A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross-section of the filament in 16 seconds would be roughly
(a) 10^20
(b) 10^16
(c) 10^18
(d) 10^23
A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross-section of the filament in 16 seconds would be roughly
(a) 10^20
(b) 10^16
(c) 10^18
(d) 10^23

Solution: Answer is (a) 1020

Given : I =1 A , t =16 s , e =1.6 \times 10^{-19} C

Step 1: The filament draws current from the power supply.

I =\frac{ Q }{ t }=\frac{ Ne }{ t } \quad <!-- /wp:paragraph -->  <!-- wp:paragraph --> ( Since Q = Ne, where N is the number of electrons crossing in time t)

    \[ <!-- /wp:paragraph --> <!-- wp:paragraph --> \Rightarrow N =\frac{ It }{ e } <!-- /wp:paragraph --> <!-- wp:paragraph --> \]

Step 2: Substituting the values in equation (1)

    \[ <!-- /wp:paragraph --> <!-- wp:paragraph --> N =\frac{ It }{ e }=\frac{1 A \times 16 s }{1.6 \times 10^{-19} C }=10^{20} <!-- /wp:paragraph --> <!-- wp:paragraph --> \]

As a result, the number of electrons passing through a cross section of the filament in 16 seconds is 10^{20}.