When two resistors of resistances $R_{1}$ and $R_{2}$ are connected in parallel, the net resistance is $3\, \Omega$. When connected in series, its value is $16\, \Omega$. Calculate the values of $R_{1}$ and $R_{2}$

When $R _{1}$ and $R _{2}$ are connected in parallel, net resistance $\left( R _{p}\right)$ is given by

$\frac{1}{ R _{ P }}=\frac{1}{ R _{1}}+\frac{1}{ R _{2}}$

Therefore, we have

$R _{ P }=\frac{ R _{1} R _{2}}{ R _{1}+ R _{2}}=3$ $...(1)$

When $R _{1}$ and $R _{2}$ are connected in series, net resistance $\left( R _{ S }\right)$ is given by

$R _{ S }= R _{1}+ R _{2}=16$ $....(2)$

From equations $(1)$ and $(2),$ we have

$\frac{R_{1} R_{2}}{16}=3$ or $R_{1} R_{2}=48$ $....(3)$

Using equations $(2)$ and $(3),$ we have

$R_{1}\left(16-R_{1}\right)=48$

Solving for $R_{1},$ we have $R_{1}=4 \Omega$ or $12 \Omega$. Therefore, the resistances of two resistors are $4 \Omega$ and $12 \Omega$