Question

In a closed circuit, the current $I$ (in ampere) at an instant of time $t$ (in second) is given by $I=4-0.08t$ . The number of electrons flowing in $50\, s$ through the cross-section of the conductor is

• A. $1.25\times {{10}^{19}}$
• B. $6.25\times {{10}^{20}}$
• C. $5.25\times {{10}^{19}}$
• D. $2.55\times {{10}^{20}}$

Answer 1

Answer: (B)

$6.25\times {{10}^{20}}$

Answer 2

$I=4-0.08tA$
Or $\frac{dq}{dt}=4-0.08tA$
Or $q=\int_{0}^{50}{(4-0.08t)dt}C$
Or $Ne=\left[ 4t-\frac{0.08{{t}^{2}}}{2} \right]_{0}^{50}=100C$
where N is number of electrons.
Or $N=\frac{100}{e}=\frac{100}{1.6\times {{10}^{-19}}}$
$=6.25\times {{10}^{20}}$