A) \[\frac{10}{e}coulomb\]
B) \[\frac{{{e}^{2}}}{10}coulomb\]
C) \[\frac{10}{{{e}^{2}}}coulomb\]
D) \[\frac{e}{10}coulomb\]
Correct Answer: A
Solution :
Key Idea: Exponential decay of charge takes place. In a \[C\text{-}R\] circuit, discharging takes place. The \[q\text{-}t\] equation is \[q={{q}_{0}}{{e}^{-t/RC}}\] where \[R\] is resistance, \[C\] is capacitance and\[{{q}_{0}}=C{{V}_{0}}\] Hence, \[q=C{{V}_{0}}{{e}^{-t/RC}}\] Given, \[{{V}_{0}}=5V,\,\,C=2F,\,\,t=12s,\,\,R=6\Omega \] \[\therefore \] \[q=2\times 5{{e}^{-12/6\times 2}}=\frac{10}{e}\]coulomb
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