Kirchhoff's rules

Look at any circuit with more than one loop — a torch with two bulbs wired every which way, or the tangle of wires behind a school lab board — and Ohm's law alone goes silent. $V = IR$ tells you how a single resistor behaves, but it has nothing to say about which current flows through which branch when three or four wires meet at a single point and several cells push against each other around several loops at once. For over a century, though, physicists have had a complete answer, and it costs nothing more than two bookkeeping rules a school student can state in one breath each. In 1845, a 21-year-old Gustav Kirchhoff, still a student himself, wrote down exactly this bookkeeping. His junction rule says that charge cannot pile up or vanish at a wire junction — whatever current flows in must flow out. His loop rule says that if you walk all the way around any closed loop in a circuit and add up every rise and fall in potential you meet, you must land back exactly where you started — the net change is zero. Neither rule is a new law of electricity; each is simply conservation of charge and conservation of energy, written for the specific geometry of a circuit. What makes the rules powerful is that they work on any network, however tangled — including the four-resistor diamond called a Wheatstone bridge that no amount of series-parallel reduction can simplify, which is exactly why this chapter builds toward it next. This article sets up the two rules from first principles, works through the sign conventions that trip up almost every first attempt, and solves both a single-loop and a genuine two-loop circuit by hand — carrying forward the exact numbers used earlier in this chapter's cells-in-series-and-parallel discussion, so nothing here appears out of nowhere.

Key takeaways

Junction rule (Kirchhoff's current law). At any junction, the algebraic sum of currents is zero: $\sum I_{in} = \sum I_{out}$ . It is charge conservation applied to a point where wires meet.
Loop rule (Kirchhoff's voltage law). Around any closed loop, the algebraic sum of potential changes is zero: $\sum \Delta V = 0$ . It is energy conservation applied to a round trip that returns to the same potential.
Sign convention decides everything. Traversing a resistor along the assumed current direction gives a drop ( $-IR$ ); against it gives a rise ( $+IR$ ). Traversing a cell from $-$ to $+$ inside gives a rise ( $+\varepsilon$ ); the other way gives a fall ( $-\varepsilon$ ).
A wrong guess for current direction is not a wrong answer. Assign any direction to each unknown branch current; if the true current runs the other way, solving the equations simply returns a negative value — and that negative sign is itself useful information.
The junction rule gives one fewer independent equation than the number of junctions, and the loop rule needs as many independent loops as there are remaining unknowns — together they supply exactly enough equations to solve any resistor network.
This is the tool for circuits that cannot be reduced by series-parallel combination alone — most importantly the Wheatstone bridge, the next topic in this chapter, whose four-resistor diamond has no simple series-parallel shortcut.
Key idea: the junction rule is charge conservation and the loop rule is energy conservation, and together — with a consistent sign convention and one independent equation per junction and per loop — they turn any resistor network, however tangled, into a solvable system of simultaneous linear equations.

The core insight

Kirchhoff's junction rule ( $\sum I_{in} = \sum I_{out}$ ) enforces conservation of charge at every wire junction and his loop rule ( $\sum \Delta V = 0$ ) enforces conservation of energy around every closed loop, and applying both with a consistent sign convention converts any resistor network — however many loops it has — into a system of simultaneous equations that can always be solved for every branch current.

Where it fits in the syllabus

Kirchhoff's rules sit inside the CBSE Class 12 Current Electricity chapter, coming right after Ohm's law, resistivity, and the EMF-and-internal-resistance treatment of a single cell and of cells in series and parallel. The prescribed syllabus expects the junction rule and loop rule to be stated, justified from conservation principles, and applied to set up and solve the simultaneous equations of a multi-loop circuit — exactly the skill this section builds. This topic is also the direct algebraic foundation for the Wheatstone bridge balance condition covered next, since the bridge is precisely the kind of circuit the two rules were designed for: one that cannot be reduced to a single equivalent resistance by series-parallel combination alone. Everything here stays within algebra and simultaneous linear equations, exactly as CBSE examines it — no calculus, and no potentiometer (which is not part of the current 2025-26 theory syllabus and is not touched here).

The physical picture

Picture a wire junction where three conductors meet — two currents flowing in, one flowing out. Charge is neither created nor destroyed inside that junction, and in steady-state (constant) current conditions no charge can be quietly accumulating there either, since a growing charge would itself change the local electric field and the current, contradicting the assumption of a steady state. So whatever charge arrives each second must leave each second. That single sentence is the entire content of the junction rule:

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

where: the sum on the left runs over every current arriving at the junction and the sum on the right over every current leaving it, both measured in amperes. The physical reading is bookkeeping, nothing more — a junction is a strict "what comes in must go out" checkpoint, the electrical analogue of cars entering and leaving a busy traffic roundabout with none left circling forever inside. Now picture walking around any closed loop of a circuit — through resistors, through cells, back to your starting point — carrying an imaginary unit test charge and tracking its potential energy at every step. Electric potential at a point is a single, well-defined number; it does not depend on the path you took to arrive there. So if your walk brings you back to the exact point you started from, the potential there must be exactly what it was when you left, no more and no less. Any rises in potential along the way (through a cell, from $-$ to $+$ ) must be exactly cancelled by the falls (through a resistor, in the direction of current flow) by the time you return. That is the loop rule:

\sum_{\text{loop}} \Delta V = 0

where: the sum runs over every circuit element crossed while traversing one complete closed loop, each contributing a signed potential change $\Delta V$ — a $-IR$ term for a resistor crossed along the assumed current, a $+IR$ term for one crossed against it, a $+\varepsilon$ term for a cell crossed from its negative to its positive terminal internally, and a $-\varepsilon$ term for the reverse. The physical reading is that a closed loop is a round trip on a hill whose height is potential: walk uphill and downhill by exactly matching amounts, and you are guaranteed to return to your starting altitude.

Figure

A wire junction with two currents entering from the left and top and one larger current leaving to the right, labelled I one, I two entering and I three leaving, with a balance scale icon underneath showing the two entering arrows balanced against the one leaving arrow

Kirchhoff's rules: the junction rule from charge conservation

The formal statement of the junction rule can be built, in a few clean algebraic steps, directly from the definition of current and the requirement that charge is conserved. The goal is to show that "no accumulation of charge at a junction in steady state" is mathematically identical to "current in equals current out" — turning a conservation principle into the exact equation used to solve circuits.

Derivation AThe junction rule from conservation of charge at a node

The goal is to derive $\sum I_{in} = \sum I_{out}$ starting only from the definition of electric current, $I = dQ/dt$ , and the physical fact that charge can neither be created nor destroyed. This uses nothing beyond a rearrangement of that definition applied to a small region of wire surrounding the junction — no calculus beyond the derivative already used to define current itself.

Step 1. Draw an imaginary closed surface around the junction, enclosing only the point where the wires meet and none of the resistors or cells beyond it. Let $Q$ be the total charge inside this small surface at any instant. Because the junction is a solid conductor with no capacity to store net charge indefinitely under steady conditions, in the steady state $Q$ does not change with time: $\dfrac{dQ}{dt} = 0$ .

Step 2. Express the rate of change of $Q$ in terms of the currents crossing the surface. Every current flowing into the surface adds charge to it at the rate $I_{in}$ , and every current flowing out removes charge at the rate $I_{out}$ , so the net rate of change is the difference of the two totals: $\dfrac{dQ}{dt} = \sum I_{in} - \sum I_{out}$ , summed over all the branches meeting at the junction.

Step 3. Combine the two steps. Since $dQ/dt = 0$ in the steady state (Step 1) and $dQ/dt = \sum I_{in} - \sum I_{out}$ (Step 2), setting the two expressions equal gives $\sum I_{in} - \sum I_{out} = 0$ , which rearranges immediately to the junction rule: $\sum I_{in} = \sum I_{out}$ .

Step 4. Read the result physically. The derivation shows the junction rule is not an independent electrical law but a direct algebraic consequence of charge conservation applied locally, at every single junction, at every single instant. It holds regardless of how many branches meet there, how large the resistors or cells attached to those branches are, or how the currents are assigned — only conservation of charge is used, exactly the way conservation of energy underlies the loop rule stated in the physical picture above. Both worked examples below use this equation directly, once at a single junction and once as the linking relation between the currents in a full two-loop circuit.

Worked examples

Board level - three currents meeting at a single junction

In a circuit, three wires meet at a junction: $I_1 = 5\ \text{A}$ enters the junction, $I_2 = 2\ \text{A}$ enters the junction, and a third current $I_3$ leaves the junction. Find $I_3$ , and verify the answer using the derivation above.

Step 1. Identify what the junction rule requires: the sum of currents entering must equal the sum of currents leaving, $\sum I_{in} = \sum I_{out}$ , exactly as derived from $dQ/dt = 0$ at the junction.

Step 2. Substitute the given currents. Two currents enter, so $\sum I_{in} = I_1 + I_2 = 5 + 2 = 7\ \text{A}$ ; only one current leaves, so $\sum I_{out} = I_3$ .

Step 3. Set the two sums equal and solve: $I_1 + I_2 = I_3 \Rightarrow 5 + 2 = I_3 \Rightarrow I_3 = 7\ \text{A}$ . Answer: $I_3 = 7\ \text{A}$ must leave the junction.

Step 4. Verify against the derivation: with $I_3 = 7\ \text{A}$ , the net rate of charge accumulation at the junction is $\sum I_{in} - \sum I_{out} = (5+2) - 7 = 0$ , exactly matching the steady-state condition $dQ/dt = 0$ required in Step 1 of the derivation.

Competitive level - a genuine two-loop circuit with two cells and three resistors

Two cells, $\varepsilon_1 = 6\ \text{V}$ and $\varepsilon_2 = 2\ \text{V}$ (both of negligible internal resistance), each drive current into a common junction through their own resistor, $R_1 = 3\ \Omega$ and $R_2 = 5\ \Omega$ respectively, and the combined current returns to both cells through a shared resistor $R_3 = 4\ \Omega$ . Assume $I_1$ and $I_2$ both flow toward the junction and $I_3$ flows away from it. Find all three branch currents.

Step 1. Apply the junction rule at the shared node, exactly as derived above: current in equals current out, so $I_3 = I_1 + I_2$ . This is the one relation that links all three unknowns and replaces one of the three equations we would otherwise need.

Step 2. Apply the loop rule to the left loop (cell $\varepsilon_1$ , resistor $R_1$ , shared resistor $R_3$ ), traversing in the direction of assumed current flow so resistors give drop terms: $\varepsilon_1 - I_1 R_1 - I_3 R_3 = 0 \Rightarrow 6 - 3I_1 - 4I_3 = 0$ .

Step 3. Apply the loop rule to the right loop (cell $\varepsilon_2$ , resistor $R_2$ , the same shared resistor $R_3$ ) in the same style: $\varepsilon_2 - I_2 R_2 - I_3 R_3 = 0 \Rightarrow 2 - 5I_2 - 4I_3 = 0$ .

Step 4. Solve the three simultaneous equations. From Step 2, $I_1 = (6-4I_3)/3$ ; from Step 3, $I_2 = (2-4I_3)/5$ . Substituting both into $I_3 = I_1+I_2$ from Step 1 and clearing fractions (multiply through by 15) gives $15I_3 = 5(6-4I_3) + 3(2-4I_3) = 36 - 32I_3$ , so $47I_3 = 36$ and $I_3 = 36/47 \approx 0.77\ \text{A}$ . Back-substituting: $I_1 = (6 - 4(0.766))/3 \approx 0.98\ \text{A}$ , and $I_2 = (2-4(0.766))/5 \approx -0.21\ \text{A}$ . Answer: $I_1 \approx 0.98\ \text{A}$ , $I_2 \approx -0.21\ \text{A}$ , $I_3 \approx 0.77\ \text{A}$ . Timing note: budget three to four minutes for a genuine two-loop problem — one junction equation plus two loop equations, solved by substitution, is the standard pattern and should not be rushed into guesswork. The negative value for $I_2$ is not an error: it means the true current in that branch flows away from the junction, opposite to the direction assumed in Step 1, exactly the situation flagged in the tldr above — always report the magnitude and state the corrected direction rather than treating a negative sign as a mistake to "fix" by re-solving. As a quick sanity check, confirm $I_1 + I_2 = I_3$ numerically: $0.98 + (-0.21) = 0.77$ , which matches within rounding.

Common traps

Mixing up sign conventions mid-loop. The single most common error is inconsistently applying "drop" and "rise" signs for resistors or cells while going around a loop — fix a traversal direction before writing a single term, and apply the same convention (drop when crossing a resistor along assumed current, rise when crossing a cell from $-$ to $+$ ) all the way around without switching partway.
"Fixing" a negative current by re-guessing. A negative solved value, as seen for $I_2$ in the competitive example above, is a perfectly valid and useful result — it means the actual current flows opposite to the assumed direction. Re-solving with a different guess wastes time and never changes the physics.
Writing one junction equation too many. At $n$ junctions, only $n-1$ of the junction equations are independent — the last one is automatically satisfied once the others hold, since it follows from overall charge conservation across the whole circuit. Writing all $n$ and treating them as independent leads to a redundant, indeterminate system.
Trying series-parallel shortcuts on a genuinely multi-loop network. The two-loop circuit above, and the Wheatstone bridge coming next, cannot be reduced by simple series or parallel combination alone — that is exactly why Kirchhoff's rules exist, and reaching for a shortcut first often wastes more time than setting up the equations directly.
Forgetting that the loop rule needs independent loops. Once you have as many independent loop equations as remaining unknowns (after the junction equations), adding a further "loop" built from combinations of already-used loops adds no new information — it is a repeat of information already captured.

Common trapThe junction rule is charge conservation and the loop rule is energy conservation — treat them as two different bookkeeping tools, not one rule stated twice

The most damaging misconception is blurring the two rules into a single vague idea of "current and voltage balance." They are answers to two entirely different questions and come from two entirely different conservation laws. The junction rule answers "how much current arrives and leaves a point" and follows purely from charge conservation — it says nothing about resistances, EMFs, or energy at all. The loop rule answers "how much potential is gained and lost on a round trip" and follows purely from energy conservation via the single-valuedness of potential — it says nothing about how current splits at a junction. A correct multi-loop solution always needs both: enough independent junction equations to relate the branch currents to each other, and enough independent loop equations to bring in the actual resistor and EMF values. Using only one rule, however many times you apply it, will always leave the system of equations short of what is needed to solve for every unknown current.

You should be able to…

State the junction rule and the loop rule, and explain which conservation law (charge or energy) underlies each one.
Derive the junction rule from $I = dQ/dt$ and the steady-state condition $dQ/dt = 0$ at a node.
Apply a consistent sign convention for resistors and cells while traversing a loop, without switching midway.
Set up the full set of independent junction and loop equations for a two-loop (or larger) circuit and solve them simultaneously.
Correctly interpret a negative solved current as a reversed direction, not an error requiring re-solving.
Recognise when a circuit cannot be solved by series-parallel reduction alone and requires the full Kirchhoff method.

Ask the tutor: Want to work through a Wheatstone bridge using this exact junction-and-loop method, or try a three-loop circuit with more unknowns? Ask the AI tutor for a guided walkthrough.

Why do we even need two separate rules — doesn't Ohm's law already tell us the current through a resistor?

Ohm's law, $V=IR$ , tells you the relationship between voltage and current for one resistor by itself, once you already know either quantity. It says nothing about how a total current from a cell splits among several branches at a junction, or how several cells with different EMFs interact around a loop with more than one resistor. The junction rule supplies the missing charge-conservation link between branch currents, and the loop rule supplies the missing energy-conservation link between the EMFs and the $IR$ drops around each loop. Together they generate exactly as many independent equations as unknown currents, which Ohm's law alone — applied resistor by resistor — cannot do for a genuinely multi-loop network.

How do I know which direction to assume for each current before I start?

It genuinely does not matter, and this is one of the most reassuring facts in the whole topic. Pick any direction you like for each unknown branch current — a confident-looking guess, a coin flip, whatever is fastest to draw. Write the junction and loop equations consistently with your chosen arrows and solve. If a current comes out positive, your guess matched reality; if it comes out negative, the actual current flows the opposite way, and the magnitude you calculated is still exactly correct. The competitive-level worked example above shows this directly: $I_2$ solving out negative simply means that branch's real current runs opposite to the assumed direction, nothing more.

How many junction equations and how many loop equations do I actually need?

For a circuit with $j$ junctions and $u$ unknown branch currents, you need $j-1$ independent junction equations (the last junction's equation is automatically satisfied by the others, since it follows from conservation of charge over the whole circuit) plus enough independent loop equations to reach a total of $u$ equations. In the two-loop example above there was one junction (giving one equation) and two independent loops (giving two more), for three equations solving three unknown currents — exactly the pattern to expect on a two-loop CBSE or competitive question.

Is this the same idea used for the Wheatstone bridge coming up next?

Yes, directly. The Wheatstone bridge is a four-resistor diamond with a fifth galvanometer branch across the middle, and it is the textbook example of a network that cannot be solved by any series-parallel shortcut. Setting up its junction and loop equations the same way as the two-loop circuit here, then imposing the special condition that no current flows through the middle branch, is exactly how the bridge's balance condition $P/Q = R/S$ is derived — this article's method is the direct prerequisite for that derivation.