Two equal resistances are connected in series and parallel combination. The ratio of resistances in parallel and series combination is:
$\left( a \right){\text{ 4:1}}$
$\left( b \right){\text{ 1:4}}$
$\left( c \right){\text{ 2:1}}$
$\left( d \right){\text{ 1:2}}$

Hint The current in every resistance has an equivalent worth. The overall fall is up to the availability voltage. The effective resistance is often over the minimum value resistance within the circuit. The voltage across every resistance is up to the supply voltage. The overall current within the circuit is up to the total of the current in an individual loop. The effective resistance is often but the minimum value resistance within the loop.

Complete Step By Step Solution So first of all we will assume$R$, will be the value for each of the resistance.
So, for the series combination, the formula for it will be
${R_s} = {R_1} + {R_2} + ...... + {R_N}$
Since we have the two resistance given which is of the value $R$of each.
Therefore, from this, the resistance in the series will be
${R_s} = R + R = 2R$
Now, for the parallel combination, it will be given by
$ \Rightarrow \dfrac{1}{{{R_P}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}}......... + \dfrac{1}{{{R_N}}}$
Since we have the two resistance given which is of the value $R$of each.
Therefore, from this, the resistance in parallel will be
$ \Rightarrow \dfrac{1}{{{R_P}}} = \dfrac{1}{R} + \dfrac{1}{R}$
On solving the above equation, we get
$ \Rightarrow {R_P} = \dfrac{{R \times R}}{{R + R}} = \dfrac{R}{2}$
So we have to find the ratio between the parallel and the series combination. i.e. ${R_P}{\text{ and }}{R_S}$
So we can write the ratios as-
$ \Rightarrow \dfrac{{{R_P}}}{{{R_S}}} = \dfrac{R}{2} \times \dfrac{1}{{2R}} = \dfrac{1}{4}$

Therefore, the option $\left( b \right)$ is correct.

Note A parallel resistance shares the same voltage however completely different current prices relying on the worth of the impedances. Have you ever detected a Christmas light with one bulb out? If all lights leave, it means that it was connected in series. If solely that one goes out, it means that it was connected in Parallel.