# Voltage Division or Voltage Divider

The principle of voltage division is defined as the source voltage v is divided among the resistors in direct proportion to their resistances the larger the resistance, the larger the voltage drop.

The basic idea behind the voltage divider is to assign a portion of the total voltage to each resistor.

VT is divided into IR voltage drops that are proportional to the series resistance values. Each resistance provides an IR voltage drop equal to its proportional part of the applied voltage.

This formula can be used for any number of series resistances because of the direct proportion between
each voltage drop V and its resistance R.

By the circuit configuration shown one can divide off any voltage desired ( Vout), less than the supply voltage E, by adjusting R1 R2 and R3

output of the voltage divider is

The voltage across any resistor Ri ( Ri i =1,2,…..n ) in a series circuit is equal to the applied voltage ( E ) across the circuit multiplied by a factor

The two resistors are in series since the same current i flows in both of them. Applying Ohm’s law to each of the resistors, we obtain

V1 = iR1 and V2=iR2

If we apply KVL to the loop (moving in the clockwise direction), we have

-V+V1+V2 = 0

V = V1 +V2 = i(R1+ R2)

It can be written as V = iReq

Implying that the two resistors can be replaced by an equivalent resistor Req that is, Req = R1+ R2

The equivalent resistance of any number of resistors connected in series is the sum of the individual resistances.

To determine the voltage across each resistor in Fig and obtain

## Series Voltage Dividers

Lets understand Series Voltage Divider through an example

Step 1 : calculate the equivalent resistance of the circuit

Req = R1+ R2 = 1k + 999k = 1000k

Step 2: Voltage across the resistance R1 is

Step 3: Voltage across the resistance R2 is

Step 3 : Lets check that the voltage obtained across the resistance R1 and R2 is correct or not

Apply KVL V1+V2 = 1V + 999V =1000V

so it is equal to the apply voltage so our KVL equation satisfied that -V+V1+V2 = 0