Electric Circuits

A circuit is a closed loop. Charge is not used up as it goes round — the same charge that leaves the battery returns to it. What the components take from the charge is energy, not charge itself. Getting this straight early saves a great deal of confusion later, because almost every wrong answer in circuit work comes from imagining that current is consumed by a lamp.

The battery does not supply the charge either. The charges are already in the wires: they are the free electrons of the metal. The battery supplies the push that makes them drift in one direction, and the energy they carry.

Conventional current is drawn from the positive terminal round to the negative. Electrons actually drift the other way. Every rule you will use — the ammeter reading, the right-hand grip rule, the motor rule — is stated for conventional current, so use it and do not worry that the electrons disagree.

I

currentA

Q

chargeC

t

times

If current is the traffic, potential difference is the drop in energy the traffic suffers on the way through. A 12 V lamp gives up 12 joules for every coulomb that passes. This is why p.d. is always measured across a component and never through it: it is a difference between two points.

Resistance measures how hard a component makes that traffic work. It is defined by R = V/I, and that definition holds for every component, whether or not it obeys Ohm's law.

Ohm's law is the narrower statement that for a metallic conductor at constant temperature, current is directly proportional to potential difference. Leave out the temperature condition and the statement is simply false — a filament lamp is the standard counter-example.

V

potential differenceV

I

currentA

R

resistanceΩ

E

energy transferredJ

Everything in circuit work follows from two pairs of rules. Learn them as a table and most questions become bookkeeping rather than thinking.

In series there is one path. The same charge must pass through every component each second, so the current is the same everywhere. The energy each coulomb carries is shared out among the components, so the potential differences add up to the supply.

In parallel the charge has a choice of routes. Each branch is connected directly across the supply, so each gets the full potential difference. The currents in the branches add up to the current drawn from the supply, because charge is conserved at a junction.

Switch between series and parallel and watch the total resistance. In parallel it always falls below the smaller of the two — add a second identical resistor and the total halves, because you have given the current a second route.

That parallel resistance comes out smaller than either resistor surprises almost everyone the first time. The arithmetic is not the point; the picture is. Adding a parallel branch is like opening a second checkout in a shop. Neither till got faster, but more people get through per minute — so the overall resistance to the flow has dropped.

This gives you a free check on every parallel calculation. If your answer is bigger than the smallest resistor in the combination, it is wrong. Two 10 Ω resistors in parallel give 5 Ω, not 20 Ω. A 4 Ω and a 12 Ω give 3 Ω, which is less than 4.

The other trap is stopping one step early. The formula gives you 1/R, not R. Inverting at the end is a separate mark on almost every mark scheme, and forgetting it is the single most common error in this topic.

A mixed circuit is solved in a fixed order, and following it turns a hard-looking question into four easy ones. Reduce every parallel block to a single resistance. Add the series parts to get the total. Use the supply voltage and the total resistance to find the main current. Then work back outwards, using V = IR on each part in turn.

The step that catches people is the last one. Once you know the current through a parallel block, you can find the potential difference across the whole block, and that same p.d. sits across both branches. From there each branch current comes from I = V/R separately.

Never try to split the current between branches by guessing a ratio. If you must, remember it goes the opposite way to intuition: the smaller resistor takes the larger current.

Two resistors in series across a supply share the voltage in proportion to their resistances. That arrangement is a potential divider, and it is how nearly every sensor circuit is read.

Replace one of the two with a component whose resistance changes — a thermistor, whose resistance falls as it warms, or a light-dependent resistor, whose resistance falls as it brightens — and the output voltage now varies with temperature or light. That varying voltage is what a control circuit actually responds to.

The reasoning to write down in an exam is always the same two steps: state which resistance has changed and in which direction, then state that the component taking a larger share of the total resistance takes a larger share of the voltage.

V_out

output voltageV

V_in

supply voltageV

R₂

resistance across the outputΩ