Answer
$3\frac{1}{5}\; \Omega$.
Work Step by Step
The given values are
$R_1=6 \;\Omega$
$R_2=12 \;\Omega$
$R_3=24 \;\Omega$
$R_3=48 \;\Omega$
Total resistance in parallel circuit:-
$\Rightarrow \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}$
Substitute all values.
$\Rightarrow \frac{1}{R}=\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}$
At the right hand LCD is $48$.
$\Rightarrow \frac{1}{R}=\frac{8}{48}+\frac{4}{48}+\frac{2}{48}+\frac{1}{48}$
Simplify.
$\Rightarrow \frac{1}{R}=\frac{8+4+2+1}{48}$
$\Rightarrow \frac{1}{R}=\frac{15}{48}$
$\Rightarrow R=\frac{48}{15}$
$\Rightarrow R=\frac{16}{5}$
$\Rightarrow R=3\frac{1}{5}\; \Omega$.
