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: