# Ohm's Law

## Voltage, current and resistance

Ohm's law states that

*"the current through a conductor between two points is directly proportional to the potential difference or voltage across the two points, **and inversely proportional to the resistance between them".*

Ohm's law can be expressed as

*I = U / R (1)*

*where *

*I = current (ampere, A)*

*U = electrical potential (volts, V)*

*R = resistance (ohms, Ω)*

### Example - Ohm's law

A *12 volt* battery supplies power to a resistance of *18 ohms*. The current in the elctrical circuit can be calculated as

I= (12 volts) / (18 ohm)

= 0.67 ampere

### Equivalent Expressions

Ohm's law *(1)* can also be expressed as

*U = R I (2)*

*or *

*R = U / I (3)*

### Example - Electric Circuit Resistance

A current of *1 ampere* is flowing through a *230 V* electric circuit. From the diagram above this indicates resistance

*R ≈ 220 Ω *

This can alternatively be calculated with Ohm's law

*R = (230 V) / (1 A)*

* = 230 Ω*

### Example - Ohm's Law and Multiples and Submultiples

Currents, voltages and resistances in electric circuits may often be very small or very large - so multiples and submultiples are often used.

The voltage required applied to a *3.3 kΩ* resistor to generate a current of *20 mA* can be calculated as

*U = (3.3 kΩ) (1000 Ω/kΩ) (20 mA) (10 ^{-3} A/mA)*

* = 66 V*

### Power

Electric power can be expressed as

*P = U I *

* = R I ^{2} *

* = U ^{2} / R (4)*

*where *

*P = electrical power (watts, W)*

#### Example - Power Consumed

The power consumed in the *12V* electrical circuit above can be calculated as

*P = (12 volts) ^{2} / (18 ohm)*

* = 8 W*

#### Example - Power and Electrical Resistance

A *100 W *electric light bulb is connected to a *230 V* supply. The current flowing can be calculated by reorganizing *(4)* to

* I = P / U *

* = (100 W) / (230 V)*

* = 0.43 ampere*

The resistance can be calculated by reorganizing *(4)* to

*R = U^{2 }/ P*

* = (230 V)^{2 } / (100 W) ^{}*

* = 529 Ω*

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