Derive the relation $R = R_{1} + R_{2} + R_{3}$ when three resistors $R_{1}, R_{2}$ and $R_{3}$ are connected in series in an electric circuit.

(N/A) Consider three resistors of resistances $R_{1}, R_{2}$ and $R_{3}$ connected in series to a cell of potential $V$ as shown in the figure. Since the three resistors are connected in series,the current $I$ flowing through each of them is the same.
By Ohm's law,the potential drop across each resistor is given by:
$V_{1} = IR_{1}, V_{2} = IR_{2}$ and $V_{3} = IR_{3}$
Since $V$ is the total potential in the circuit,by the conservation of energy,we have:
$V = V_{1} + V_{2} + V_{3}$ $...(1)$
Substituting the values of $V_{1}, V_{2}$ and $V_{3}$ in equation $(1)$,we get:
$V = IR_{1} + IR_{2} + IR_{3}$ $...(2)$
If $R_{S}$ is the equivalent resistance of the series combination,then by Ohm's law:
$V = IR_{S}$ $...(3)$
Comparing equations $(2)$ and $(3)$,we have:
$IR_{S} = IR_{1} + IR_{2} + IR_{3}$
Dividing both sides by $I$,we get:
$R_{S} = R_{1} + R_{2} + R_{3}$ $...(4)$
Thus,in a series combination,the equivalent resistance is the sum of the individual resistances.