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In the given circuit, AB is a wire of resistance $6 Ω$ , length 10 cm, number of electrons per unit volume in material of wire is $10^{29} m^{- 3}$ . If cross-sectinal area of wire is $1 m m^{2}$ , then which of the following options are correct?

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Ohmic loss in wire is 6J/s

Energy absorbed by electrons is approximately $2 \times 10^{- 17} J$

Ohmic loss in wire is 16J/s

Energy absorbed by electrons is approximately $2 \times 10^{- 16} J$

answer is A, C.

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Detailed Solution

$P = i^{2} R = \frac{V^{2}}{R} = \frac{V^{2}}{R} = \frac{6 \times 6}{6} = 6 J / S$

$V_{d} = \frac{i}{n e A} = \frac{1}{(10^{29}) \times (1.6 \times 10^{- 19}) \times (10^{- 6})} = \frac{10^{- 4}}{1.6} m / s$

$K E o f o n e e l e c t r o n = \frac{1}{2} m {v_{d}}^{2} = \frac{1}{2} \times 9.1 \times 10^{- 31} \times \frac{10^{- 8}}{2.56} J$

Number of electrons in the wire, $N = n A l = 10^{29} \times 10^{- 6} \times 0.1 = 10^{22}$

So, energy of all electrons, $E = 1.77 \times 10^{- 17} J \approx 2 \times 10^{- 17} J$

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