Answer
$R_2=4{\Omega}$
Work Step by Step
As total resistance is $3{\Omega}$, which is less than $12{\Omega}$, this indicates that the two resistors are connected in parallel. Therefore,
$\frac{1}{R_{total}}=\frac{1}{R_1}+\frac{1}{R_2}$
$\frac{1}{3}=\frac{1}{12}+\frac{1}{R_2}$
$\frac{1}{3}=\frac{R_2+12}{12R_{2}}$
By cross multiplication, we get
$12R_2=3(R_2+12)$
$12R_2-3R_2=36$
$9R_2=36$
$R_2=4{\Omega}$
