Question

calculate the equivalent resistance in both parallel and series 6ohm,12 ohm,18 ohm ​

Answer 1

Explanation:

parallel 6+12+18=36/2=18

series = 6+12+18=36

Answer 2

Answer: 36Ω in series and 11/36Ω in parallel.

Given:

$\sf Resistor\:1(R_1)=6\Omega$

$\sf Resistor\:2(R_2)=12\Omega$

$\sf Resistor\:3(R_3)=18\Omega$

To find:

$\sf\text{Equivalent resistance in both series and parallel.}$

Explanation:

$\sf\underline{\textbf{Equivalent Resistance in Series:}}$

$\sf\text{We know that,}$

$\boxed{\sf R_{eq(series)} = R_1+R_2+R_3+...}$

$\sf\text{Substituting the values in the above formula, we get:}$

$\sf R_{eq(series)} = 6\Omega+12\Omega+18\Omega.$

$\boxed{\sf R_{eq(series)} = {36\Omega}}$

$\sf\underline{\textbf{Equivalent Resistance in Parallel:}}$

$\sf\text{We know that,}$

$\boxed{\sf R_{eq(parallel)} = \frac{1}{R_1} +\frac{1}{R_2} +\frac{1}{R_3} +...}$

$\sf\text{Substituting the values in the above formula, we get:}$

$\large\sf R_{eq(parallel)} = \frac{1}{6\Omega} +\frac{1}{12\Omega} +\frac{1}{18\Omega}$

$\large\sf R_{eq(parallel)} = \frac{6+3+2}{36}$

$\boxed{\sf R_{eq(parallel)} = \frac{11}{36}\Omega}$

$\sf\text{Therefore, equivalent resistance in} \:series\: is\: 36 \Omega\:and\: \frac{11}{36}\Omega\: in\: parallel.$