Kirchhoff's Loop Rule: Study Guide (Physics 2 11.6)

Energy carried by moving charge

In Topic 10.5 you met electric potential difference as energy per unit charge. In a circuit this gives a simple way to track energy. When a charge q moves between two points with a potential difference ΔV between them, its electric potential energy changes by

ΔU_E = qΔV, or equivalently ΔV = ΔU_E / q

Positive charges (the carriers of conventional current) gain electric potential energy when they move to a higher potential and lose it when they move to a lower potential.

• Inside a battery, the battery does work on the charges, moving them from the negative terminal to the positive terminal. Non-electric energy (for example chemical energy) becomes electric potential energy. Each coulomb gains ℰ joules.
• In a resistor, charges move from high potential to low potential. Electric potential energy becomes thermal energy (or light, in a bulb).
• Along an ideal wire, there is no change in potential, so there is no energy change.

For example, when 0.40 C passes through a 1.5 V cell, it gains (0.40 C)(1.5 V) = 0.60 J of electric potential energy.

The loop rule

Pick any point in a circuit and give it a value of electric potential. Now follow a closed loop and come back to the same point. You must arrive at the same potential you started with, because a point has only one value of potential. So the rises and falls along the way must cancel:

ΣΔV = 0 around any closed loop

This is Kirchhoff’s loop rule. It is a statement of conservation of energy. Multiply every term by a test charge q: the charge gains energy in the batteries and loses energy in the resistors, and when it returns to its starting point its electric potential energy is back where it started. The energy given to it equals the energy it gave out.

Two features matter in practice:

• The rule applies to every closed loop, not only the one containing the battery. A loop through two parallel branches has no emf, so the potential differences across the branches must cancel: parallel branches have equal potential differences, which is the rule you used in Topic 11.5.
• The rule says nothing about which loop you choose or where you start. Any closed path gives a correct equation.

Sign rules

Choose a direction to go round the loop. Then add a term for each element you cross:

You cross	In this direction	Term in ΣΔV
Battery (emf ℰ)	− terminal to + terminal	+ℰ
Battery (emf ℰ)	+ terminal to − terminal	−ℰ
Resistor R with current I	same direction as the current	−IR
Resistor R with current I	against the current	+IR
Internal resistance r	treat it like any resistor	−Ir or +Ir
Ideal wire or ideal ammeter	either	0

The resistor rule follows from the direction of conventional current: it flows from high potential to low potential through a resistor, so going with the current you go downhill in potential.

Potential–position graphs

A useful picture of the loop rule is a graph of electric potential against position as you go round a loop. Choose one point (often the negative terminal of the battery) as V = 0. Then:

• The graph rises by ℰ across each ideal emf crossed from − to +.
• It falls by IR across each resistor crossed in the direction of the current. A larger resistance in series gives a larger fall.
• It is flat along ideal wires.
• It ends at the same value it started at. That is the loop rule.

Figure 2 shows the graph for Worked example 1.

Worked example 1: one loop, with a potential graph

Question. In Figure 1 a battery of emf 12 V and internal resistance 1.0 Ω is connected to R₁ = 3.0 Ω and R₂ = 8.0 Ω in a single loop. (a) Use the loop rule to find the current. (b) Taking point a (the negative terminal) as 0 V, find the potential at b and c, and sketch the graph of potential against position. (c) How much energy does 2.0 C of charge gain and lose in one trip round the loop?

1. (a) Go round a → b → c → a, in the direction of the current. Cross the emf from − to + (+ℰ), then r, R₁ and R₂ each in the direction of the current: +12 V − I(1.0 Ω) − I(3.0 Ω) − I(8.0 Ω) = 0
2. So 12 V = I(12 Ω), and I = 1.0 A.
3. (b) Start at a: V_a = 0. Crossing the emf raises the potential to 12 V. Crossing r lowers it by (1.0 A)(1.0 Ω) = 1.0 V, so V_b = 11 V (this is the terminal voltage, ℰ − Ir).
4. Crossing R₁ lowers it by (1.0 A)(3.0 Ω) = 3.0 V, so V_c = 8.0 V.
5. Crossing R₂ lowers it by (1.0 A)(8.0 Ω) = 8.0 V, back to 0 V at a. The loop closes, as the rule requires.
6. (c) Using ΔU = qΔV with q = 2.0 C: the charge gains (2.0 C)(12 V) = 24 J in the emf. It loses 2.0 J in r, 6.0 J in R₁ and 16 J in R₂, a total of 24 J.

Answer. (a) 1.0 A. (b) V_b = 11 V, V_c = 8.0 V; graph as in Figure 2. (c) It gains 24 J and loses 24 J.

Check. The larger resistor has the larger fall in potential, because the current is the same in every element of a single loop. A voltmeter between b and a reads 11 V, not 12 V: the terminal voltage is less than the emf when there is a current. If R₁ and R₂ swapped places, the graph would fall by 8.0 V first and then by 3.0 V, and V_c would be 3.0 V. The shape changes, but the size of each fall, and the return to 0 V at a, do not.

Worked example 2: a loop with no battery

Question. In Figure 3 a battery (ℰ = 9.0 V, r = 0.50 Ω) is in series with R₁ = 2.0 Ω and two parallel branches between junctions P and Q. Branch X contains R₂ alone. Branch Y contains R₃ = 2.0 Ω in series with an unknown resistor Rₓ. Ammeters (not drawn) show 2.0 A in the battery, 1.2 A in branch X and 0.80 A in branch Y. Find the terminal voltage, R₂ and Rₓ.

1. Terminal voltage. Cross the battery from − to +: ℰ − Ir = 9.0 V − (2.0 A)(0.50 Ω) = 8.0 V.
2. Loop through the battery and branch X. Going with the current: +9.0 V − (2.0 A)(0.50 Ω) − (2.0 A)(2.0 Ω) − ΔV_X = 0, so ΔV_X = 9.0 − 1.0 − 4.0 = 4.0 V.
3. R₂. R₂ = ΔV_X / I_X = 4.0 V ÷ 1.2 A = 3.3 Ω (3.33 Ω).
4. Loop with no battery: P → X → Q → Y → P. Go down through X with its current (−4.0 V), then up through Y against its current (+IR for each resistor): −4.0 V + (0.80 A)(2.0 Ω) + (0.80 A)Rₓ = 0.
5. So (0.80 A)Rₓ = 4.0 V − 1.6 V = 2.4 V, and Rₓ = 3.0 Ω.

Answer. Terminal voltage 8.0 V; R₂ = 3.3 Ω; Rₓ = 3.0 Ω.

Check. Branch Y has 1.6 V + 2.4 V = 4.0 V across it, the same as branch X, as the no-battery loop requires. The branch currents 1.2 A + 0.80 A add to the 2.0 A in the battery, which you will justify formally with the junction rule in Topic 11.7.

Common misconceptions

• “The first resistor after the battery uses up most of the voltage.” The share of each resistor in a single loop depends on its resistance, not on its position. The current is the same everywhere in the loop.
• “The loop rule only works for the loop with the battery.” It holds for every closed loop. Loops without a source are often the quickest way to an unknown.
• “Potential is used up, so the current gets smaller round the loop.” The potential falls round the loop; the current in a single loop does not change.
• Mixing sign conventions half-way round. Choose a direction once. Each resistor term is −IR going with its current and +IR going against it.
• Forgetting internal resistance. It is a resistor in the loop and needs its own −Ir term.
• “The terminal voltage always equals the emf.” Only with no current in the battery.
• Using a different zero of potential part-way through. Only differences matter, but you must keep one reference point for the whole problem.

Where this leads

The loop rule is one of two tools for analysing any circuit. Topic 11.7 adds the junction rule, from conservation of charge, so you can set up and solve equations for circuits with several unknown currents. Continue with Topic 11.7, Kirchhoff’s Junction Rule. First test yourself with the practice questions, then use the revision notes and the topic checklist to consolidate.