Answer
$18\frac{6}{13}\; \Omega$.
Work Step by Step
The given values are
$R_1=40 \;\Omega$
$R_2=60 \;\Omega$
$R_3=80 \;\Omega$
Total resistance in parallel circuit:-
$\Rightarrow \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$
Substitute all values.
$\Rightarrow \frac{1}{R}=\frac{1}{40}+\frac{1}{60}+\frac{1}{80}$
At the right hand LCD is $240$.
$\Rightarrow \frac{1}{R}=\frac{6}{240}+\frac{4}{240}+\frac{3}{240}$
Simplify.
$\Rightarrow \frac{1}{R}=\frac{6+4+3}{240}$
$\Rightarrow \frac{1}{R}=\frac{13}{240}$
$\Rightarrow R=\frac{240}{13}$
$\Rightarrow R=18\frac{6}{13}\; \Omega$.
