This tool calculates the voltage across each of five (5) resistors in series.

Enter the Input Voltage **Vin**, and resistor values **R1, R2, R3**, **R4 **and **R5**. Use the drop down menu to select the units.

**Formula**

**V1=Vin*R1/(R1+R2+R3+R4+R5)**

**V2=Vin*R2/(R1+R2+R3+R4+R5)**

**V3=Vin*R3/(R1+R2+R3+R4+R5)**

**V4=Vin*R4/(R1+R2+R3+R4+R5)**

**V5=Vin*R5/(R1+R2+R3+R4+R5)**

The largest voltage drop will be across the largest resistor value. Conversely, the smallest voltage drop will be across the smallest.

If you want to calculate the voltage drop across three resistors in series, set R4=0, R5=0. In the equations above V4 and V5 will be zero (no voltage drop across a short circuit) and effectively this is reduced to a three resistor divider network.

????♀️The series resistors have the same current flowing through them. What about resistors that are connected in parallel?

**Example Calculation**

With R1=1Ω, R2=2Ω, R3=1Ω, R4=7Ω, R5=10Ω, and Vin=12V, the voltages

- V1 = 0.57 V
- V2 = 1.14 V
- V3 = 0.57 V
- V4 = 4 V
- V5 = 5.71 V

The voltage drop is proportional to the resistance. V2 is twice V1 as the impedance R2 = 2*R1.

The sum of the 5 voltages = the input voltage. As a check you can see that 0.57+1.14+0.57+4+5.71 = 12 V

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