Kirchhoff's Laws: Junction and Loop Rules with Examples

When Series and Parallel Rules Are Not Enough

Many circuits can be solved by combining resistors in series and parallel and then using Ohm's law. But as soon as a circuit has two batteries in different branches, or a network that cannot be reduced to simple series and parallel parts, those rules run out. Two laws stated by Gustav Kirchhoff in 1845 solve any circuit, however complicated. This tutorial covers Kirchhoff's laws as taught in PHY 102 and needed for WAEC and JAMB physics: the junction rule and the loop rule, the physics behind them, sign conventions, a step-by-step method, worked examples with two sources and with internal resistance, and how to interpret negative answers. It builds on the tutorial on current electricity.

Kirchhoff's First Law: the Junction Rule

The sum of the currents entering any junction equals the sum of the currents leaving it.

ΣI(in) = ΣI(out)

This is the conservation of charge. Charge cannot pile up at a junction or vanish from it, so whatever flows in each second must flow out.

Worked example 1

At a junction, currents of 3 A and 2 A flow in. One wire carries 4 A out. What current flows in the remaining wire, and in which direction?

In: 3 + 2 = 5 A. Out: 4 A + I. So I = 1 A, flowing out of the junction.

The familiar rule for parallel circuits, that the total current equals the sum of the branch currents, is simply the junction rule applied to the point where the branches split.

Kirchhoff's Second Law: the Loop Rule

Around any closed loop in a circuit, the sum of the e.m.f.s equals the sum of the potential drops (IR).

ΣE = ΣIR

This is the conservation of energy. A charge that travels all the way round a loop returns to its starting potential, so the energy it gains from the sources must equal the energy it gives up in the resistors.

The rule for series circuits, that the p.d.s across the components add up to the supply e.m.f., is the loop rule applied to a single loop.

Sign Conventions

Choose a direction to travel round each loop (clockwise, say), and assume a direction for every current. Then:

An e.m.f. is positive if travelling the chosen way takes you through the source from its negative terminal to its positive terminal, and negative if you pass through it from positive to negative.
An IR term is positive if you travel through the resistor in the same direction as the assumed current, and negative if you travel against it.

You do not need to guess the current directions correctly. If you assume the wrong way, the answer simply comes out negative, meaning that the current actually flows the opposite way, and its size is still correct.

A Step-by-Step Method

Label the current in each branch (I₁, I₂, …) with an assumed direction.
Apply the junction rule to reduce the number of unknowns, for example I₃ = I₁ + I₂.
Choose enough independent loops: one equation is needed for each remaining unknown.
Write the loop rule for each loop, using the sign conventions.
Solve the simultaneous equations.
Interpret any negative values, then find the p.d.s or powers that the question asks for.

Worked Example 2: Two Sources in Parallel Branches

A circuit has three branches connected between two junctions, X (top) and Y (bottom):

The left branch contains a 12 V cell (positive terminal towards X) in series with a 2 Ω resistor.
The right branch contains a 6 V cell (positive terminal towards X) in series with a 4 Ω resistor.
The middle branch contains a 6 Ω resistor.

Neglect the internal resistance of the cells. Find the current in each branch.

Step 1. Let I₁ flow up through the left branch towards X, and I₂ flow up through the right branch towards X. Both then flow down through the middle resistor, so by the junction rule, the middle current is I₃ = I₁ + I₂.

Step 2: the left loop (the left branch and the middle branch). Going up through the 12 V cell, from − to +, then down through the 6 Ω resistor in the direction of I₃:

12 = 2I₁ + 6(I₁ + I₂), so 12 = 8I₁ + 6I₂ … (1)

Step 3: the right loop (the right branch and the middle branch), in the same way:

6 = 4I₂ + 6(I₁ + I₂), so 6 = 6I₁ + 10I₂ … (2)

Step 4: solve. From (1), I₁ = (12 − 6I₂) ÷ 8. Substituting into (2) gives 6 = 6(12 − 6I₂)/8 + 10I₂, so 48 = 72 − 36I₂ + 80I₂, which gives 44I₂ = −24.

So I₂ = −6/11 ≈ −0.55 A, and then I₁ = (12 + 36/11) ÷ 8 = 21/11 ≈ 1.91 A, and I₃ = 15/11 ≈ 1.36 A.

Step 5: interpret. I₂ is negative, so the current in the right branch actually flows downwards, into the 6 V cell's positive terminal. The stronger 12 V cell is charging the 6 V cell. That is exactly what happens when a car battery is connected to a charger, or a flat battery is jump-started.

Check with the left loop: 2(1.909) + 6(1.364) = 3.82 + 8.18 = 12.0 V.

The p.d. across the middle resistor is 6 × 15/11 ≈ 8.18 V, so junction X is 8.18 V above junction Y.

Worked Example 3: Opposing E.M.F.s in One Loop

A 14 V charger with an internal resistance of 0.4 Ω charges a 12 V battery with an internal resistance of 0.1 Ω. The connecting leads have a total resistance of 0.5 Ω. The positive terminals are connected together. Find the charging current, and the power delivered by the charger that is stored in the battery.

Going round the loop in the direction of the charger's push, the 14 V counts as positive and the 12 V as negative, because the current is forced into the battery's positive terminal:

14 − 12 = I(0.4 + 0.1 + 0.5), so I = 2 ÷ 1.0 = 2 A.

Power stored in the battery as chemical energy = E × I = 12 × 2 = 24 W. The charger supplies 14 × 2 = 28 W in total, and the remaining 4 W is wasted as heat in the resistances, I²R = 4 × 1.0 = 4 W.

Internal Resistance and Terminal P.D.

A real cell of e.m.f. E and internal resistance r, delivering current I, has a terminal p.d. V = E − Ir. Kirchhoff's loop rule includes this automatically if you draw the internal resistance as a separate resistor r in series with an ideal cell. When a cell is being charged, the current flows into it through its positive terminal, and its terminal p.d. is larger than its e.m.f.: V = E + Ir. In example 3, the battery's terminal p.d. while charging is 12 + 2 × 0.1 = 12.2 V.

Why the Laws Always Work

The junction rule follows from the conservation of charge, and the loop rule from the conservation of energy, two of the most fundamental principles in physics.
In a network with b branches, j junctions and therefore l = b − j + 1 independent loops, the junction rule gives j − 1 independent equations and the loop rule gives l, which together are enough to find all b currents.
Electrical engineers solve circuits with thousands of components using the same ideas, organised as matrix equations and solved by computer (as in the SPICE simulation program).

Common Mistakes with Kirchhoff's Laws

Inconsistent signs for e.m.f.s and IR terms. Decide the direction of travel for each loop, and apply the conventions every time.
Using the same current in two different branches. Each branch between two junctions has its own current.
Writing too many loop equations that are not independent, or too few to find every unknown.
Treating a negative answer as an error. It means only that the current flows the other way.
Forgetting internal resistance when the question gives it.

Kirchhoff's Laws: Practice Questions

Four wires meet at a junction. Currents of 5 A and 1.5 A flow in, and 2 A flows out in a third wire. Find the current in the fourth wire.
State the physical principle on which each of Kirchhoff's laws is based.
Repeat worked example 2 with a 10 V cell and 1 Ω resistor in the left branch, a 5 V cell and 1 Ω resistor in the right branch, and 2 Ω in the middle branch. Find all three currents.
A 9 V battery with internal resistance 0.5 Ω and a 3 V battery with internal resistance 0.5 Ω are connected in series, opposing each other, with a 5 Ω resistor. Find the current.
A cell of e.m.f. 1.5 V and internal resistance 0.2 Ω is charged with a current of 0.5 A. Find its terminal p.d.

Answers: 1. 5 + 1.5 − 2 = 4.5 A, flowing out. 2. The first law is based on the conservation of charge, and the second on the conservation of energy. 3. The loop equations are 10 = 3I₁ + 2I₂ and 5 = 2I₁ + 3I₂, giving I₁ = 4 A, I₂ = −1 A (so 1 A flows down through the 5 V cell, charging it) and a middle current of 3 A. 4. (9 − 3) ÷ (0.5 + 0.5 + 5) = 1 A. 5. V = 1.5 + 0.5 × 0.2 = 1.6 V.

Conclusion

Kirchhoff's laws solve any circuit. The junction rule, ΣI(in) = ΣI(out), comes from the conservation of charge, and the loop rule, ΣE = ΣIR around any closed loop, comes from the conservation of energy. Label each branch current, apply the junction rule, write one loop equation for each remaining unknown using consistent signs, and solve the simultaneous equations. A negative current simply flows the other way, often showing that one source is charging another. The rules for series and parallel circuits are special cases of these two laws.