## Resistors in Series

In this article we are talking about the resistors in series, how to calculate the equivalent resistance and about the voltage divider. Consider the following picture:

resistors in series and the equivalent resistance (click to enlarge)

In circuit 1 we have 2 resistors R1 and R2 connected in series and in circuit 2 is shown the equivalent resistor Rs. I is the current that flows through the resistors and V_{B} is the voltage of the battery or power supply.

Using the Kirchhoff’s voltage law (KVL) in circuit 1:

Vand_{R1}+ V_{R2}= V_{B}I = I=>_{R1}= I_{R2}V_{B}= I * (R1 + R2)

Now we can find the voltage on R1 or R2

Vand_{R1}= I * R1V_{R2}= I * R2

Using the Ohm’s Law in circuit 2:

V=>_{B}= I * R_{s}V=>_{B}= V_{s}R1 + R2 = R_{s}

**The equivalent total resistance (Rs) for resistors in series is the sum of the individual resistors.** In our case Rs = R1 + R2, but we can extend it for any number of them. Let’s assume that we have n resistors:

R_{s}= R1 + R2 + R3 + … + Rn

## Voltage Divider Rule

Note in circuit 1 that:

Vand_{R1}= I * R1I = Vs/RswhereRs = R1+R2

Vx = (Vs * Rx)/Rs(voltage divider rule)

where Vx is the voltage on Rx.

## Example of how to calculate resistors in series

As you can see in circuit 1 where the resistor R1 and resistor R2 are connected in series the current I has the same value thru R1 and R2. Only the voltages are different depending on the values of the resistors. For example let’s consider:

- R1 = 1KΩ and R2 = 12KΩ
- V
_{B}= 9 Volt from a battery

I = V/R=>I = V=>_{B}/R1+R2I = 9V/1KΩ+12KΩ

we convert KΩ in Ω

I = 9V/1000Ω+12000Ω=>I = 9/13000 = 0.00069A = 0.69mA

and now we can calculate the voltage on R1 and R2

V_{R1}= 1000Ω * 0.0006923A = 0.6923V

V_{R2}= 12000Ω * 0.0006923A = 8.3076V

Using the voltage divider rule

V=>_{R1}= (Vs * R1)/RsV_{R1}= (9 * 1000)/13000 = 0.69V

V=>_{R2}= (Vs * R2)/RsV_{R2}= (9 * 12000)/13000 = 8.30V