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Q.

A technician has only two resistance coils. By using them singly in series or in parallel he is able to obtain the resistance 3,4,12 and 16 ohms. The resistance of two coils are

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a

4 ohm and 16 ohm

b

7 ohm and 9 ohm

c

6 ohm and 10 ohm

d

4 ohm and 12 ohm

answer is B.

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Detailed Solution

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$\begin{array}{l} R_{p} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = least \\ R_{p} = \frac{6 \times 10}{6 + 10} = \frac{60}{16} = 3 .75 \dots \dots . (1) \\ R_{p} = \frac{4 \times 12}{4 + 12} = \frac{48}{16} = 3 \dots \dots (2) \\ R_{p} = \frac{7 \times 9}{7 + 9} = \frac{63}{16} = 3 .93 \dots \dots . (3) \\ R_{p} = \frac{4 \times 16}{4 + 16} = \frac{64}{20} = 3 .2 \dots \dots \dots \dots . (4) least (2) \end{array}$

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