## Resistors in Series

6 September 2013 - Basic Electronics, Passive Components, Resistor - Hypsine Electronics - No Comments

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 VB is the voltage of the battery or power supply.

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

VR1 + VR2 = VB   and   I = IR1 = IR2   =>   VB = I * (R1 + R2)

Now we can find the voltage on R1 or R2

VR1 = I * R1 and VR2 = I * R2

Using the Ohm’s Law in circuit 2:

VB = I * Rs   =>   VB = Vs   =>   R1 + R2 = Rs

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:

Rs = R1 + R2 + R3 + … + Rn

## Voltage Divider Rule

Note in circuit 1 that:

VR1 = I * R1 and I = Vs/Rs where Rs = 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
• VB = 9 Volt from a battery I = V/R => I = VB/R1+R2 => I = 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

VR1 = 1000Ω * 0.0006923A = 0.6923V

VR2 = 12000Ω * 0.0006923A = 8.3076V

Using the voltage divider rule

VR1 = (Vs * R1)/Rs => VR1 = (9 * 1000)/13000 = 0.69V

VR2 = (Vs * R2)/Rs => VR2 = (9 * 12000)/13000 = 8.30V