A) \[\frac{10}{e}\text{coulomb}\]
B) \[\frac{{{e}^{2}}}{10}\text{coulomb}\]
C) \[\frac{10}{{{e}^{2}}}\text{coulomb}\]
D) \[\frac{e}{10}\text{coulomb}\]
Correct Answer: A
Solution :
Key Idea: Exponential decay of charge takes place. In a C-R circuit, discharging takes place. The q-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, \[{{V}_{0}}=5V,\,C=2F,t=12s,R=6\,\Omega \] \[\therefore \] \[q=2\times 5{{e}^{-12/6\times 2}}=\frac{10}{e}\text{coulomb}\]You need to login to perform this action.
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