Series connection

Place two resistors one after another. The first heats up and receives energy. Must the current become smaller after it, or does equal charge per second pass through both?

Trace the wire from the source: charge passes the first element, then the second, with no junction where it can split. This is a series connection. In steady operation, charge does not keep accumulating between elements. If more charge continually entered a section than left it, the accumulation would change the field and currents. After a brief settling period, the same charge flow therefore passes every element on the single path.

In the same time, equal net charge passes the cross-sections before the first resistor, between the resistors and after the second. Dividing by that time gives I = I₁ = I₂, in amperes. If the common current increases, it increases in every element. A lamp receives energy and may heat up, but does not destroy electrons. Equal current is therefore consistent with energy being transferred to each successive element.

Consider the electric field's work in transferring one coulomb through both resistors. The field does work in the first section and then the second; these works add. Thus work per coulomb, or voltage, adds: U = U₁ + U₂, in volts. The voltages need not be equal. At a fixed total U, a larger share for one section leaves a smaller share for the other. An ideal connecting wire has zero voltage drop, so contributes nothing to the sum.

Replace the pair by one resistance that carries the same I at the same U. For each ohmic resistor, U₁ = IR₁ and U₂ = IR₂. Hence U = I(R₁ + R₂), giving equivalent R = R₁ + R₂, in ohms. Adding a positive resistance increases the total; two identical resistors give twice the resistance of one. A connection with negligible resistance contributes practically zero.

The measurement follows the reasoning. Insert an ammeter in series at different points: it reads the same current. Connect a voltmeter across the first resistor, then the second, then both: the individual readings add to the total. The model assumes almost resistance-free wires and meters that hardly affect the circuit. If wire resistance is significant, voltage also drops across the wires and must be included in the sum.

Imagine cutting the wire anywhere in the chain. The path through both source terminals disappears, so steady current stops in every element. Changing one resistance also changes the common current and affects the whole chain. This is why independent household appliances cannot be supplied along one series path. Modern strings of lights can be more complex: having many lamps does not prove a purely series connection.

R₁ = 4 Ом, R₂ = 6 Ом, with a total voltage of 12 V. Find the current and voltages across the resistors.

First replace the two resistors by one to find the common current: R = 4 + 6 = 10 Ом. Then I = 12/10 = 1,2 А.

This current is the same in both elements. Thus U₁ = 1,2·4 = 4,8 В and U₂ = 1,2·6 = 7,2 В.

Check energy per coulomb: 4,8 + 7,2 = 12 В. At a common current, the larger resistance takes the larger voltage share.

I = 1,2 А; U₁ = 4,8 В, U₂ = 7,2 В. There is one current; the energy per coulomb is shared between two resistors.

Equal currents follow from the absence of branches and of continual charge accumulation.

Voltage addition expresses addition of the field's work; together with Ohm's law it gives resistance addition.

Breaking the single path switches off the whole chain; changing one R changes the common current.

Giving charge several paths creates a parallel connection with different rules.