# What happens to the total resistance when a fourth resistor is connected in a series with three resistors?

Mar 16, 2015

Well we know that when a resistor is connected in series

${R}_{n} = {R}_{1} + {R}_{2} + {R}_{3.} \ldots$

So i am taking that the forth resistor has the same resistance as the first 3

$i . e . {R}_{1} = {R}_{2} = {R}_{3} = {R}_{4}$

Okay so lets say the increase % = Increase $/$original $\cdot 100$

$= {R}_{4} / \left({R}_{1} + {R}_{2} + {R}_{3}\right) \cdot 1 00$

given that

${R}_{1} = {R}_{2} = {R}_{3} = {R}_{4}$

We can rewrite as
$= {R}_{4} / \left(3 {R}_{4}\right) \cdot 100$
$= \frac{1}{3} \cdot 100$

$\therefore$Resistance increases by 30.333..... %