Physics › Electricity › Circuits and Kirchhoff's laws
Circuits and Kirchhoff's laws
Every circuit rule you have ever used is one of two conservation laws wearing a disguise. Charge is conserved at every junction and energy is conserved round every loop, and series, parallel, and power all follow from those two sentences.
Builds on Current, charge and the direction problem.
IN THIS TOPIC
- Apply conservation of charge at junctions and conservation of energy round loops.
- Use the series and parallel rules for current, pd and resistance, including cells in series and identical cells in parallel.
- Choose and use the energy and power equations, E = IVt and the three forms of P.
WHAT YOU PROBABLY THINK
Current gets used up on the way round.
Two conservation laws in disguise
Circuit analysis rests on two statements, known as Kirchhoff's laws, and AQA words them as the conservation principles they are. Conservation of charge: charge cannot pile up or vanish at a junction, so the total current flowing in equals the total flowing out. Conservation of energy: each coulomb returns from a 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 these two dressed for a specific situation.
Series circuits
In a series loop there is 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 is the cleanest possible refutation of current being consumed.
Conservation of energy round the loop means the pds across the components add up to the supply's emf, and dividing that statement by the shared current gives the resistance rule:
Cells in series behave the same way: their emfs add, which is why 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.
The resistance rule follows from adding the branch currents at the shared pd:
The combined resistance always comes out smaller than the smallest branch, because adding a branch adds another path for current without removing any. Identical cells in parallel give the emf of a single cell, but they share the current between them, so each works less hard and the battery lasts longer.
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
and the rate of transfer, the power, comes in three interchangeable forms via R = V/I:
Choose the form built from the quantities the circuit shares. Series components share a current, so I2R compares them directly: the biggest resistance dissipates the most. Parallel branches share a pd, so V2/R does the comparing, and there the smallest resistance dissipates the most. Same physics, opposite winners.
THE EXAM BIT
- Current is not used up. In a series circuit an ammeter reads the same at every position, and saying so, with the conservation-of-charge reason, is a standard mark.
- A parallel combination's total resistance is less than the smallest branch. Any answer above it has slipped, usually by forgetting the final reciprocal.
- The parallel rule computes 1/RT. Pressing the reciprocal button once more at the end is the step exam scripts most often miss.
- Pick the power form by the shared quantity: series shares I, so use I2R; parallel shares V, so use V2/R.
- E = IVt is not on the data sheet, unlike the power forms. It is one of the few electricity equations you must carry in your head.
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: 1R = 16.0 + 13.0 gives R = 2.0 Ω, smaller than either branch, as it must be.
Total resistance: 2.0 + 4.0 = 6.0 Ω, so the supply current is I = 12 / 6.0 = 2.0 A.
The pd across the pair: V = IR = 2.0 × 2.0 = 4.0 V, leaving 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.
No animated video for this topic yet; these notes stand alone. InkPhysics on YouTube.