Ohm's Law — Physica

Equivalent forms

I = \frac{V}{R}

R = \frac{V}{I}

\vec{J} = \sigma \vec{E}

The simplest equation in electronics — three quantities, one line, and the entire foundation of circuit analysis.

Symbol	Name	SI unit	Dimension	Range
$V$	Voltageoutput Potential difference across the resistor	V	M·L²·T⁻³·I⁻¹	0 – 240
$I$	Current Electric current through the resistor	A	I	0 – 20
$R$	Resistance Electrical resistance of the component	Ω	M·L²·T⁻³·I⁻²	0.1 – 10000

Unit systems

Symbol	SI	CGS	Imperial
$V$	V	statV	V
$I$	A	statA	A
$R$	Ω	s/cm	Ω

Where it holds

Valid for ohmic materials at constant temperature. Non-ohmic devices (diodes, transistors, superconductors) do not obey this law. Resistance itself varies with temperature in real materials.

Dimensional analysis

$[V] = [I]\cdot [R] \to A\cdot \Omega = A\cdot (V$ /A $) = V \checkmark$

Discovery

Georg Simon Ohm · 1827

Ohm published his law in 'Die galvanische Kette', initially met with skepticism. It took decades before the physics community recognized the fundamental relationship between voltage, current, and resistance.

Try this

Why does a thin wire heat up more than a thick one carrying the same current?

A 12 V battery drives a current through a 4 Ω resistor. Find the current and power dissipated.

Research status: stable

Real-world applications

Every circuit analysis from phone chargers to power grids
Resistance-based temperature sensors (RTDs, thermistors)
Current-limiting resistors in LED circuits
Fuse and circuit breaker sizing

Common misconceptions

Ohm's law is not universal — it is an empirical law valid only for ohmic materials
Resistance is not always constant; it depends on temperature, frequency, and material state
Current does not get 'used up' in a resistor — it is the same entering and leaving

Experimental verification

Ohm verified using thermocouple voltage sources and a torsion galvanometer. Modern verification spans from microelectronics (MOSFET characterization) to superconducting transitions where R abruptly drops to zero.

Derivation

From the microscopic Drude model: electrons accelerate under E, collide with lattice ions with mean free time

\tau

.

Drift velocity

v_d = eE\tau /m_e

.

Current density

J = nev_d = (ne^{2}\tau /m_e)E = \sigma E

.

Integrating over a wire of length L and cross-section A gives

V = IR

where

R = L/(\sigma A)

.

Limiting cases

R \to 0

⟶ Short circuit

(I \to \infty for

finite V)Zero resistance means unlimited current — a dangerous short circuit in practice.

R \to \infty

⟶

I \to 0

(open circuit)Infinite resistance blocks all current flow.

V \to 0

⟶

I \to 0

No driving voltage means no current through a resistor.

What if…

What if the resistance doubles?

Current halves: $I = 12/8 = 1.5$ A. Power drops to $P = 1.5 \times 12 = 18\,\mathrm{W}$ — half the original power.

What if the wire heats up (resistance increases 20%)?

R becomes $4.8 \Omega$ , I drops to 2.5 A, $P = 30\,\mathrm{W}$ . Real resistors have temperature coefficients that cause this effect.

What if you use a superconductor

(R \to 0)

?

Current becomes limited only by the source's internal resistance. In an ideal circuit, $I \to \infty$ — a short circuit. Superconductors require careful current limiting.

1

Current through a resistor

Given ·

V:: 12
R:: 4

Find · I and power P

Steps

Identify: $V = 12\,\mathrm{V}$ , $R = 4 \Omega$
Apply Ohm's law: $I = V/R = 12/4 = 3$ A
Power dissipated: $P = IV = 3 \times 12 = 36\,\mathrm{W}$
Alternatively: $P = I^{2}R = 9 \times 4 = 36\,\mathrm{W} \checkmark$
Or: $P = V^{2}/R = 144/4 = 36\,\mathrm{W} \checkmark$

Result ·

I = V/R = 12/4 = 3

A;

P = IV = 3 \times 12 = 36\,\mathrm{W}

2

Voltage divider

Given ·

V in:: 9
R1:: 3000
R2:: 6000

Find · V_out across

R_{2}

Steps

Total resistance: R_total $= R_{1} + R_{2} = 3000 + 6000 = 9000 \Omega$
Current: $I = V$ _in / R_total $= 9 / 9000 = 1\,\mathrm{mA}$
Voltage across $R_{2}$ : $V_out = IR_{2} = 0.001 \times 6000 = 6\,\mathrm{V}$
Equivalently: $V_out = V$ _ $in \times R_{2}/(R_{1} + R_{2}) = 9 \times 2/3 = 6\,\mathrm{V}$
The voltage divides in proportion to resistance

Result ·

V_out = V

_

in \times R_{2}/(R_{1} + R_{2}) = 9 \times 6000/9000 = 6\,\mathrm{V}

joule heatingkirchhoffs voltage lawkirchhoffs current law