InterviewSolution
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InterviewSolution
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A charged capacitor of capacitance C is discharged through a resistance R. A radioactive sample decays with an average-life `tau`.Find the value of R for which the ratio of the electrostatic field energy stored in the capacitor to the activity of the radioactive sample remains constant in time. |
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Answer» `bar(T) = tau` `lambda = (1)/(bar(T)) = (1)/(tau)` Activity of radioactive smaple at time `t` `A = A_(0)e^(-lambdat) = A_(0)e^(-t//tau) ….(i)` In discharging `RC` circuit `q = q_(0)e^(-t//RC)` Energy stored in capacitor `U = (q^(2))/(2C) = (q_(0)^(2))/(2C)e^(-(2t)/(RC))` `(U)/(A) =((q^(2))/(2C)e^(-(2t)/(RC)))/(A_(0)e^(-t//tau))` If `U//A` is independent of time, `e^(-2t//RC) = e^(-t//tau)` `(2)/(RC) = (1)/(tau)` `R =(2tau)/(C )` |
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