Question

show how would you join three resistors each of resistance 9 ohm so that the equivalent resistance of the combination 3.5 ohm and 6 Ohm ​

Answer 1

Have a look  at the picture below:

Answer 2

We know that

If two resistors, $R_{1}$ and $R_{2}$ are connected in a series combination then the equivalent resistance is the summation of their resistance.

$R=R_{1}+R_{2}$

If two resistors, $R_{1}$ and $R_{2}$ are connected in a parallel combination then the reciprocal of their equivalent  resistance equals the sum of the reciprocals of their individual resistances.

$\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$

To obtain the resultant resistance value of R=13.5(As shown in picture) we use the flowing combination as shown in figure-1 as attachment.

As per the picture first find the resultant resistance of $R_{1}$ and $R_{2}$ connected in parallel combination

$\frac{1}{R'}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\\\\\Rightarrow\frac{1}{R'}=\frac{1}{9}+\frac{1}{9}\\\\\Rightarrow\frac{1}{R'}=\frac{2}{9}\\\\\Rightarrow{R'}=\frac{9}{2}=4.5\Omega$

Now find the resultant resistance of $R'$ and $R_{3}$ connected in series combination

$\therefore R=R'+R_{3}=4.5+9=13.5$

To obtain the resultant resistance value of R=6 we use the flowing combination as shown in figure-2 as attachment.

As per the picture first find the resultant resistance of $R_{1}$ and $R_{2}$ connected in series combination.

$R'=R_{1}+R_{2}=9+9=18$

Now find the resultant resistance of $R'$ and $R_{3}$ connected in parallel combination

$\frac{1}{R}=\frac{1}{R'}+\frac{1}{R_{3}}\\\\\Rightarrow\frac{1}{R}=\frac{1}{18}+\frac{1}{9}\\\\\Rightarrow\frac{1}{R}=\frac{1+2}{18}\\\\\Rightarrow{R}=\frac{18}{3}=6\Omega$[/tex]