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

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Solution

Highest will be by series = 4+12+8+24=48 ohms

Lowest will be by parallel $= \frac{1}{R} = \frac{1}{4} + \frac{1}{8} + \frac{1}{24} + \frac{1}{12} = \frac{12}{24} = \frac{1}{2}$
so R = 2

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

Highest will be by series = 4+12+8+24=48 ohms Lowest will be by parallel =1R=14+18+124+112=1224=12 so R = 2

Highest will be by series = 4+12+8+24=48 ohms

Lowest will be by parallel $= \frac{1}{R} = \frac{1}{4} + \frac{1}{8} + \frac{1}{24} + \frac{1}{12} = \frac{12}{24} = \frac{1}{2}$
so R = 2