Current, resistance and Ohm's law

An unloaded battery shows one voltage; connecting a load gives a lower one. Charge does not disappear along the way. We will see how the source supplies energy and why some is transferred inside the source itself.

Particles in a metal undergo thermal motion even without a battery, but opposite transfers of charge cancel on average. The field adds directed motion of free charge carriers: electric current. In a metal the carriers are electrons; conventional current follows positive charge motion and is opposite to electron drift. Sustained current in an ordinary circuit needs a closed path and an energy source. An open path cannot accumulate charge at its end indefinitely.

Imagine a counting surface across the wire: how much charge passes through it in time t? With q in C and t in s, steady current I=q/t is in amperes. Doubling the charge transferred in the same time doubles current; doubling the time for the same q halves average current. For varying current, q/t gives the interval average, not necessarily the value every second. Here q is charge transferred through a cross-section, not all the charge in the wire.

Two wires may carry the same current although one is much thinner. To compare current through equal areas, define current density j=I/S, in A/m². This is the uniform value when current is evenly distributed and perpendicular to the cross-section; otherwise it is the average normal component. At fixed I, halving S doubles j. Current counts the total transfer; density describes how tightly it is distributed over area.

As charge passes through a resistor, it transfers energy through the field's action. Voltage U=A/q, in volts, is the electric field's work per unit transferred charge. A larger U means more energy transferred per coulomb. The resistor does not consume charge: in a steady circuit as much arrives as leaves. A fall in electrical energy therefore cannot be explained by fewer coulombs passing along a series circuit.

Apply different voltages to a conductor and measure current. If R=U/I stays constant under unchanged conditions, the conductor is ohmic. R is resistance; 1\,\text{Ом}=1\,\text{В/А}. At a given U, greater R means smaller current. Constancy of the ratio is an experimental property of this operating regime, not an automatic consequence of the definition. Strong heating, for example, changes the conductor's state and can make I(U) cease to be a straight line.

For an ohmic circuit section without an emf source, the proportionality is Ohm's law I=U/R. Keeping temperature and state unchanged, double U and current doubles; at fixed U, double R and current halves. At zero voltage, this section sustains no directed current. The formula relates quantities; it does not require fixed I. Constant resistance fixes U/I, so a source can change current by changing voltage.

Lengthen a uniform wire without changing its cross-section: charge travels through a longer section, requiring greater voltage for the same current. Add an identical wire alongside it to provide another path and reduce resistance. These observations give R=\rho l/S, with l in m, S in m² and resistivity ρ in ohm metres. ρ depends on material and temperature. Here the letter denotes an electrical property, not mass density.

The formula uses area, so doubling wire thickness does not automatically double S. For a circular cross-section of diameter d, S=\pi d^2/4. Double d at unchanged length and material: S becomes four times greater and R four times smaller. Doubling length alone doubles R. Such imagined changes help predict the result and catch substitution of diameter for area.

A table may give ρ in ohm mm²/m. Then use mm² for cross-section and still m for length. Since 1\,\text{м}^2=10^6\,\text{мм}^2, we obtain 1\,\text{Ом}\cdot\text{м}=10^6\,\text{Ом}\cdot\text{мм}^2/\text{м}. This is the same material property in different units. Before calculating, check that area units cancel in ρl/S; otherwise a hidden factor of a million spoils the result despite the correct physics formula.

An electrostatic field alone cannot repeatedly supply charge with energy around a complete circuit: its work around a closed path is zero. Chemical processes in a battery separate charges against their tendency to equalise. Denote their non-electrostatic work by A_{\text{ст}}. Electromotive force \mathcal E=A_{\text{ст}}/q measures energy supplied by the source per unit charge and explains how sustained transfer to the load is possible.

The name ‘electromotive force’ can mislead: its definition divides joules by coulombs, so its unit is the volt, not the newton. A source with an emf of 6 V supplies 6 J per coulomb passing through it. Doubling transferred charge doubles the source's work, but need not double its emf. A voltage value alone does not specify the current: we must also know the circuit carrying the energy.

In a simple closed circuit, the same current passes through external resistance R and source internal resistance r, both in ohms. The source's energy per coulomb is shared between them: \mathcal E=IR+Ir. Thus I=\mathcal E/(R+r), Ohm's law for a complete steady circuit with one source. At fixed resistances, doubling emf doubles current; increasing either resistance reduces it. The source is part of the electrical path too.

The external load voltage is U=IR=\mathcal E-Ir. The difference between emf and U is the internal drop Ir: energy transferred inside the source per coulomb. At fixed emf and r, increasing current lowers U. At very small current, Ir almost vanishes and voltage approaches emf; the ideal model r = 0 makes them equal. Voltage sag under load therefore reflects energy distribution, not lost charge.

If external R becomes negligible, the model current tends to I_{\text{кз}}=\mathcal E/r. Internal resistance limits it; it can be large while external voltage IR is small. As r → 0, the formal value grows without bound. This exposes the ideal-source model's failure in this limit, not an infinite real current. We treat emf and r as constant; substantial heating or changes in the source's state invalidate that condition.

A source has emf 6 V and internal resistance 1 ohm. It is connected to a 5-ohm resistor. Find the current and voltage across the resistor.

Current passes through both resistances: R+r=5+1=6 Ом. Use Ohm's law for the complete circuit.

I=ε/(R+r)=6/6=1 А.

Across the external resistor, U=IR=1×5=5 В.

Check the balance: the internal drop is Ir=1 В; 5 В+1 В=6 В.

I=1 А, U=5 В.

Current measures charge transfer; voltage is energy per charge. A resistor in a steady circuit does not consume coulombs.

Ohm's law with constant R describes an ohmic section under unchanged conditions; wire geometry enters through length and area.

Emf specifies source energy per coulomb; U = ЭДС − Ir gives the part available to the external load.

Next we will calculate how quickly a circuit transfers energy and how loads can be connected.