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Kirchhoff's laws

The two laws

Kirchhoff's laws are conservation of charge and conservation of energy applied to circuits. They handle circuits that cannot be reduced by simple series and parallel combination.

Kirchhoff's current law (the junction rule):

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

The total current into any junction equals the total current out of it.
This is conservation of charge — charge cannot accumulate at a point or vanish.

Kirchhoff's voltage law (the loop rule):

$\sum \varepsilon = \sum IR$

Around any closed loop, the sum of the emfs equals the sum of the potential drops.
This is conservation of energy — a coulomb of charge that returns to its starting point must have given up exactly as much energy as it gained.

Applying the loop rule

The bookkeeping is where marks are won and lost, so keep it mechanical:

Draw the circuit and mark a current direction in every branch. It does not matter if you guess wrong — a negative answer simply means that current flows the other way.
Label each junction and choose your loops so that every component appears in at least one.
Travel around each loop in a consistent direction, and apply the sign rules:
- passing through a source from − to +: add the emf,
- passing through a source from + to −: subtract the emf,
- passing through a resistor with the current: subtract $IR$ (a drop),
- passing through a resistor against the current: add $IR$ .
Write one equation per loop, plus junction equations, until you have as many equations as unknowns.
Solve the simultaneous equations.

How many equations you need

You need as many independent equations as unknown currents.
A circuit with two loops and three branches has three unknown currents, so you need three equations: one junction equation and two loop equations.
Adding more loops beyond that gives equations that are combinations of the ones you already have — they add no new information.

Checking your answer

Substitute back into an equation you did not use to derive the answer.
Check the junction rule holds numerically at each junction.
A negative current is a valid answer — report it as flowing in the opposite direction to the arrow you drew, with its size as a positive number.
Check the power balance if you have time: the total power supplied by the sources must equal the total dissipated in the resistors.

Worked ExampleA two-loop circuit

Two cells are connected in a circuit. Cell 1 has emf $12$ V and internal resistance $1.0$ Ω; cell 2 has emf $6.0$ V and internal resistance $0.50$ Ω. They are connected in parallel (both positive terminals joined) across an external resistor of $4.0$ Ω. Find the current in each branch.

Step 1 — Define the currents

Let $I_1$ flow out of cell 1, $I_2$ flow out of cell 2, and $I_3$ flow through the $4.0$ Ω resistor, all toward the junction and then out through the resistor.

Step 2 — Junction rule

$I_1 + I_2 = I_3 \qquad \text{(1)}$

Step 3 — Loop containing cell 1 and the resistor

$12 = I_1(1.0) + I_3(4.0) \qquad \text{(2)}$

Step 4 — Loop containing cell 2 and the resistor

$6.0 = I_2(0.50) + I_3(4.0) \qquad \text{(3)}$

Step 5 — Substitute (1) into (2) and (3)

From (2): $12 = I_1 + 4(I_1 + I_2) = 5I_1 + 4I_2$ … (2′) From (3): $6.0 = 0.5I_2 + 4(I_1 + I_2) = 4I_1 + 4.5I_2$ … (3′)

Step 6 — Solve the pair

From (2′): $I_1 = \dfrac{12 - 4I_2}{5} = 2.4 - 0.8I_2$

Substituting into (3′):

$6.0 = 4(2.4 - 0.8I_2) + 4.5I_2 = 9.6 - 3.2I_2 + 4.5I_2$ $6.0 - 9.6 = 1.3I_2$ $I_2 = \frac{-3.6}{1.3} = -2.77 \text{ A}$

$I_1 = 2.4 - 0.8(-2.77) = 2.4 + 2.22 = 4.62 \text{ A}$ $I_3 = I_1 + I_2 = 4.62 - 2.77 = 1.85 \text{ A}$

Step 7 — Interpret the signs

$I_1 = 4.6 \text{ A}, \quad I_2 = 2.8 \text{ A (flowing INTO cell 2)}, \quad I_3 = 1.9 \text{ A}$

The negative $I_2$ means current flows into cell 2 rather than out of it — the $12$ V cell is strong enough to drive current backwards through the $6.0$ V cell, charging it.

Worked ExampleChecking a solution

For the circuit above, verify the answer using the power balance.

Step 1 — Power supplied by cell 1

$P = \varepsilon I_1 = (12)(4.62) = 55.4 \text{ W}$

Step 2 — Power absorbed by cell 2

Current flows into it, so it absorbs energy:

$P = \varepsilon I_2 = (6.0)(2.77) = 16.6 \text{ W}$

Step 3 — Power dissipated in the resistances

$P_{r1} = I_1^2 r_1 = (4.62)^2(1.0) = 21.3 \text{ W}$ $P_{r2} = I_2^2 r_2 = (2.77)^2(0.50) = 3.84 \text{ W}$ $P_R = I_3^2 R = (1.85)^2(4.0) = 13.7 \text{ W}$

Step 4 — Balance

$\text{supplied} = 55.4 \text{ W}$ $\text{accounted for} = 16.6 + 21.3 + 3.84 + 13.7 = 55.4 \text{ W} \;\checkmark$

$\text{Energy balances — the solution is consistent}$