# Given the resistance of $1\Omega,2 \Omega$ and $3 \Omega$ how will be combine them to get an equivalent resistance of $\large\frac{11}{5}$$\Omega \begin{array}{1 1} \text{R_1 and R_2 in series and the resultant in parallel with R_3} \\ \text{R_1 and R_3 in series and the resultant in parallel with R_2} \\ \text{R_2 and R_3 in parallel and the resultant series in with R_1} \\ \text{R_1 and R_2 in parallel and the resultant series with R_3} \end{array} ## 1 Answer Comment A) Solution : Let R_1= 1\Omega,R_2= 2 \Omega and R_3 =3 \Omega If R_2 and R_3 are in parallel. then \large\frac{1}{R_p} =\frac{1}{R_2} +\frac{1}{R_3} \large\frac{1}{2}+\frac{1}{3} =\frac{3+2}{6} \qquad= \large\frac{5}{6} R_p =\large\frac{6}{5}$$\Omega$
$R_p$ and $R_1$ are in series . hence
$R= R_p +R_1$
$\quad= \large\frac{6}{5} $$+1 \qquad= \large\frac{11}{5}$$\Omega$
Answer : $\text{$R_2$and$R_3$in parallel and the resultant series in with$R_1$}$