Question

What will be the value of each resistance when the two resistances are connected in parallel gives the value of $2\Omega$ and when they are connected in series gives the value of $9\Omega$?

Answer 1

Hint We know that when the resistances are connected in series, we get their equivalent resistance by adding both of them.
${R_{eq}} = {R_1} + {R_2} + \ldots + {R_n}$
So, we will use this equation for series combination and for parallel combination we use the equation-
$\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \ldots \dfrac{1}{{{R_n}}}$
Then, compare both the equations and find out the values of two resistances.

Complete step by step answer:
According to the question, it is given that
When two resistances are connected in parallel, then the equivalent resistance is $2\Omega$.
And when two resistances are connected in series, then the equivalent resistance is $9\Omega$.
We know that, when $n$ number of resistances are connected in series, the equivalent resistance is
${R_{eq}} = {R_1} + {R_2} + \ldots + {R_n}$
where, ${R_{eq}}$ is equivalent resistance
when $n$ number of resistances are connected in parallel, the equivalent resistance is
$\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \ldots \dfrac{1}{{{R_n}}}$
Let the first resistance be ${R_1}$ and the second resistance be ${R_2}$.
Therefore, when two resistances are connected in series, we get

And when the two${R_1} + {R_2} = 9 \\\ {R_1} = 9 - {R_2} \cdots (1) \\\$ resistances are connected in parallel, we get
$\dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} = \dfrac{1}{2} \\\ \dfrac{{\left( {{R_1} + {R_2}} \right)}}{{{R_1}{R_2}}} = \dfrac{1}{2} \cdots (2) \\\$
Using the value of ${R_1}$ from equation $(1)$ in equation $(2)$
$\dfrac{{(9 - {R_2} + {R_2})}}{{(9 - {R_2}){R_2}}} = \dfrac{1}{2} \\\ \dfrac{9}{{9{R_2} - R_2^2}} = \dfrac{1}{2} \\\$
By cross-multiplication method
$9{R_2} - R_2^2 = 18$
By transposition method
$\Rightarrow R_2^2 - 9{R_2} + 18 = 0 \\\ \Rightarrow R_2^2 - 6{R_2} - 3{R_2} + 18 = 0 \\\ \Rightarrow {R_2}({R_2} - 6) - 3({R_2} - 6) = 0 \\\ \Rightarrow ({R_2} - 6)({R_2} - 3) = 0 \\\ \Rightarrow {R_2} = 6, 3 \\\$
So, we got the two values of second resistance
Therefore, when ${R_2} = 6\Omega$ then, ${R_1} = 9 - 6 = 3\Omega$

Hence, we got the all values of each resistance.

Note When the ${R_2} = 3\Omega$ resistors are connected in parallel combination, each resistance has the same potential drop across it, and current in each resistor is different. But in series combination, the output of the first resistor flows into the input of the second resistor, therefore, the current is the same in each resistor.