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

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× 1019  (B)  6.25× 1020  (C)  5.25× 1019  (D)  2.55× 1020

Tardigrade Admin · Accepted Answer

I=4-0.08tA Or (dq/dt)=4-0.08tA Or q=∫050(4-0.08t)dtC Or Ne=[ 4t-(0.08t2/2) ]050=100C where N is number of electrons. Or N=(100/e)=(100/1.6× 10-19) =6.25× 1020

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

$1.25 \times 10^{19}$

$6.25 \times 10^{20}$

$5.25 \times 10^{19}$

$2.55 \times 10^{20}$

$4.25 \times 10^{20}$

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

Q. 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 50s through the cross-section of the conductor is

1.25×1019

6.25×1020

5.25×1019

2.55×1020

4.25×1020

I=4−0.08tA Or dtdq​=4−0.08tA Or q=∫050​(4−0.08t)dtC Or Ne=[4t−20.08t2​]050​=100C where N is number of electrons. Or N=e100​=1.6×10−19100​ =6.25×1020