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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 $ 50\, s $ through the cross-section of the conductor is

KEAMKEAM 2007Current Electricity

Solution:

$ 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}} $