A Question on Varying Currents

The following MCQ which appeared in KEAM 2007 (Engineering) question paper can claim to be one different from the usual type:

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 (e) 4.25×10²⁰

To obtain the electrons flowing in 50 s, you have to calculate the total charge flowing in 50 s.

Total charge Q = ∫Idt = ∫(4–0.08t)dt = [4t–(0.08t²/2)].

Since the limits of integration are 0 and 50 seconds, Q = 4×50–(0.08×50²/2) = 100 coulomb.

Since the electronic charge is 1.6×10^–19 coulomb, the number of electrons flowing in 50 s is 100/(1.6×10^–19) = 6.25×10²⁰.