Question

Why is resistance less in parallel circuits?

Answer 1

Hint: In electrical circuits, resistors can either be connected in one of the two possible ways. One is a series combination and the other is a parallel combination. Let us check the net resistances in both cases and then compare which is less and which is more.


Complete step by step answer:
In series combination, let us say there are n number of resistors connected in series having resistances of ${R_1},{R_2},{R_3}....{R_n}$ and let ${R_{series}}$ denotes the net resistance in series combination then, their net resistance is calculated as
${R_{series}} = {R_1} + {R_2} + {R_3} + .... + {R_n}$ which is simply the sum of each individual resistance.
In parallel combination, let us say there are n number of resistors connected in parallel having resistances of ${R_1},{R_2},{R_3}....{R_n}$ and let ${R_{parallel}}$ denotes the net resistance in parallel combination then, their net resistance is calculated as
$\dfrac{1}{{{R_{parallel}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + .... + \dfrac{1}{{{R_n}}}$ and we know that for any numerical value of x,
$\dfrac{1}{x} < x$ which shows that
In parallel combination, net resistance is not the sum of individual resistance but it’s the sum $\dfrac{1}{R}$ of each resistances which implies that,
${R_{parallel}} < {R_{series}}$
Hence, resistance is less in parallel circuit as compared to series combination because of the net resistance formula as $\dfrac{1}{{{R_{parallel}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + .... + \dfrac{1}{{{R_n}}}$ whereas in series case its ${R_{series}} = {R_1} + {R_2} + {R_3} + .... + {R_n}$ .

Note:It should be remembered that, the less resistance in parallel combination of resistors gives the most important advantage in electrical circuits that every component of resistance will get the same amount of voltage from the battery while in series combination the current is the same but not voltage.