Question

derive the equation for resultant resistance of resistors in series

Answer 1

An arrangement of resistors arranged in series is described as a set of resistors attached one after the other in an electric circuit. The current  flowing through them is same. The voltage differs depending on the value of resistance of each resistor. As the current flowing through each of the resistors is the same, the total resistance offered by the resistors is calculated by the sum of the individual resistances.

Let $R_{s}$ be the Resultant resistance. 
Let $R_{1}$, $R_{2}$, $R_{3}$, ......, $R_{n}$  be the resistors arranged in series 

Thus the resultant resistance offered by the individual resistors is : -

$R_{s}$ =  $R_{1}$ + $R_{2}$ + $R_{3}$ + ......+ $R_{n}$

This the equation of resistors arranged in series

Answer 2

The combined resistance of a number of resistance connected in series is calculated by using the law of combination of resistances in series.
According to law of combination of series, the combined resistance of any number of resistances connected in series is equal to the sum of the individual resistances.

Derivation:

Let us consider two resistances R1 and R2 connected in series.
A battery of V volts has been applied to the ends of this series connection.
Let us assume the potential difference across resistance R1 is V1 and R2 be V2 .
Let V be the applied voltage.
So, the potential difference across the two resistance will be :
V=V1+V2
by ohm's law
V=IR------(equation 1 )
Since same current flows I, flows through the circuit when resistors are connected in series:
I=I1=I2
By applying Ohm's separately to both the resistances we get :
V1=I xR1
V2=I X R2
now by substituting the values of V1 and V2 in equation 1 we get,
IxR=IxR1+IxR2
IxR=I[R1+R2]
cancelling I from both the sides , we get
R=R1+R2

Hence derived.