# Ohm's Law

Summary: Ohm's Law describes the relationship between resistance, voltage, and current.

See the Video. |

There is a good description of Ohm's Law on Wikipedia.

## Ohm's Law

**Ohm's Law** is a statement of the relationship between Voltage, Current, and Resistance in an electrical circuit. It can be represented by this diagram:

Note: You may also see Voltage represented by the letter **E**. This stands for "Electromotive Force," an older name for what is now called Voltage. So, you may see the equations as:

- E =I × R (The animation below uses the European "
**U**" for Voltage) - I =E ÷ R
- R =V ÷ I

### Examples

One common question is: "How much resistance is needed for an LED?

To determine this, you begin with the data sheet for the LED in question. Specifically, the values for the continuous forward current, I_{FWD}, and forward voltage V_{FWD}.

For example, the data sheet lists I_{FWD} as 30mA maximum, V_{FWD} at 20mA is from 3 to 3.6V.

The power source is 16VDC. The maximum voltage across the LED, V_{FWD}, is 3.6V. For safety, the V_{FWD} will be 3.3V. The series resistor (R_{series}) will be determined by the 3.3V at 20mA rating. R_{series} will have the difference between 16 and 3.3V across it.

VR_{series} = 16 − 3.3 = 12.7V

Using R = V÷I, calculate the value:
12.7 ÷ I_{FWD} = 12.7 ÷ 20mA

- = 635 Ohms

- Using I= E ÷ R: 12.7 ÷ 635 = 0.02A
- Using E = I × R, 0.02 × 635 = 12.7V

There is no 635-ohm resistor available on the market, so the closest value would be 680 ohms.

Using 680 ohms, the current is 12.7 ÷ 680 = 19mA.

### Resistors in Series and Parallel

#### Series

Total Resistance equals the **sum** of the resistances.

#### Parallel

The total resistance of resistors connected in parallel is the reciprocal of the sum of the reciprocals of the individual resistors.

1/(1/R_{1} + 1/R_{2} … + 1/R_{n})

When resistors are wired in parallel, the total resistance will be less than that of the lowest resistor's value.

### Power Dissipation

The series resistor must be able to handle the power that will flow through it. Using the above calculations, the result is found as follows:

Watts = I^{2} × R
= 0.2^{2} × 635
= .0004 × 635
= 0.25 W

In this case, a ¼W resistor or better will suffice.

## Ohm's Law Explained