What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?

Asked by Vishal kumar | 1 year ago |  199

Solution :-

(a) If the four resistors are connected in series, their total resistance will be the sum of their individual resistances, and it will be the highest. The total equivalent resistance of the resistors connected in series will be 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω.

(b) If the resistors are connected in parallel, then their equivalent resistances will be the lowest. Their equivalent resistance connected in parallel is

$$R = \frac{1}{\frac{1+1+1+1}{4+8+12+24}}=\frac{24}{12}=2\,\Omega$$

Hence, the lowest total resistance is 2 Ω

Answered by Shivani Kumari | 1 year ago

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