Question

How can three resistors of resistances $2\Omega ,3\Omega$ and $6\Omega$ be connected to give a total resistance of
(A) $4\Omega$
(B) $1\Omega$

Answer 1

Hint : The equivalent resistance of a parallel configuration of resistors is always lower than the resistance of each resistor being combined. The equivalent resistance of a series resistor is always higher than each resistance of each resistor that are being combined.

Formula used: In this solution we will be using the following formula;
${R_s} = {R_1} + {R_2} + .... + {R_n}$ where ${R_s}$ is the equivalent resistance of a series arrangement of resistors, and ${R_1}...{R_n}$ are the resistance of the individual resistors being combined.
$\dfrac{1}{{{R_p}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + ... + \dfrac{1}{{{R_n}}}$ where ${R_p}$ is the equivalent resistance of a parallel arrangement of resistors.

Complete step by step answer
To create a particular resistance out of some other resistance, one has to connect the resistors in some form of arrangement, either series or parallel or a combination of both.
For a $4\Omega$ ohms resistance, we can connect the $6\Omega$ resistance and the $3\Omega$ resistance together in parallel to give a $2\Omega$ resistance, then combine these combination in parallel to the $2\Omega$ resistance, as in:
$\dfrac{1}{{{R_{p36}}}} = \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{{6 + 3}}{{18}}$ , ( since parallel arrangement is given as $\dfrac{1}{{{R_p}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + ... + \dfrac{1}{{{R_n}}}$ where ${R_p}$ is the equivalent resistance of a parallel arrangement of resistors).
Hence, by computing and inverting we have
${R_{p36}} = \dfrac{{18}}{9} = 2\Omega$
Then combining by series arrangement with $2\Omega$ we have
${R_{p22}} = 2 + 2 = 4\Omega$
This gives us the first equivalent resistance.
To get a $1\Omega$ resistance, we combine all three in parallel arrangement as in:
$\dfrac{1}{{{R_{p236}}}} = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{{9 + 6 + 3}}{{18}}$
Hence, by computing and inverting we have
${R_{p236}} = \dfrac{{18}}{{18}} = 1\Omega$
This gives us the second equivalent resistance.

Note
Alternatively, for the 1 ohms resistance, we could say that the 2 ohms resistor was added in parallel to the 2 ohms equivalent resistance of the 3 and 6 ohms parallel combination. Then the mathematics becomes,
$\dfrac{1}{{{R_{p22}}}} = \dfrac{1}{2} + \dfrac{1}{2} = \dfrac{{1 + 1}}{2}$
Hence, by computing and inverting we have
${R_{p22}} = \dfrac{2}{2} = 1\Omega$ . This is identical to the solution above.