In this lesson, you will create a smaller voltage from a bigger voltage with series resistors.

## Resistors in series When two resistors, R1 and R2, are connected in series, the total resistance is:

$Rtotal = R_{1} + R_{2}$

For example, when $R_{1}$ is 1 ohm, and $R_{2}$ is 2 ohm, then $Rtotal$ is:

$Rtotal = R_{1} + R_{2} = 1 ohm + 2 ohm = 3 ohm$

When you have $n$ resistors, simply add all the values:

$Rtotal = R_{1} + R_{2} + R_{3} + ... + R_{n}$

## Current and series resistors

When a voltage is applied to series resistors, the current that flows the resistors is same. This is true even when values of resistors are different, or the number of resistors increases.

$Itotal = I_{1} = I_{2}$ $Itotal = I_{1} = I_{2} = I_{3} = ... = I_{n}$

## Voltages in series resistors

When $R_{1}$ is 10 ohm and $R_{2}$ is 20 ohm, and 3.3 V is applied to the series resistors, the current is:

$I = { V \over Rtotal } = { V \over { R_{1} + R_{2} } } = { 3.3 \over { 10 + 20 } } = 0.11 (A)$

110 mA, $I_{1}$, is flowing through each resistor. $R_{1}$ is 10 ohm. The voltage across $R_{1}$, $V_{1},$ is:

$V_{1} = I_{1} R_{1} = 0.11 \times 10 = 1.1 (V)$

110 mA, $I_{2}$, is flowing through each resistor. $R_{2}$ is 20 ohm. The voltage across $R_{2}$, $V_{2},$ is:

$V_{2} = I_{2} R_{2} = 0.11 \times 20 = 2.2 (V)$

The total of voltage drops is:

$V_{1} + V_{2} = 1.1 + 2.2 = 3.3 (V)$

## Other lessons

Other lessons in Electronics Basic Course: