In short

Kirchhoff's current law (KCL) — the junction rule. At any junction in a circuit, the sum of currents flowing in equals the sum flowing out:

\boxed{\;\sum I_\text{in} \;=\; \sum I_\text{out} \quad\Leftrightarrow\quad \sum I_\text{at junction} = 0.\;}

This is conservation of charge applied to a tiny volume: charge does not pile up at a point in a steady-state circuit.

Kirchhoff's voltage law (KVL) — the loop rule. Around any closed loop in a circuit, the sum of potential changes is zero:

\boxed{\;\sum_\text{loop} \Delta V \;=\; 0.\;}

This is single-valuedness of potential: if you walk all the way around a loop and return to your starting point, your potential energy must end where it began.

Sign conventions when you walk a loop.

  • Through a resistor in the direction of the assumed current: potential drops by IR, contribute -IR.
  • Through a resistor against the assumed current: potential rises by IR, contribute +IR.
  • Through a cell from - to + terminal: potential rises by \varepsilon, contribute +\varepsilon.
  • Through a cell from + to - terminal: potential drops by \varepsilon, contribute -\varepsilon.

Together, KCL and KVL solve every resistor-and-cell network — including bridge networks, mesh circuits, and anything else that defeats series-parallel reduction. The systematic recipe: assign current variables, write as many independent KCL and KVL equations as you have unknowns, solve the linear system.

Turn the ignition key on a Maruti Swift. Four distinct electrical things happen in about half a second: the 12-volt battery delivers a roaring 150-ampere burst to the starter motor; the ignition coil fires the spark plugs; the instrument cluster lights up; and the fuel pump primes the line. All of this is happening simultaneously, through a tangled web of wires, fuses, and relays running under the dashboard. Every one of those currents has a value — specific, predictable, measurable — and none of it can be worked out using just "series" and "parallel". The starter branch is not simply in series with the spark coil, and the dashboard is not simply in parallel with the fuel pump. Currents are branching, merging, crossing, and sharing paths all at once.

How, then, does the engineer at the Gurgaon factory know exactly how thick to make each wire? How does the spec sheet say the alternator must deliver 70 amps rather than 50? They use two rules — Gustav Kirchhoff wrote them down in 1845 when he was 21, applying conservation of charge at a point and single-valuedness of potential around a loop. The rules look almost trivially simple. They are trivially simple. And together, they reduce any circuit problem — even one with fifteen branches — to a system of linear equations you could solve on paper during an afternoon cricket rain delay.

This article is that reduction, step by step. By the end, you will be able to take a two-loop circuit with two cells and three resistors, write four equations, and find every current exactly. That is the skill every Class 12 CBSE board paper and every JEE Main and Advanced electricity question leans on.

Why series and parallel are not enough — the bridge that defeats them

Every network in the previous chapter yielded to the move "collapse a clearly-series or clearly-parallel pair, repeat." But that only works when the network's topology is a tree of series-parallel nestings. The moment you draw a circuit with a diagonal connection — a bridge — the trees fail.

Five-resistor bridge network defeating series-parallel reductionA diamond of four resistors R1, R2, R3, R4 with a fifth resistor R5 bridging the two middle nodes C and D. Power supply between A (left) and B (right). No pair of resistors is clearly series or clearly parallel because of the bridge R5. ACDB R₁ (AC)R₂ (AD)R₃ (CB)R₄ (DB)R₅ (CD, the bridge)
Five resistors wired as a bridge. Look at any pair: $R_1$ and $R_2$ share node A but not the second node — not parallel, not series. $R_1$ and $R_3$ share only node C — not parallel. The diagonal bridge $R_5$ prevents the usual reduction. This network needs Kirchhoff's laws.

Try to reduce this. Pick any pair — say R_1 and R_3. They meet at node C, but nothing taps off between them on one side (good, that hints at series) — except that node C is also the end of R_5. That third connection at the midpoint kills the series pattern. No matter which pair you try, either a bridge branch taps off the middle (killing series) or the two ends don't coincide (killing parallel). Series-parallel reduction hits a dead wall.

Yet there is a single answer for the equivalent resistance between A and B, and Kirchhoff's laws will hand it to you. But first the laws themselves.

Kirchhoff's current law — conservation of charge at a point

Statement

At any junction (node) in an electrical circuit, the algebraic sum of currents is zero. Equivalently, the total current flowing into the junction equals the total current flowing out.

\sum_{\text{at junction}} I \;=\; 0, \qquad \text{or} \qquad \sum I_\text{in} \;=\; \sum I_\text{out}.

Derivation — directly from conservation of charge

Pick any junction in a steady-state circuit. Draw a tiny imaginary closed surface around the junction — a small sphere that cuts every wire meeting the junction. Let Q(t) be the total charge inside the sphere at time t.

Step 1. The rate of change of charge inside equals the rate of charge flowing in:

\frac{dQ}{dt} \;=\; \sum I_\text{in} \;-\; \sum I_\text{out}.

Why: this is just the accounting statement of conservation of charge — charge can't appear or disappear. What is inside the sphere at time t + dt is what was inside at time t, plus what flowed in, minus what flowed out.

Step 2. In a steady-state circuit, all currents are constant in time and all node voltages are constant in time. In particular, no charge builds up anywhere. The amount of charge Q inside the imaginary sphere is constant:

\frac{dQ}{dt} \;=\; 0.

Why: a steady-state circuit is defined by stationary currents and voltages. If charge were piling up at the junction, the voltage at that junction would be rising — that would not be steady state.

Step 3. Combine.

\sum I_\text{in} \;=\; \sum I_\text{out}.

Why: from step 1, with the left side set to zero by step 2. The total current in equals the total current out. This is KCL.

Using KCL — sign conventions for writing \sum I = 0

Choose a sign convention — say, currents flowing in are positive, flowing out are negative. Then \sum I = 0 at every junction. If you've assigned current arrows to each branch, a current whose assumed arrow points into the junction contributes +I, and one pointing out contributes -I.

The key pragmatic point: you don't need to know the actual direction of the current in advance. Guess a direction for each branch, write KCL consistently, solve. If your guess was wrong, the value of the current comes out negative — and that sign itself is the correction.

Three-wire junction showing KCLA junction point where three wires meet. Current I1 flows in from the left, I2 flows in from below, and I3 flows out to the right. KCL says I1 + I2 = I3. I₁ = 3 A I₂ = 2 A I₃ = ?KCL: I₁ + I₂ = I₃, so I₃ = 5 A
A simple junction. Two currents (3 A and 2 A) flow in, one flows out. KCL gives the outgoing current immediately: $I_3 = I_1 + I_2 = 5$ A. No voltage, no Ohm's law — just charge conservation.

Kirchhoff's voltage law — single-valuedness of potential

Statement

Around any closed loop in a circuit, the algebraic sum of potential changes is zero:

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

Walking around a loop, if you tally all the rises and drops in potential and you come back to the same point, the tally must be zero.

Derivation — from single-valuedness of potential

In an electrostatic field, the potential \varphi(P) at a point P is a single real number. If you walk from point A back to point A along any path, the change in potential must be zero — because you are at the same place, and the potential has one value there.

Step 1. Write the potential change along a path as a sum.

Along any path from A back to A, the total potential change is the sum of small changes \sum \Delta V along each piece of the path.

\sum \Delta V \;=\; \varphi(\text{end}) - \varphi(\text{start}) \;=\; \varphi(A) - \varphi(A) \;=\; 0.

Why: the potential is a property of the point, not the path. Same point, same potential. Add and subtract your way around and the sum is forced to zero.

Step 2. Identify the contributions.

In a DC circuit, the only things that change the potential as you walk along a wire are (i) a cell, which pumps charge and raises potential from - to + by \varepsilon, and (ii) a resistor, which drops potential by IR in the direction of current flow.

Step 3. Combine.

The total around a loop is the sum of cell EMFs (with signs) and resistor drops (with signs), and that total equals zero.

The sign conventions, in detail

The most common source of wrong answers in Kirchhoff problems is sloppy signs. Fix the convention once and use it everywhere.

  1. Assign a current direction to each branch — just pick an arrow. It doesn't matter if you're right; the math will correct you.

  2. Pick a direction to walk around the loop — clockwise or counter-clockwise. Either works; just stick with it for the whole loop.

  3. Through a resistor:

    • If you walk in the same direction as the assumed current → the potential drops by IR → contribute -IR to the sum.
    • If you walk against the assumed current → the potential rises by IR → contribute +IR.

  4. Through a cell (EMF \varepsilon):

    • If you walk from the negative terminal to the positive terminal (i.e., in the direction the cell pumps charge) → the potential rises by \varepsilon → contribute +\varepsilon.
    • If you walk from the positive terminal to the negative terminal → the potential drops by \varepsilon → contribute -\varepsilon.

Write these four rules once on the first page of your rough notebook. Every Kirchhoff problem uses them.

KVL sign convention diagramA loop with a cell and a resistor. Walking direction is clockwise (shown by a green arrow). Assumed current I is clockwise. Walking through the cell from minus to plus contributes plus epsilon. Walking through the resistor in the direction of current contributes minus IR. Sum equals zero. +−ε R walking CW I cell: walk − → + means +εresistor (with I): −IR
Walking clockwise around a single-loop circuit. At the cell you go from $-$ to $+$ (gain $+\varepsilon$); at the resistor you go with the current (lose $IR$). KVL: $+\varepsilon - IR = 0 \Rightarrow I = \varepsilon/R$. Which is just Ohm's law — KVL contains Ohm's law as the single-loop case.

Putting KCL and KVL together — the recipe

For a network with B branches (edges carrying currents), you need exactly B current values to fully describe it. Collect equations until you have B independent linear equations, then solve the linear system.

The recipe.

  1. Label each branch with an assumed current direction. Just pick — don't agonise.
  2. Identify the junctions. A network with N junctions has N - 1 independent KCL equations (the N-th is a redundant combination of the others).
  3. Identify the independent loops. You need B - (N - 1) = B - N + 1 independent KVL equations. A useful heuristic for planar circuits: one loop per "window" in the layout.
  4. Write the equations. N - 1 KCL + (B - N + 1) KVL = B equations, exactly matching the B unknowns.
  5. Solve the linear system. For two or three unknowns, substitute by hand. For more, use matrix methods (reserved for the going-deeper section).

The critical point: KCL alone is not enough (it only tells you what's conserved at junctions; it doesn't involve resistances or EMFs). KVL alone is not enough (it only applies to closed loops, and circuits can have branches that aren't in any single loop). Together they exactly determine everything.

Worked examples

Example 1: The starter circuit in a Maruti Swift (simplified)

A 12\ \text{V} car battery has a small internal resistance of 0.05\ \Omega. It drives two loads in parallel: the starter motor with effective resistance 0.08\ \Omega, and the dashboard/lights combined with effective resistance 2.0\ \Omega. Find the current from the battery, the current through the starter, and the current through the dashboard — using Kirchhoff's laws directly (no shortcut to parallel).

Maruti starter and dashboard circuitA 12 V battery with internal resistance 0.05 ohm feeds a junction node C. From C two branches in parallel return to the battery: a 0.08 ohm starter branch and a 2 ohm dashboard branch. Currents I1, I2, I3 labelled. 12 V r = 0.05 Ω C starter, 0.08 Ω dash, 2 Ω D II₂I₁
The car's 12 V battery drives the starter and dashboard through a small internal resistance. Two branches in parallel, with a single-source cell — classic two-loop network.

Step 1. Assign currents and directions.

Let I = current out of battery's + terminal (total battery current), I_1 = current through the starter branch, I_2 = current through the dashboard branch. All directions chosen clockwise in their respective loops.

Why: just pick directions. If any comes out negative, that branch's actual current flows opposite to the assumed arrow.

Step 2. Apply KCL at junction C.

Currents in: I (from the battery side). Currents out: I_1 (to starter), I_2 (to dash).

I \;=\; I_1 + I_2. \tag{1}

Why: charge balance at C. Whatever comes in must go out, and with our direction choices, the two branch currents leaving add up to the incoming total.

Step 3. Apply KVL to the top loop (battery → internal resistance → starter → back).

Walk clockwise. Start at the battery's - terminal, go up through the cell (+\varepsilon), through the internal resistance r in the direction of current I (-Ir), continue through the starter in the direction of I_1 (-I_1 R_\text{starter}), and back to start.

+12 - I(0.05) - I_1(0.08) \;=\; 0. \tag{2}

Why: cell gain, then two resistor drops (both in the direction of their assumed currents). The net change around a loop is zero.

Step 4. Apply KVL to the bottom loop (battery → internal resistance → dash → back).

Similarly for the dashboard branch:

+12 - I(0.05) - I_2(2.0) \;=\; 0. \tag{3}

Why: same structure as equation (2), with the dashboard resistance instead of the starter.

Step 5. Solve the three equations for I, I_1, I_2.

From (2): \;0.05 I + 0.08 I_1 = 12.

From (3): \;0.05 I + 2.0 I_2 = 12.

Subtract: 0.08 I_1 - 2.0 I_2 = 0 \Rightarrow I_1 = 25 I_2.

Using (1): I = 26 I_2.

Plug into (3): 0.05 (26 I_2) + 2 I_2 = 12 \Rightarrow 1.3 I_2 + 2 I_2 = 12 \Rightarrow 3.3 I_2 = 12 \Rightarrow I_2 \approx 3.64\ \text{A}.

Then I_1 = 25 \times 3.64 \approx 90.9\ \text{A}, and I = 90.9 + 3.64 \approx 94.5\ \text{A}.

Why: a clean three-equation linear system. The massive ratio between the two branch currents (about 25:1) is real — the starter has a resistance 25 times smaller than the dashboard, so it draws 25 times the current.

Step 6. Sanity check.

Voltage drop across the internal resistance: I r = 94.5 \times 0.05 \approx 4.7\ \text{V}. So the terminal voltage of the battery (what the dashboard and starter actually see) is 12 - 4.7 = 7.3\ \text{V}. That fits with what you measure at the battery terminals when cranking — the voltage sags hard the moment the starter kicks in. This is the reason your headlights dim when you turn the key.

Result. Total current I \approx 94.5 A, starter I_1 \approx 90.9 A, dashboard I_2 \approx 3.64 A.

What this shows. KCL + KVL reproduces the parallel-combination answer (as expected — two parallel resistors seen by a single cell). But writing out the three equations makes explicit a detail that the series-parallel trick hides: during cranking, the internal resistance r = 0.05\ \Omega drops nearly 5 V, so the dashboard is momentarily running at 7 V, not 12 V. That sag is responsible for the "headlight dip" you see every time someone starts a car — the headlights are dimmed by the starter's current through the battery's own internal resistance.

Example 2: Two cells facing each other (charging circuit)

A 12\ \text{V} cell with internal resistance 0.5\ \Omega and a 10\ \text{V} cell with internal resistance 0.3\ \Omega are connected in a loop with a 2\ \Omega external resistor. The two cells are wired so that their positive terminals face each other (typical charging configuration — the stronger cell drives current through the weaker one). Find the current and its direction.

Two cells in series opposition with a resistorSingle loop with two cells whose positive terminals face each other, and a 2 ohm resistor. Cell A is 12 V with internal resistance 0.5 ohm on the left. Cell B is 10 V with internal resistance 0.3 ohm on the right. A 2 ohm resistor sits on the top rail between them. Assumed current I flows clockwise. 2 Ω 0.3 Ω +−10 V +−12 V 0.5 Ω I (assumed)
A single loop with two EMF sources and three resistors (two internal, one external). The 12 V cell on the left *pushes* current clockwise; the 10 V cell on the right *opposes* clockwise current because its $+$ terminal is on the clockwise-leaving side. Net EMF = 12 − 10 = 2 V drives the clockwise current.

Step 1. Assign a current direction.

Assume I flows clockwise (up through cell A, across the top, down through cell B, back along the bottom).

Step 2. Apply KVL walking clockwise around the loop.

Walk clockwise starting from the bottom-left corner. Go up through cell A (from - to +, gain +12 V). Go up through its internal resistance 0.5\ \Omega with the current (drop -0.5 I). Go right across the top through the 2 Ω resistor with the current (drop -2 I). Go right through cell B's internal resistance 0.3\ \Omega with the current (drop -0.3 I). Go down through cell B — but its + terminal is on top (the clockwise-leaving side means you enter from + and exit from -), so this is a drop: -10 V.

+12 - 0.5 I - 2 I - 0.3 I - 10 \;=\; 0.
2 - 2.8 I \;=\; 0.
I \;=\; \frac{2}{2.8} \;\approx\; 0.714\ \text{A}.

Why: two cells in opposition, so their EMFs subtract. The net driving EMF is 2 V; the total loop resistance is 0.5 + 2 + 0.3 = 2.8\ \Omega; the current is the ratio. The positive sign confirms our assumed direction is correct — current really does flow clockwise, from the stronger cell.

Step 3. Interpret the result.

Current flows clockwise at about 0.714 A. The 12 V cell is discharging (current leaves its + terminal). The 10 V cell is being charged (current enters its + terminal from outside, reversing its normal discharge direction). The 2 Ω resistor dissipates power I^2 R = 0.714^2 \times 2 \approx 1.02 W.

Step 4. Check with terminal voltages.

Terminal voltage of cell A: \varepsilon_A - I r_A = 12 - 0.714(0.5) = 11.64 V.

Terminal voltage of cell B: \varepsilon_B + I r_B = 10 + 0.714(0.3) = 10.21 V (the plus sign is because current flows into cell B's + terminal — it is being charged, which raises its terminal voltage above its EMF).

Voltage drop across external resistor: 11.64 - 10.21 = 1.43 V. Check: I \times 2 = 0.714 \times 2 = 1.43 V. ✓

Why: Ohm's law on the external resistor must match the terminal voltages of the two cells. It does. This is how you sanity-check every Kirchhoff solution — does the potential at each node work out consistently?

Result. Current I \approx 0.714 A clockwise (12 V cell discharging, 10 V cell being charged). External 2 Ω resistor carries the same current and dissipates about 1.0 W.

What this shows. Kirchhoff's voltage law handles multi-cell circuits just as cleanly as single-cell circuits — you just sign each EMF according to the walking direction. The oppositely-placed cells give a net EMF equal to the difference, which matches Indian Railways' multi-source traction systems, where two locomotive power units in opposition would charge each other through the overhead line. The same algebra you did here would scale to the whole section.

Common confusions

If you came here to solve CBSE-board or JEE-Main Kirchhoff problems, you already have the tools. What follows is for readers who want (i) the matrix form of Kirchhoff, (ii) the mesh-analysis shortcut, and (iii) the Maxwell connection — what Kirchhoff's laws actually are, in the grown-up electromagnetic picture.

Matrix form — setting up any circuit as a linear system

Kirchhoff's laws always reduce to a linear system Ax = b, where x is the vector of unknown currents. For a network with B branches, N nodes, and L = B - N + 1 independent loops:

  • The first N - 1 rows of A are the KCL equations, with entries \pm 1 for each current meeting a junction (and a 0 right-hand side).
  • The last L rows of A are the KVL equations, with entries \pm R_j for each branch in the loop (and right-hand side equal to the net EMF around that loop).

This (B \times B) matrix always has a unique solution for a network with at least one cell. Circuit-simulation software like SPICE and PSpice builds A and b automatically from the netlist and solves it numerically. For hand calculations with B \le 4, you just write the equations out and reduce.

Mesh analysis — one current per "window", not one per branch

For planar circuits, mesh analysis cuts the number of variables almost in half. Instead of assigning a current I_j to each branch and writing KCL at every junction, you assign a mesh current i_k to each independent loop (each "window") — and you write only KVL equations, one per loop. KCL becomes automatic.

Rules. Each branch's actual current equals the algebraic sum of mesh currents that flow through it. For a branch shared between mesh k and mesh m, and walking in the direction of i_k, the voltage drop across a resistor R is

V_R \;=\; R(i_k - i_m).

For a two-loop planar circuit this gives two equations in two unknowns, not three or four. The mesh current is conceptually an imaginary current circulating in the loop; the real branch current emerges from subtracting adjacent mesh currents on shared branches.

Example — the five-resistor bridge. For the bridge from the chapter opening, three independent loops exist (the top triangle, the bottom triangle, and the outer diamond). Assign mesh currents i_1, i_2, i_3 (say, all clockwise) and write three KVL equations. The diagonal bridge resistor R_5 carries mesh current i_1 - i_2 (shared between the two inner triangles). Solving the 3 \times 3 system gives every current in the bridge in about a third of the work of branch-by-branch KCL+KVL.

This is the technique electrical engineers use in practice; it is also the way JEE Advanced expects you to handle bridge circuits (though they rarely say so explicitly).

Kirchhoff's laws are Maxwell's equations, in disguise

In full electromagnetic theory, the two Kirchhoff laws are specific limits of Maxwell's equations.

KCL is the time-integrated statement of the continuity equation \nabla \cdot \mathbf{J} + \partial \rho/\partial t = 0, specialised to steady state (\partial \rho/\partial t = 0). Integrating \nabla \cdot \mathbf{J} = 0 over a small volume enclosing a junction and applying the divergence theorem gives

\oint_{\partial V} \mathbf{J} \cdot d\mathbf{A} \;=\; 0 \;\Longleftrightarrow\; \sum I = 0.

That last step is KCL.

KVL is Faraday's law \oint \mathbf{E} \cdot d\boldsymbol{\ell} = -d\Phi_B/dt specialised to the case where d\Phi_B/dt = 0 (no changing magnetic flux through the loop). When that is true,

\oint \mathbf{E} \cdot d\boldsymbol{\ell} \;=\; 0 \;\Longleftrightarrow\; \sum \Delta V = 0.

That is KVL. For AC circuits with inductors, d\Phi_B/dt \ne 0, and KVL must be generalised to \sum \Delta V = -L\,dI/dt — you treat the changing flux as an effective back-EMF source and write KVL with it. This is why an inductor "acts like" a voltage source in AC analysis: the -L\,dI/dt term from Faraday's law is moved to the "source" side of KVL and called V_L.

So the two humble laws you just met are genuinely special cases of two of Maxwell's four equations — charge conservation (continuity) and Faraday's law — restricted to DC steady state. Nothing you will do in Class 12 electricity goes beyond this, but it is worth knowing where the laws come from.

A systematic shortcut: the superposition principle

For linear circuits (resistors, cells, capacitors), currents and voltages obey the superposition principle: turn on one cell at a time (replace the others with their internal resistances only), compute the currents it produces alone, and sum the results. This works because Kirchhoff's laws are linear in currents and EMFs.

Superposition doesn't reduce total algebra — you solve the circuit as many times as you have sources — but it cleanly decomposes "which source is responsible for what current", which matters when you're diagnosing a malfunctioning multi-source circuit (like a car with a dodgy alternator and a weak battery). It is also the cleanest way to see why the two cells in Example 2 can be analysed as a difference of EMFs: turn off the 10 V cell → 12 V drives 12/2.8 = 4.29 A; turn off the 12 V cell → 10 V drives -10/2.8 = -3.57 A (opposite direction); net = 4.29 - 3.57 = 0.71 A. That matches our full-Kirchhoff answer of 0.714 A.

Where this leads next