (12 V)(135 Ω) / (135 + 61) Ω = 8.26 V
Using the junction and two loop equations, solve for I₃:
I₃ = R₁ / [R₁R₃ + (R₁ + R₃)(R₂ + R₆)] × [V₂ - V₁(1 + (R₂ + R₆)/R₁)]
I₃ = (61) / [(61)(50) + (111)(282)] × [12 - 18(1 + 282/61)]
I₃ = -0.158 A
The negative sign means the actual current flows opposite to the assumed arrow direction.
I₂ = V₁/R₁ + I₃(R₁ + R₃)/R₁
I₂ = 18/61 + (-0.158)(111)/61 = 0.007 A (approximately 7 mA)
From the junction rule: I₁ = I₂ - I₃
I₁ = 0.007 - (-0.158) = 0.165 A
V(a) - V(b) = I₂ · R₆ = (0.0068 A)(141 Ω) = 0.959 V
Formula | Use |
|---|---|
ΣI_in = ΣI_out | Junction rule at any node |
ΣΔV = 0 (around a loop) | Loop rule for any closed path |
V = IR | Voltage drop across a resistor |
V_R = V_source · R / (R₁ + R₂) | Voltage divider (series resistors sharing one source) |
Kirchhoff's rules are how electrical engineers analyse power grids, where dozens of sources and loads share a complex network. They are also the foundation for circuit simulation software like SPICE, which uses the same node and loop equations internally to predict voltages and currents in everything from smartphone circuits to satellite power systems.
"A negative current means I did something wrong." It does not. A negative result simply means the actual current flows opposite to the direction you assumed. Your maths is fine.
"I need to pick the 'correct' current direction before I start." You do not. Pick any direction. The algebra corrects for a wrong guess via a negative sign.
"Every node gives me a new independent equation." Not quite. For N nodes, only N - 1 junction equations are independent. The last one is always a combination of the others.
"The loop rule only works if I go clockwise." The traversal direction is arbitrary. Going anticlockwise simply flips all the signs, and the resulting equation is equivalent.
⚠️ Sign errors are the number-one source of lost marks. Be methodical: write the sign convention for batteries and resistors at the top of your working before you start.
⚠️ Exams often give a circuit that looks complicated but has one branch you can solve by inspection (like the R₄/R₅ voltage divider above). Spot these first to save time.
⚠️ If a question says "a positive value is defined in the direction of the arrow," a negative answer is valid and expected when current flows the other way.
⚠️ Double-check your junction equations by substituting your final currents back in. If ΣI_in ≠ ΣI_out at any node, you have an error.
True or false: Kirchhoff's loop rule is a consequence of conservation of charge. False. The loop rule comes from conservation of energy. The junction rule comes from conservation of charge.
Fill in the blank: For a circuit with N nodes, the maximum number of independent junction equations is ___. N - 1.
True or false: If you traverse a resistor against the direction of current, the voltage change is negative. False. Traversing against the current gives a positive voltage change (+IR).
Fill in the blank: When you walk through a battery from - to + during a loop traversal, the voltage change is ___. +V (positive EMF).
Q: In a two-loop circuit, you write a junction equation I₂ = I₁ + I₃ and obtain I₃ = -0.2 A. What does the negative sign tell you?
A: The actual current through R₃ flows in the direction opposite to the arrow you assumed when setting up the problem. The magnitude is 0.2 A.
Q: A branch contains R₄ = 135 Ω and R₅ = 61 Ω in series, powered by a 12 V source. What is the voltage across R₄?
A: V₄ = 12 × 135 / (135 + 61) = 8.27 V (voltage divider).
Q: Why can you not write an independent junction equation at every node in a circuit?
A: Because the last node's equation is always a linear combination of the others (by conservation of charge, if all other nodes balance, the last one must too). So N nodes yield only N - 1 independent equations.
Q: You traverse a 18 V battery from + to - in your chosen loop direction. What voltage change do you write?
A: -18 V.
This material connects directly to series and parallel resistor reduction (Chapter on DC circuits): for branches that do not share nodes with other loops, you can simplify first and reduce the number of unknowns before applying Kirchhoff's rules. It also leads into mesh analysis and nodal analysis, which are formalised versions of Kirchhoff's rules used in more advanced circuit courses. The same conservation principles reappear in AC circuit analysis with impedances replacing resistances.
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