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Which of the following ratios is the highest to the lowest total resistance that can be secured by a combination of four coils of resistance?

$4 Ω, 8 Ω, 12 Ω, 24 Ω$

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2 Ω; 48 Ω

12 Ω : 48 Ω

48 Ω : 2 Ω

None of the above

answer is C.

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Detailed Solution

The highest to the lowest total resistance that can be secured a combination of four coils of
resistance

4 Ω, 8 Ω, 12 Ω, 24 Ω

is given by

48 Ω : 2 Ω

.
Connecting resistors in series always gives maximum resistance and parallel gives minimum
resistance.
(a) The highest total resistance is given by

R = R₁ + R₂ + R₃ + R₄

= 4 Ω + 8 Ω + 12 Ω + 24 Ω

= 48 Ω

(b) The lowest total resistance is given by

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

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

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

\Rightarrow \frac{1}{R} = \frac{12}{24}

\Rightarrow 24 = 12 R

\Rightarrow R = \frac{24}{12}

= 2 Ω

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