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(A) The power $P$ of the bulb is given by $P = V \cdot I$.Given $P = 110 \,W$ and $V = 220 \,V$.Substituting the values,$110 = 220 	imes I$.Therefore

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$31.25 	imes 10^{17}$

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(A) The power $P$ of the bulb is given by $P = V \cdot I$.Given $P = 110 \,W$ and $V = 220 \,V$.Substituting the values,$110 = 220 	imes I$.Therefore,the current $I = \frac{110}{220} = 0.5 \,A$.The current $I$ is defined as the rate of flow of charge,$I = \frac{q}{t} = \frac{n \cdot e}{t}$,where $n$ is the number of electrons and $e$ is the elementary charge.We need to find the number of electrons per second,which is $\frac{n}{t} = \frac{I}{e}$.Substituting $I = 0.5 \,A$ and $e = 1.6 	imes 10^{-19} \,C$:$\frac{n}{t} = \frac{0.5}{1.6 	imes 10^{-19}} = \frac{5}{16} 	imes 10^{19} = 0.3125 	imes 10^{19} = 31.25 	imes 10^{17}$ electrons per second.

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

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