The lowest resistance that can be secured by a combination of four coils of resistance 4 Ω, 8 Ω, 12 Ω and 24 Ω is [[1]] Ω.

Solution:

The lowest resistance that can be secured by a combination of four coils of resistance

4 Ω

8 Ω

12 Ω

and

24 Ω

2 Ω

.
It is known that the lowest resistance is observed when the resistances are connected in parallel. It is so because, in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of all the individual resistances. Therefore, the lowest resistance is secured by combining all four coils of resistance in parallel.

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

On substituting the given values, we get the lowest resistance as

\frac{1}{R} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{24}

\frac{1}{R} = \frac{6 + 3 + 2 + 1}{24} = \frac{12}{24}

R = 2 Ω

Hence, the lowest resistance is

2 Ω

The lowest resistance that can be secured by a combination of four coils of resistance $4 Ω$ , $8 Ω$ , $12 Ω$ and $24 Ω$ is [[1]] $Ω$ .

Solution:

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