Question

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

Answer 1

HELLO THERE!

The four resistances given are 4Ω, 8Ω, 12Ω, 24Ω.

The highest resistance will be obtained on combining these four resistances in series.

The lowest resistance will be obtained on combining these four resistances in parallel.

(i) Hence, for highest resistance, combine these in series, and the effective resistance (R) = 4 + 8 + 12 + 24 = 48Ω.

Hence, the highest resistance is 48Ω.

(ii) For the lowest resistance, combine these in parallel, and the effective resistance (R) is:

$\frac{1}{R} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{24} \\\\\implies \frac{1}{R} = \frac{6 + 3 + 2 + 1}{24} \\\\\implies \frac{1}{R} = \frac{12}{24} \\\\\implies \frac{1}{R} = \frac{1}{2} \\\\\implies R = 2$

Hence, the lowest resistance obtained is 2Ω.

HOPE MY ANSWER IS SATISFACTORY...

THANKS!

Answer 2

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