Kirchhoff's laws

A single loop with one cell and one resistor yields to Ohm's law in one line. Give a circuit a second loop, a second EMF, and a junction where currents split three ways, and Ohm's law alone runs out of road — there is no single "the current" left to solve for. Kirchhoff's laws rescue exactly this situation. The first law is charge conservation applied to a junction: whatever current flows in must flow out. The second is energy conservation applied to a loop: go all the way around back to your starting point and the net change in potential must be zero. Together they turn any network of resistors and cells, however tangled, into a set of simultaneous algebraic equations that always has a definite solution — the systematic method underlying the Wheatstone bridge and potentiometer topics that follow this one.

The core insight

At every junction the currents in equal the currents out (charge conservation), and around every closed loop the EMFs equal the sum of the $IR$ drops (energy conservation), and applying both together turns any multi-loop circuit into solvable simultaneous equations.

The core idea

The junction rule (Kirchhoff's first law). At any point where three or more wires meet, charge cannot pile up or vanish in steady state, so the total current arriving must equal the total current leaving:

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

where: the sum runs over every branch meeting at that one junction, and each $I$ is the current in that branch. This is nothing more than conservation of charge written for a single point in the circuit.

The loop rule (Kirchhoff's second law). Trace any closed path in the circuit — through cells and resistors — back to where you started, and the algebraic sum of EMFs must equal the algebraic sum of $IR$ drops around that path:

\sum \varepsilon = \sum IR

where: $\varepsilon$ is the EMF of each cell encountered and $IR$ is the potential drop across each resistor carrying current $I$ , both summed with signs that depend on the direction of travel around the loop. Because electric potential is single-valued at every point, returning to the start after a full circuit of the loop must give zero net change — which is exactly what this equation enforces.

Sign conventions. Assign an arbitrary current direction to every branch first; a wrong guess simply comes out negative later, which is not an error. While tracing a loop in one consistent sense (say, clockwise), a resistor crossed along the assumed current direction counts as a drop ( $-IR$ ) and crossed against it counts as a rise ( $+IR$ ); a cell crossed from its negative to positive terminal counts as a rise ( $+\varepsilon$ ), and from positive to negative as a drop ( $-\varepsilon$ ).

How many equations you need. For a circuit with $b$ branches (unknown currents) and $j$ junctions, only $j-1$ of the junction equations are independent, and exactly $b-(j-1)$ independent loop equations fill the rest — together giving $b$ equations for $b$ unknowns, always solvable by simultaneous elimination.

Figure

A two loop circuit with a shared branch, showing current arrows at the junction and the two independent loop paths traced around it

Derivation ASolving a two-loop circuit with the junction and loop rules together

The goal is to find both unknown currents in a circuit with two loops sharing one branch — the standard multi-loop problem — by combining one junction equation with two loop equations into a solvable pair of simultaneous linear equations, then checking the answer is physically sensible even when a current comes out negative.

Step 1. Label the unknowns and apply the junction rule. Let $I_1$ flow through the left branch (cell $\varepsilon_1$ , resistor $R_1$ ) and $I_2$ flow through the right branch (cell $\varepsilon_2$ , resistor $R_2$ ), both arriving at the shared junction and leaving together through the shared branch $R_3$ . The junction rule fixes the current in $R_3$ immediately, with no further unknowns needed: $I_3 = I_1 + I_2$ .

Step 2. Apply the loop rule to the left loop. Tracing clockwise through $\varepsilon_1$ , then $R_1$ , then the shared branch $R_3$ (carrying $I_1+I_2$ ), and back: $\varepsilon_1 = I_1R_1 + (I_1+I_2)R_3$ .

Step 3. Apply the loop rule to the right loop. Tracing its own closed path through $\varepsilon_2$ , $R_2$ , and the same shared branch $R_3$ : $\varepsilon_2 = I_2R_2 + (I_1+I_2)R_3$ .

Step 4. Solve the two linear equations from Steps 2 and 3 simultaneously (elimination or substitution) for $I_1$ and $I_2$ , then recover $I_3 = I_1+I_2$ from Step 1. If either solved current is negative, it means only that the true direction is opposite the one assumed in Step 1 — the magnitude from the algebra is still correct, and the junction relation $I_3=I_1+I_2$ still holds with the signed value.

Worked example

Two loops share a common branch $R_3 = 4\ \Omega$ . The left loop has a $10\ \text{V}$ cell (negligible internal resistance) in series with $R_1 = 2\ \Omega$ ; the right loop has a $4\ \text{V}$ cell in series with $R_2 = 1\ \Omega$ . Let $I_1$ flow through $R_1$ and $I_2$ through $R_2$ , so by the junction rule $I_3 = I_1+I_2$ flows through the shared branch. Find $I_1$ , $I_2$ and $I_3$ .

Step 1. Write the left-loop equation, substituting $I_3 = I_1+I_2$ : $10 = 2I_1 + 4(I_1+I_2) = 6I_1 + 4I_2$ … (1)

Step 2. Write the right-loop equation the same way: $4 = 1\cdot I_2 + 4(I_1+I_2) = 4I_1 + 5I_2$ … (2)

Step 3. Eliminate $I_2$ . Multiply (1) by 5 and (2) by 4: $50 = 30I_1+20I_2$ and $16=16I_1+20I_2$ . Subtracting, $34 = 14I_1$ , so $I_1 = 2.43\ \text{A}$ .

Step 4. Back-substitute into (1): $10 = 6(2.43) + 4I_2 = 14.57+4I_2$ , so $4I_2 = -4.57$ and $I_2 = -1.14\ \text{A}$ . The negative sign only means the true current in that branch runs opposite to the direction assumed at the start; it is not an error and is not discarded. Finally $I_3 = I_1+I_2 = 2.43-1.14 = 1.29\ \text{A}$ .

Answer: $I_1 = 2.43\ \text{A}$ (as assumed), $I_2 = -1.14\ \text{A}$ (actually flows the opposite way, magnitude $1.14\ \text{A}$ ), and $I_3 = 1.29\ \text{A}$ through the shared branch.

Competitive stretch

Check the answer against the simplest case: at a junction where $I_1 = 3\ \text{A}$ and $I_2 = 5\ \text{A}$ both flow in, a third wire carries current out — find it. By the junction rule alone, $I_3 = I_1+I_2 = 3+5 = 8\ \text{A}$ flows out, with no loop equation needed at all — a reminder that the junction rule by itself already solves any purely current-splitting question, and the loop rule is only needed once unknown EMFs or resistances enter the picture.

Common traps

Treating a negative solved current as a mistake. It only flips the direction from the one arbitrarily assumed; the magnitude from the algebra stands.
Mixing sign conventions mid-loop. Switching between "drop" and "rise" rules for resistors or cells partway around one loop is the most common source of wrong answers — fix a traversal direction and a rule before writing the equation, and hold it for the whole loop.
Writing redundant equations. Only $j-1$ independent junction equations and $b-(j-1)$ independent loop equations are needed; extra equations are not wrong but add confusion, not information.
Tracing a path that is not actually closed, or repeating a branch. The loop rule only applies to a path that returns to its exact starting point without retracing any branch twice.

Ask the tutor: paste your own multi-loop circuit and ask for help setting up its junction and loop equations step by step.

Why can I choose the current directions arbitrarily before solving anything?

Because the algebra self-corrects. You assign a direction to each unknown current purely as a bookkeeping choice — a label, not a claim about reality. When you solve the simultaneous equations, a positive value confirms your guess was right, and a negative value tells you the real current runs the other way, with the magnitude still correct. Either way, the final physical answer is unaffected by which direction you happened to start with.

Do I need to apply the loop rule to every single loop I can trace in the circuit?

No, and doing so wastes time. Only $b-(j-1)$ loop equations are independent for a circuit with $b$ branches and $j$ junctions; any additional loop equation you write can be derived from the ones you already have and adds no new information. Pick the smallest, simplest closed paths that together cover every branch at least once, combine them with the independent junction equations, and stop there.

How is the loop rule different from just using Ohm's law on each resistor?

Ohm's law, $V=IR$ , only relates the current through one resistor to the voltage across that one resistor — it says nothing about how several resistors and cells interact around a loop. The loop rule is the additional statement, from energy conservation, that the EMFs and IR drops around any closed path must sum to zero. You still use Ohm's law to write each individual IR term; Kirchhoff's loop rule is what tells you how to combine those terms correctly for a multi-loop circuit that Ohm's law alone cannot solve.

Kirchhoff's laws

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads