1ya its B but please post the solution clearly yaar the symbols u use in the post r bouncer for me.
In a closed circuit, the current I ( in amp.) at an instant of time t( in secs) is given by I = 4 - 0.08t. The no. of electrons flowing in 50 s trough the cross section of the conductor is
A) 1.25 x 1019 B) 6.25 x 1020 C) 5.25 x 1019 D) 2.55 x 1020
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5 Answers
62here you have to take the integral from time 0 to time 50 of I dt..
THat will give you the total charge in coulumbs..
Divide by the charge of one electon .. you get the answer...
current flow will be 100 coulombs
100/ (1.6x10^-19) = 10^21/1.6
gives answr B i guess
1
1ITS AN EASY 1
\int_{0}^{50}(4-0.08t){dt}=\int_{0}^{q}{}dq
this gives q=100..then divide by 1 charge
ans is b
62\\\int_{0}^{50}{(4-0.08t)dt}=100 \\\text{no of electrons}=\frac{100}{1.6\times 10^{-19}}=6.25\times10^20