Junctions Count Charge

Imagine a steady DC circuit after the first brief switch-on moment has passed. At a junction, charge carriers arrive and leave continuously. If more charge arrived than left, charge would build up at the junction; if more left than arrived, the junction would become depleted. Neither is the steady situation we are modelling.

So for a junction:

This is a charge-conservation statement. Current is the rate of flow of charge, so conserving charge at a junction means conserving the flow rate into and out of the junction.

For example, if a current of $0.75\ \text{A}$ reaches a junction and one branch carries $0.20\ \text{A}$, the other branch must carry

The tempting wrong idea is that current is "used up" as it moves through components. A resistor transfers energy from the charges to the surroundings, but it does not remove charge from the circuit. In a single unbranched path, the current is the same everywhere because every coulomb that enters a component must leave it.

If you were checking this with meters, an ammeter placed before a junction should read the sum of ammeters placed in the outgoing branches. The meter readings are evidence for charge conservation, not a separate rule to memorise.

Loops Count Energy

Potential difference is energy transferred per unit charge. A $12\ \text{V}$ supply gives each coulomb $12\ \text{J}$ of energy. As that same coulomb moves around a complete loop, it must transfer that $12\ \text{J}$ to components before returning to the supply.

So for a complete loop:

This is an energy-conservation statement. It is often called the loop rule, but the physics is simply that energy per unit charge cannot disappear or appear from nowhere.

Consider two resistors in series across a $12\ \text{V}$ supply. There is one route for the charge, so the same current passes through both resistors. The supply p.d. is shared:

If the measured p.d.s are $4.0\ \text{V}$ and $8.0\ \text{V}$, the sum is $12.0\ \text{V}$, so the readings fit energy conservation. The larger p.d. is not because more current has gone into that component; in series the current is the same. The larger p.d. means each coulomb transfers more energy in that component.

Now compare two branches in parallel across the same supply. Each branch connects between the same two points of the circuit, so each complete route from the positive terminal to the negative terminal includes the full supply p.d. If the supply is $12\ \text{V}$, each parallel branch has $12\ \text{V}$ across it.

The tempting wrong idea is that p.d. behaves like a flow and gets split at every junction. It does not. Current splits at a junction because charge flow has alternative routes. P.d. is energy transferred per unit charge between two points, so it depends on which two points the component is connected across.

Series Resistance From One Current

Two resistors are in series when the same current must pass through each one. The equivalent resistance is the single resistance that would take the same current from the same supply p.d. as the whole combination.

Start with conservation before using the formula. For two series resistors:

because energy per unit charge is shared around the loop. The current $I$ is the same in both resistors because there is no junction between them where charge can split.

Using $R = V/I$, or $V = IR$:

where $R_\text{s}$ is the equivalent series resistance. Divide by $I$:

For more than two series resistors:

Here is the same reasoning in a numerical circuit. A $12.0\ \text{V}$ supply is connected to a $4.0\ \Omega$ resistor and an $8.0\ \Omega$ resistor in series.

The first decision is the topology: there is one path, so this is series. That tells us to conserve charge as the same current through both resistors, and to conserve energy as p.d.s that add around the loop.

The p.d. across each resistor comes from $V = IR$:

The result makes physical sense: the same current passes through both resistors, and the larger resistance gets the larger share of the energy transfer per coulomb. The p.d.s add to $12.0\ \text{V}$, matching the supply.

In an Edexcel "derive" response, the conservation step matters. Starting with $R_\text{s} = R_1 + R_2$ only quotes the result. A derivation shows why that result follows from $V = V_1 + V_2$, the same current in series, and $R = V/I$.

Parallel Resistance From Shared P.D.

Two resistors are in parallel when they are connected across the same two points. That means each branch has the same p.d. across it. The current splits between the branches, and the total current is the sum of the branch currents.

For two parallel resistors:

because charge is conserved at the junctions. Each branch has the same p.d. $V$, so using $I = V/R$:

where $R_\text{p}$ is the equivalent parallel resistance. Divide by $V$:

For more than two parallel branches:

A $6.0\ \Omega$ resistor and a $3.0\ \Omega$ resistor are connected in parallel across a $12.0\ \text{V}$ supply.

The first decision is that each branch is across the same two supply terminals, so each branch has $12.0\ \text{V}$ across it.

Charge conservation gives the supply current:

The equivalent resistance is therefore:

That $2.0\ \Omega$ value is less than either branch resistance. This is not a calculation accident. Adding a parallel branch gives charge an extra route. For the same p.d., the total current increases, so the equivalent resistance must decrease.

A quick guided check: if $12.0\ \Omega$ and $6.0\ \Omega$ are connected in parallel, the branch with $6.0\ \Omega$ has the larger current because both branches have the same p.d. The reciprocal equation gives

so

That passes the sense check: $4.0\ \Omega$ is less than the smallest branch resistance, $6.0\ \Omega$.

Choosing The Conservation Route

When a circuit combines series and parallel sections, do not begin by hunting for a memorised formula. Begin by asking what is conserved where.

Use this route:

1. Find any unbranched sections. In those sections, current is the same because charge has only one path.
2. Find components connected across the same two points. Across those branches, p.d. is the same because each branch has the same energy-per-charge drop between the same two points.
3. Combine parallel groups with the reciprocal equation.
4. Combine series groups by adding resistances.
5. Return to the physical quantities: total current from $I = V/R$, branch p.d.s from energy conservation, and branch currents from charge conservation.

Suppose a $2.0\ \Omega$ resistor is in series with a parallel pair of $6.0\ \Omega$ and $3.0\ \Omega$, all connected to a $12.0\ \text{V}$ supply.

The parallel pair is handled first:

Now the circuit is equivalent to $2.0\ \Omega$ in series with $2.0\ \Omega$:

So the supply current is:

The series $2.0\ \Omega$ resistor has:

That leaves $6.0\ \text{V}$ across the parallel pair, because p.d.s around the loop add to the $12.0\ \text{V}$ supply. Each parallel branch has $6.0\ \text{V}$ across it, so the branch currents are $1.0\ \text{A}$ through $6.0\ \Omega$ and $2.0\ \text{A}$ through $3.0\ \Omega$. They add to the $3.0\ \text{A}$ supply current, so the charge-conservation check works.

If this were an unfamiliar practical circuit, the same reasoning tells you where to place meters. Use ammeters in series with the supply and branches to test current sums. Use voltmeters across components or branch pairs to test p.d. sums. A stable DC circuit should give readings that agree within sensible meter resolution; large mismatches would suggest a wiring error, a non-ohmic or changing component, or an unsteady supply.