# Question #efb97

Oct 23, 2016

Let the three resistances (resistors) ${R}_{1} , {R}_{2} \mathmr{and} {R}_{3}$ be connected in series as shown in the figure below Let ${R}_{s}$ be the resistance of the combination.

We see that as there is only one path for the current to flow. Therefore, the current through each of the resistors is the same.

${I}_{1} = {I}_{2} = {I}_{3} = I$

By Ohm’s law, the potential differences across the three resistors is,

${V}_{1} = I {R}_{1} , \text{ " V_2 = IR_2, " } {V}_{3} = I {R}_{3}$

Also, the voltage drops across the resistors must add up to the total voltage supplied by the battery, we have

$V = {V}_{1} + {V}_{2} + {V}_{3}$
$\implies V = I {R}_{1} + I {R}_{2} + I {R}_{3}$
$\implies V = I \left({R}_{1} + {R}_{2} + {R}_{3}\right)$ ........(1)

As Ohm's Law must also be satisfied for the complete circuit, we have
$V = I {R}_{s}$ ......(2)

Comparing (1) and (2) we have

${R}_{s} = {R}_{1} + {R}_{2} + {R}_{3}$

In general, the equivalent resistance of resistors when connected in series is the sum of all resistances.

${R}_{\text{equivalent}} = \sum {R}_{i}$