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Q.

The resistance of a metal wire is 10 $Ω$ . A current of 30 mA is flowing in it at 20⁰ C. If p.d. across its ends is constant, then its temperature is increased to 120⁰ C, then the current flowing in the wire will be in mA $(α = 5 \times 10^{- 3} /^{0} C)$

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a

10

b

20

c

15

d

40

answer is A.

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Detailed Solution

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$\begin{array}{l} R_{1} = 10 Ω as V is constant \\ i_{1} = 30 mA; i_{1} R_{1} = i_{2} R_{2} \\ t_{1} = 20^{0} C; i_{2} = \frac{i_{1} R_{1}}{R_{2}} \\ t_{2} = 120^{0} C; i_{2} = \frac{i_{1} R_{1}}{R_{1} (1 + α Δt)}; i_{2} = \frac{i_{1}}{1 + α Δt} \\ = \frac{30}{1 + 5 \times 10^{- 3} \times 100} = \frac{30}{1.5} = 20 mA \end{array}$

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