Question

A technician has only two resistances coils. By using them singly, in series or in parallel, he is able to obtain the resistance $3,4,12$ and $16$ ohms. The resistance of the two coils are

• A. $6$ and $10$ ohms
• B. $4$ and $12$ ohms
• C. $7$ and $9$ ohms
• D. $4$ and $16$ ohms

Answer 1

Answer: (B)

$4$ and $12$ ohms

With two resistance $R_1$and $R_2$ we have the three possible combinations possible

$\left(1\right)R_1$ or $R_2$ if they are used individually

let $R_2$ be the larger of the two.

$\left(2\right)\frac{R_1{R}_2}{R_1+R_2}$ if used in parallel

This is the smallest value. It is $<R_1$ and $<R_2$

$\left(3\right)(R_1+R_2)$ if used in series.

This is the largest value. So it is $>R_1$ and $>R_2$

Hence from the given value

$(R_1+R_2)=16\Omega$

$\frac{R_1{R}_2}{(R_1+R_2)}=3\Omega$

So, $R_1$ and $R_2$ are $4\Omega$ and $12\Omega$ respectively.