Question

A resistor of 8 $\Omega$ is connected in parallel with another resistor X. The resulting resistance of the combination is 4.8 $\Omega$. What is the value of the resistor X?

Answer 1

Given:
$R_1$ = 8 $\Omega$
$R_{eq}$ = 4.8 $\Omega$
$R_2$ = ?

We know that,

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$

$\Rightarrow \frac{1}{4.8}=\frac{1}{8}+\frac{1}{X}$

$\Rightarrow \frac{10}{48}=\frac{1}{8}+\frac{1}{X}$

$\Rightarrow \frac{1}{X}=\frac{10}{48}-\frac{1}{8}$

$\Rightarrow \frac{1}{X}=\frac{10-6}{48}$

$\Rightarrow \frac{1}{X}=\frac{4}{48}$

$\Rightarrow \frac{1}{X}=\frac{1}{12}$

$\therefore X=12\Omega$