InterviewSolution
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InterviewSolution
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LetC be the capacitance of a capacitor discharging through a resistor R. Suppose `t_1` is the time taken for the energy stored in the capacitor to reduce to half its initial value and `t_2` is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio `t_1//t_2` will beA. `1`B. `1//2`C. `1//4`D. `2` |
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Answer» Correct Answer - B `U_(0) = (q_(0)^(2))/(2C)U = (q_(0)^(2)e^(-2t_(1)//tau))/(2C) = (U_(0))/(2) = (q_(0)^(2))/(4C)` `e^(-2t_(1)//tau) = (1)/(2)` `t_(1) = (tau)/(2)ln2 ….(1)` and `q = q_(0)e^(-t_(2)//tau)` `(q_(0))/(4) = q_(0)e^(-t_(2)//tau),` `e^(-t_(2)//tau) = (1)/(4)` `t_(2) = 2tau In2 ...(2)` `(t_(1))/(t_(2)) =(1)/(4)` |
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