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

($ Since $Q = Ne$, where $N$ is the number of electrons crossing in time t)

$$

\Rightarrow N =\frac{ It }{ e }

$$

Step 2: Substituting the values in equation (1)

$$

N =\frac{ It }{ e }=\frac{1 A \times 16 s }{1.6 \times 10^{-19} C }=10^{20}

$$

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