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.

V_{T} 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 ( V_{out}), less than the supply voltage E, by adjusting R_{1} R_{2} and R_{3}

**output of the voltage divider** is

The voltage across any resistor R_{i} ( R_{i} 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

V_{1} = iR_{1} and V_{2}=iR_{2}

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

-V+V_{1}+V_{2} = 0

V = V_{1} +V_{2} = i(R_{1}+ R_{2})

It can be written as V = iR_{eq}

Implying that the two resistors can be replaced by an equivalent resistor R_{eq} that is, R_{eq} = R_{1}+ R_{2}

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

R_{eq} = R_{1}+ R_{2} = 1k + 999k = 1000k

Step 2: Voltage across the resistance R_{1} is

Step 3: Voltage across the resistance R_{2} is

Step 3 : Lets check that the voltage obtained across the resistance R_{1} and R_{2} is correct or not

Apply KVL V_{1}+V_{2} = 1V + 999V =1000V

so it is equal to the apply voltage so our KVL equation satisfied that -V+V_{1}+V_{2} = 0