Question

Three unequal resistors in parallel are equivalent to a resistance $1\Omega$ . If two of them are in the ratio $1:2$ and if no resistance value is fractional, what will be the largest of the three resistance in ohm?

Answer 1

Answer: 6

Let the resistances be $R_1,R_2$ and $R_3$
$\therefore \frac{R_1}{R_2}=\frac{1}{2}⟹R_1=k,R_2=2k$
In parallel $\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$
$\frac{1}{1}=\frac{1}{k}+\frac{1}{2k}+\frac{1}{R_3}$
$\frac{1}{R_3}=1-\frac{3}{2k}$
$=\frac{2k-2-1}{2k}=\frac{2k-3}{2k}$
$R_3=\frac{2k}{2k-3}$
If $k=1$ , then $R_3$ is found to be negative, which is impossible.
If $k=2$ , then $R_1=2,R_2=4R_3=4$
$R_2=R_3$ , not satisfying the condition of the question that all resistance are unequal.
$Ifk=3,then{R}_1=3,R_2=6$
$R_3=2\Omega$
$\therefore$ Largest resistance $=6\Omega$