The number of electrons flowing per second in the filament of a 110 W bulb operating at 220 V is

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Solution

Q.
The number of electrons flowing per second in the filament of a 110 W bulb operating at 220 V is (Given $e = 1.6 \times 10^{- 19} C$ ) [2024]

1 $1.25 \times 10^{19}$

2 $6.25 \times 10^{17}$

3 $6.25 \times 10^{18}$

4 $31.25 \times 10^{17}$

Ans.
(4)

Given: P = 110 W, V = 220 V

Power, $P = V . I \Rightarrow 110 = (220) (I) \Rightarrow I = 0.5 A$

$I = \frac{q}{t} = \frac{n e}{t} \Rightarrow 0.5 = (\frac{n}{t}) \times 1.6 \times 10^{- 19}$

$(\frac{n}{t}) = \frac{0.5}{1.6 \times 10^{- 19}} = 31.25 \times 10^{17}$

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