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Is it true that the minimum resistance of the resistors $4 Ω$ , $8 Ω$ , $12 Ω$ , and $24 Ω$ is $2 Ω$ ?

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True

False

answer is A.

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Detailed Solution

The given statement is True.
When the resistance connected in parallel their total resistance can be calculated by equation,

\frac{1}{R_{P}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} + \frac{1}{R_{4}}

………………………. (i)
So, from equation (i) we conclude that the reciprocal of the equivalent resistance

R_{P}

of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances. So if the four resistance are connected in series then we get minimum total resistance.
Let

R_{1} = 4 Ω

R_{2} ​ = 8 Ω, R_{3} = 12 Ω​

and

R_{4} = 24 Ω

Put the above values in equation (ii) we get,

= > \frac{1}{R_{P}} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{24}

= > \frac{1}{R_{P}} = \frac{6 + 3 + 2 + 1}{24}

= > \frac{1}{R_{P}} = \frac{12}{24}

= > \frac{1}{R_{P}} = \frac{1}{2}

= > R_{P} = 2 Ω

The lowest resistance of the resistors is

2 Ω .

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