Current voltage resistance and Ohm law

Current $I$ is the charge passing through a conductor's cross-section per unit time: $I=\dfrac{\Delta q}{\Delta t}$; its unit is the ampere (A).

Voltage $U$ across a circuit section equals the work done by the electric field in moving a unit positive charge: $U=\dfrac{A}{q}$; its unit is the volt (V).

Resistance $R$ describes a conductor's opposition to current and depends on the material, length, and cross-sectional area: $R=\rho\dfrac{L}{S}$; its unit is the ohm ($\Omega$).

For a circuit section containing no EMF sources, Ohm's law relates these three quantities:

$$I=\frac{U}{R},\qquad U=IR,\qquad R=\frac{U}{I}.$$

If resistance is constant, the current–voltage characteristic $I(U)$ is a straight line through the origin: a larger slope corresponds to a smaller resistance.

For a series connection $R_{\text{total}}=R_1+R_2+\dots$, and for a parallel connection $\dfrac{1}{R_{\text{total}}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dots$.

A voltage $U=12\ \text{V}$ is applied across a resistor of resistance $R=4\ \Omega$. Find the current $I$ through the resistor.

A copper wire has length $L=20\ \text{m}$ and cross-sectional area $S=0{,}5\ \text{mm}^2$. The resistivity of copper is $\rho=1{,}7\cdot10^{-8}\ \Omega\cdot\text{m}$. Find the wire's resistance and the current in it at a voltage $U=6\ \text{V}$.

A wire of resistance $R$ is redrawn so that its length doubles while its volume and resistivity stay unchanged. Determine the new resistance and by what factor the current changes at the same voltage.

Simulation (safe format). Open an interactive Ohm's law model (for example, PhET "Ohm's Law" or the Falstad Circuit Simulator) and build a circuit section with a voltage source, a $10\ \Omega$ resistor, an ammeter (in series), and a voltmeter (in parallel with the resistor).

1. Vary the voltage from $0$ to $10\ \text{V}$ in $1\ \text{V}$ steps and record the current in a table.
2. Plot $I(U)$ and check that the points lie on a straight line through the origin.
3. Find the slope $k=\Delta I/\Delta U$ and compare $R=1/k$ with the resistor's nominal value.
4. Repeat for $R=20\ \Omega$ and explain how and why the slope changed.

Take all measurements only in the simulation or with battery power up to $12\ \text{V}$; never connect to a $220\ \text{V}$ mains supply.

Show how the quantities are connected: the circuit diagram fixes the meaning of $U$, $I$, and $R$, while the $I(U)$ graph makes Ohm's law visible and reveals the role of resistance.

On the left, a sketch of a circuit section: a voltage source, a resistor $R$, an ammeter in series with the resistor, and a voltmeter in parallel with it. On the right, a graph: voltage $U$ on the horizontal axis and current $I$ on the vertical axis; two straight lines through the origin with slopes $1/R_1$ and $1/R_2$, where $R_2>R_1$, so the second line is less steep.

A charger outputs $U=5.0\ \mathrm{V}$. During navigation the phone draws $I=1.5\ \mathrm{A}$; the equivalent resistance of the charging circuit is $R=U/I=5.0/1.5\approx3.3\ \Omega$. If at the same voltage the current drops to $0.5\ \mathrm{A}$, the circuit resistance becomes $R=5.0/0.5=10\ \Omega$.

This is a direct use of Ohm's law for a circuit section: at a given voltage the current is set by resistance, and a current drop lets you estimate increased resistance of the cable or contacts.

A car battery supplies $U=12\ \mathrm{V}$. The lamp filament at operating temperature has resistance $R=3.0\ \Omega$. By Ohm's law the current is $I=U/R=12/3.0=4.0\ \mathrm{A}$; in $t=30\ \mathrm{s}$ the charge through the filament is $Q=It=4.0\cdot30=120\ \mathrm{C}$.

The relation $I=U/R$ is used for the metal filament: battery voltage and filament resistance set the current. Heating changes the filament resistance, so Ohm's law gives an operating estimate rather than a constant value.

A kettle is plugged into a $U=230\ \mathrm{V}$ mains. The heating element resistance is $R=24\ \Omega$. The current is $I=U/R=230/24\approx9.6\ \mathrm{A}$. If the voltage drops to $220\ \mathrm{V}$ at the same $R$, the current becomes $I=220/24\approx9.2\ \mathrm{A}$.

This exercises Ohm's law for a circuit section: with nearly unchanged resistance, current is proportional to voltage, so a small mains voltage drop reduces the heater current.