When a number of resistance are connected in _______, their combined resistance is less than the smallest individual resistance.

Question

Question

This question was previously asked in

RRB ALP CBT I 31 Aug 2018 Shift 2 Official Paper

Answer 1

The correct answer is parallel.

Key Points

Concept:

F2 J.K 1.6.20 Pallavi D5

Addition in Series \(R_{eq}\) = R1 + R2 + R3.

Addition in Parallel \(\frac{1}{{{R_{eq}}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} + \frac{1}{{{R_3}}}\).

Calculation:

Let 4 resistances are connected in parallel.

R1 = 2 \(\Omega\), R2 = 4 \(\Omega\) , R3 = 5 \(\Omega\) ,R4 = 20 \(\Omega\).

\(\frac{1}{{{R_{eq}}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} + \frac{1}{{{R_3}}} + \frac{1}{{{R_4}}}\)

For R1||R2

\(\frac{1}{{{R_{eq_1}}}} = \frac{1}{{{2}}} + \frac{1}{{{4}}} \) = \(\frac{4}{3}\)\(\Omega\)

For R2||R3

\(\frac{1}{{{R_{eq_2}}}} = \frac{1}{{{5}}} + \frac{1}{{{20}}} \) = 4 \(\Omega\),

Req1 ||Req2

\(\frac{1}{R_{eq}} = \frac{\frac{4}{3}\times4}{\frac{4}{3} +4} \) = 1 \(\Omega\).

Hence \({R_{eq}} = 1 \Omega\)

which is less than the smallest resistance of the circuit that it \(2 \Omega\)