Question

Two resistances X and Y are connected turn by turn (i) in parallel, and (ii) in series. In which case the resultant resistance will be less than either of the individual resistances?

Answer 1

Answer:

Equivalent resistance of 2 resistances is least when it is connected in parallel (case ii).

Explanation:

R(eq) in parallel = R1*R2/(R1 + R2)

R(eq) in series = R1 + R2

Hope this helps! Do hit the thanks button and mark as brainliest!

Answer 2

Explanation:

• Let us take a simple example, i.e., let $X= 2$ ohm and $Y= 4$ ohm
• When X and Y are connected in parallel, then the resultant resistance according to the law of combination of resistances in parallel,

         $\text { R Resultant }=\frac{1}{R}=\frac{1}{X}+\frac{1}{Y}$

         $=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$

        $R=\frac{4}{3}=1.33 \mathrm{ohm}$

• Now, when X and Y are connected in series, then the resultant resistance according to the law of combination of resistances in series,
• R Resultant $= X + Y = 2 + 3 = 5 ohm$.
• Therefore, clearly the resultant resistance will be less than either of the individual resistances in the case of parallel connection.