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Circuits and Kirchhoff's laws

Every circuit rule you have ever used comes out of two conservation laws. Charge is conserved at every junction. Energy is conserved round every loop. Series, parallel and power all follow from those two sentences.

Year 12AQA 3.5.1.4 CIE 9.2.3, 10.1.1, 10.1.2, 10.2.1, 10.2.2, 10.2.3, 10.2.4, 10.2.5, 10.2.6, 10.2.7 OCR A 4.1.1(g), 4.2.1(a), 4.2.1(b), 4.2.5(a), 4.2.5(b), 4.3.1(a), 4.3.1(b), 4.3.1(c), 4.3.1(d), 4.3.1(e), 4.3.1(f)EDEXCEL 34, 35, 36, 37

Builds on Current, charge and the direction problem.

IN THIS TOPIC

Recall the standard circuit symbols, and draw and interpret a circuit diagram built from them.
Apply conservation of charge at junctions and conservation of energy round loops.
Handle the series and parallel rules for current, pd and resistance, cells in series and identical cells in parallel included.
Analyse a circuit carrying more than one source of e.m.f., adding the emfs that drive the same way round and subtracting the one that opposes.
Pick the right energy or power equation, E = IVt and the three forms of P, from what the circuit shares.

COMMON MISCONCEPTION

Current gets used up on the way round.

The symbols, before the laws

Every circuit in this unit arrives as a diagram, and a diagram is a language with a fixed vocabulary. CIE prints a list of twenty-seven symbols in its syllabus and examines two things about it: recalling the symbols, and drawing and interpreting circuits built from them. OCR sets the same list, taken from the ASE booklet. AQA and Edexcel never test the symbols on their own, and then use them in every circuit question they set. The vocabulary is worth an hour whichever board you sit.

FIG. 1The sources, the wiring and the components that go in the loop. A cell is one long thin plate and one short thick one, and the long thin plate is the positive terminal; a battery is two or more cells in a row, and a d.c. power supply is drawn the same way with its terminals marked. CIE prints a second form of the battery as well, a single long and short pair with a break drawn in the middle, standing for the same thing.

Three pairs on that chart are worth separating deliberately. A variable resistor has two terminals and an arrow across it, and it sets a current; a potentiometer has three, with the arrow coming in square to the element, and it sets a share of a pd. A thermistor takes a line across the element ending in a flat foot; an LDR takes two arrows pointing in at it, light arriving. A diode's triangle points the way conventional current is let through; an LED adds two arrows pointing away, light leaving.

FIG. 2The meters, and the components a circuit drives. Each meter is a circle with a letter inside it; the oscilloscope gets a screen and a trace instead, and both of its leads come out of the same side because it is connected across a pair of points.

A meter is a circle with a letter in it. A for an ammeter, V for a voltmeter, G for a galvanometer, and M for a motor, which is no meter at all but is drawn from the same family. A generator also takes a G, so nothing in the glyph separates it from a galvanometer. Where it sits does. A generator sits where a source would, driving the circuit; a galvanometer sits in a branch, detecting currents too small for an ammeter to notice.

Then the conventions for drawing one. Wires are straight lines meeting at right angles, and components sit on the straight runs rather than at the corners. A filled blob marks wires that are joined, and wires that cross without a blob are not connected. An ammeter goes in series with whatever it is counting, and is taken as having zero resistance; a voltmeter goes across whatever it is comparing, and is taken as having infinite resistance. Label every source with its emf and every component with its value.

FIG. 3One circuit, read. The ammeter is part of the loop, so the 0.30 A it reads is the current in the lamp. The voltmeter hangs on a branch of its own between two blobs either side of the lamp, so the 6.0 V it reads is the pd across the lamp, and the ratio of the two readings is the lamp's 20 Ω.

WORKED EXAMPLE

Where the meters go

A lamp runs from a battery, and both the current through it and the pd across it are to be measured. Say where each meter goes, and what happens if the two are swapped.

An ammeter measures what passes, so it goes in series with the lamp, in the loop itself. Taken as having zero resistance, it changes nothing about the circuit it has joined.

A voltmeter compares two points, so it goes in parallel with the lamp, across it. Taken as having infinite resistance, it draws no current from the circuit it is connected across.

Swap them and the ammeter, with no resistance, short circuits the lamp, while the voltmeter, with infinite resistance, blocks the loop. The lamp goes out and neither reading means anything. Placement is part of the measurement, as much as the reading itself.

The two conservation laws underneath

Circuit analysis rests on two statements known as Kirchhoff's laws, and both are best remembered as the conservation principles they are. Take conservation of charge first. Charge cannot pile up or vanish at a junction, so the total current flowing in equals the total flowing out.

Then conservation of energy. Each coulomb comes back from its trip round a loop with nothing left over, so the energy per coulomb supplied by the sources equals the energy per coulomb dropped across the components. Every series and parallel rule below is one of those two applied to a particular arrangement, written in the symbols from the charts above.

Series circuits

A series loop offers exactly one path, so conservation of charge leaves the current nowhere to divide. The same current flows at every point. An ammeter reads the same wherever you insert it, which settles the question of whether current gets used up on the way round.

FIG. 4One loop, one current. The pds across the components add up to the emf of the cell: 2.0 V plus 4.0 V accounts for all 6.0 V.

Conservation of energy round the loop means the pds across the components add up to the supply's emf. Divide that statement by the shared current and the resistance rule appears.

R_{T} = R_{1} + R_{2} + R_{3} + \ldots

ON THE AQA DATA SHEET

Cells in series behave the same way. Their emfs add, so two 1.5 V cells make a 3 V battery.

Parallel circuits

Parallel branches connect the same two points, so every branch sits across the same pd. The current, meanwhile, obeys the junction rule: it splits between the branches and recombines, with the totals matching exactly.

FIG. 5The current splits at the junction, 3.0 A into 2.0 A and 1.0 A, and the branches share one pd. Charge in equals charge out.

Adding the branch currents at the shared pd gives the resistance rule.

\frac{1}{R_{T}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} + \ldots

ON THE AQA DATA SHEET

The combined resistance always comes out smaller than the smallest branch, since adding a branch adds another path for the current without removing any. Identical cells in parallel give the emf of a single cell, no more, but they share the current between them, so each supplies less current and the battery lasts longer.

WORKED EXAMPLE

Combining a parallel pair

Find the combined resistance of 12 Ω and 6.0 Ω in parallel.

For two resistors, product over sum is quickest. R = (12 × 6.0)/(12 + 6.0) = 72/18 = 4.0 Ω.

Check it against the rule for parallel combinations. A parallel combination always comes out smaller than the smallest branch, and 4.0 Ω sits below 6.0 Ω.

An answer of 18 Ω (the sum) or 9 Ω (the average) fails that check instantly, so internalise the check even when the arithmetic feels easy.

More than one source in the loop

Every circuit so far has had one source in it, and nothing in either law says it must. Conservation of energy round a loop counts joules per coulomb, so the rule is that the emfs met on a trip round the loop, taken with their signs, equal the pds dropped across the components on the same trip. The sign comes from the terminal you reach first. Walk into a source at its negative terminal and come out at its positive one and it is lifting each coulomb, so its emf counts positive; walk into it the other way round and it is taking energy back out, so it counts negative.

Two cells nose to tail with both pushing the same way round the loop therefore add, which is why two 1.5 V cells make 3.0 V. Turn one of them round and it opposes the other, and the loop is driven by the difference of the two emfs. Nothing about that is a wiring fault. A cell driven backwards by a larger emf is a cell being charged, and the energy the current forces into it is stored chemically rather than dissipated, which is why it never appears in any I²R term.

WORKED EXAMPLE

A charger and the battery it is charging

A 12 V charger is connected to a 9.0 V rechargeable battery with their emfs opposing, through a 3.0 Ω protective resistor. Find the current, and say where the charger's power goes.

The two emfs oppose, so the loop is driven by 12 − 9.0 = 3.0 V, and that is the only pd available to push a current through the resistor.

I = 3.0/3.0 = 1.0 A, driven in the direction set by the net emf, so it enters the battery at its positive terminal. That is the direction that charges it.

Now the energy, per second. The charger supplies IV = 1.0 × 12 = 12 W. The resistor turns I²R = 1.0 × 3.0 = 3.0 W into heat, and the remaining 9.0 W, which is 9.0 V times 1.0 A, goes into the battery's chemistry. Kirchhoff's second law is that sentence divided through by the current.

Sources that are not in the same loop take one more step. Put two cells side by side, each with a resistor of its own, both feeding a shared load, and no single equation covers the circuit. Label an unknown current in each branch, use the junction rule to write the third current as the sum of the other two, and then write one loop equation for each independent loop. That leaves as many equations as unknowns, and solving them is the analysis.

GUIDED PRACTICE

Two unequal cells, one load

A 6.0 V cell in series with a 1.0 Ω resistor and a 3.0 V cell in series with a 1.0 Ω resistor are wired side by side across a shared 2.0 Ω load. Taking I₁ and I₂ out of the 6.0 V and 3.0 V branches, find both currents and the pd across the load.

Show the working

The junction rule sends I₁ + I₂ through the load, so its pd is V = 2.0(I₁ + I₂). Each branch then gives a loop equation: 6.0 = 1.0I₁ + V and 3.0 = 1.0I₂ + V.

Substituting I₁ = 6.0 − V and I₂ = 3.0 − V into the first line gives V = 2.0(9.0 − 2V), so 5V = 18 and V = 3.6 V.

Then I₁ = 2.4 A and I₂ = −0.6 A. The minus sign is the answer, not an error: 0.6 A is running into the 3.0 V cell, which the stronger cell is charging while it drives 1.8 A through the load. Guessing a current's direction wrongly does not matter, because the algebra returns it as a sign.

One warning about mixing unequal sources, since the algebra above hides it. The stronger cell drives a current backwards through the weaker one whether or not the weaker one is designed to take it, and a dry cell that cannot be charged simply heats up. Real sources also have a resistance built into them that no circuit diagram shows, and that resistance is the subject of the emf and internal resistance lesson two on from here.

Energy and power

A pd of V drives energy through a component at a rate set by the current. Over a time t the energy transferred is

E = IVt

NOT ON THE AQA DATA SHEET: LEARN IT

and the rate of transfer, the power, comes in three interchangeable forms via R = V/I:

P = IV = I^{2}R = \frac{V^{2}}{R}

ON THE AQA DATA SHEET

Choose the form built from the quantities the circuit shares. Series components share a current, so $I^{2}R$ compares them directly, and the biggest resistance dissipates the most. Parallel branches share a pd, so $V^{2}/R$ does the comparing, and there the smallest resistance dissipates the most. Same physics, opposite conclusion.

GUIDED PRACTICE

Inside a kettle's rating plate

A kettle is rated 3.0 kW at 230 V. Choose the power formula that uses what you are given, then find the element's resistance and the current drawn. No looking until you have both numbers.

Show the working

P and V are what you have, so use P = V²/R. That gives R = 230²/3000 = 18 Ω.

Then I = P/V = 3000/230 = 13 A, which puts a kettle near the top of what a domestic plug is rated for. Choosing the formula that matches the givens beats rearranging through ones that do not.

INDEPENDENT PRACTICE

The same heaters, rewired

Two identical 6.0 Ω heating elements, whose resistance you may take as constant, are connected to a 12 V supply, first in series, then in parallel. Find the power of each element in both arrangements.

Show the working

In series the total is 12 Ω, so 1.0 A flows everywhere and each element dissipates I²R = 1.0 × 6.0 = 6.0 W.

In parallel each element sees the full 12 V, so each dissipates V²/R = 144/6.0 = 24 W.

Four times the power in parallel, and the factor of four leans on that constant resistance. Filament lamps do not behave that way: a filament running cooler in series drops in resistance, so real lamps in series dissipate more than the fixed-R quarter, though parallel still gives far more power. Household wiring is parallel for that reason, and old fairy lights, wired in series, glowed gently and all went out when one filament broke.

ASSESSMENT FOCUS

CIE and OCR examine the symbols themselves, and they do it both ways round. Given the glyph, name it; given the name, draw it. Learn the chart in both directions.
A blob means joined and no blob means crossing. In a circuit you have drawn yourself, an ammeter in the loop and a voltmeter across the component are marked as often as the answer they lead to.
Current is not used up. An ammeter reads the same at every position in a series circuit, and saying so with the conservation-of-charge reason is a standard mark.
A parallel combination comes to less than its smallest branch. Any answer above that has slipped, almost always at the final reciprocal, the keypress exam scripts miss more than any other.
With two sources in one loop, settle their directions before anything else. Emfs driving the same way round add, opposing emfs subtract, and the current flows in the direction set by the net emf. A source the current is pushed backwards through is storing energy rather than dissipating it, so give it its own EI term and keep it out of the I²R heating.
When sources sit in different branches, do not look for a series or parallel shortcut that is not there. Name a current in each branch, use the junction rule once, then write one loop equation per loop and solve. A negative answer is a direction guessed backwards and nothing worse.
Pick the power form by the shared quantity. Series shares I, so reach for $I^{2}R$ ; parallel shares V, so reach for $V^{2}/R$ .
The sheet gives you the power forms. E = IVt you assemble yourself from a power and a time, so carry it in your head with the rest of the electricity equations.
AQA will not set a circuit needing simultaneous equations for the currents. If your working heads that way, you have missed a series or parallel simplification.

CHECK YOURSELF

A 6.0 Ω and a 3.0 Ω resistor are connected in parallel, and the pair is in series with a 4.0 Ω resistor across a 12 V supply. Find the current drawn from the supply and the pd across the parallel pair.

Show a hint

Collapse the parallel pair first, then treat the circuit as a simple series loop.

Show the answer

Parallel pair first. $\frac{1}{R} = \frac{1}{6.0} + \frac{1}{3.0}$ gives R = 2.0 Ω, smaller than either branch, as it must be.

Total resistance is now 2.0 + 4.0 = 6.0 Ω, so the supply delivers I = 12 / 6.0 = 2.0 A.

Across the pair, V = IR = 2.0 × 2.0 = 4.0 V, which leaves 8.0 V across the 4.0 Ω resistor. The loop's 12 V is fully accounted for.

Charge is conserved at every junction.

Energy is conserved round every loop.

Series shares the current.

Parallel shares the pd.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

20 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

Or read them with their mark schemes on the circuits and kirchhoff's laws questions page.

6 flashcards on this topicDefinitions, off-sheet equations and a spot-the-error card, scheduled by spaced repetition in your browser.Revise with flashcards

CHECK YOUR PROGRESS

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Recall the standard circuit symbols, and draw and interpret a circuit diagram built from them.
Apply conservation of charge at junctions and conservation of energy round loops.
Handle the series and parallel rules for current, pd and resistance, cells in series and identical cells in parallel included.
Analyse a circuit carrying more than one source of e.m.f., adding the emfs that drive the same way round and subtracting the one that opposes.
Pick the right energy or power equation, E = IVt and the three forms of P, from what the circuit shares.

Open the full revision checklist to track your progress across the whole unit.

Circuits and Kirchhoff's laws

The symbols, before the laws

The two conservation laws underneath

Series circuits

Parallel circuits

More than one source in the loop

Energy and power

Question workbook

Question workbook

Answer book