Question

A Battery is giving constant current 1.5Amp for 5 microseconds.
(a) How many electrons are really given by battery?
(b) Calculate charge given by battery to develop current.

Answer 1

$$\begin{gathered} \mathbf{Charge}\mathbf{:} \\ \mathrm{Q}=\mathrm{It} \\ \mathrm{where} \\ \mathrm{I}=\mathrm{Current} \\ \mathrm{t}=\mathrm{Time}\mathrm{of}\mathrm{flow}\mathrm{of}\mathrm{current} \\ \mathrm{Hence}, \\ \mathrm{Q}=1.5\times \left(5\times {10}^{-6}\right) \\ =7.5\times {10}^{-6}\mathrm{C} \\ \left(\mathrm{a}\right) \\ \mathbf{No}\mathbf{.}\mathbf{}\mathbf{of}\mathbf{}\mathbf{electrons}\mathbf{:} \\ \mathrm{n}=\frac{\mathrm{Q}}{\mathrm{e}}[\mathrm{Here},\mathrm{e}=\mathrm{charge}\mathrm{of}\mathrm{an}\mathrm{electron}] \\ =\frac{7.5\times {10}^{-6}}{1.6\times {10}^{-19}} \\ =\mathbf{4}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{13}}\mathbf{} \\ \mathbf{\left(}\mathbf{b}\mathbf{\right)} \\ \mathbf{Charge}\mathbf{}\mathbf{given}\mathbf{}\mathbf{by}\mathbf{}\mathbf{the}\mathbf{}\mathbf{battery}\mathbf{:} \\ \mathbf{Q}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{}\mathbf{C}\mathbf{}\mathbf{\left(}\mathbf{or}\mathbf{\right)}\mathbf{}\mathbf{7}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{\mu C} \end{gathered}$$