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A capacitor of 1 muF and resistance 82 kOmegaare connected is series with a d.c. source of 100 volt. Calculate the magnitude of energy and the time in which energy stored in the capacitor will reach half of its maximum value. |
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Answer» Solution :i) Maximum energy stored `=(CV_0^2)/(2)` Energy stored `ALPHA V_0^2` Half of maximum will be stored when voltage across capacitor is `V=(100//sqrt2)` volt = 70.7 Volt Energy stored `=1/2 CV^2 = 1/2 xx(1xx10^(-6)) xx (70.7)^2 = 0.0025 J` II) `(V_0)/(sqrt2) = V_0 (1- E^(-t//RC))` or , `70.7 =100 [1-e^(-t//(82xx10^3 xx 10^(-6)))]` or , `70.7 = 100[1-e^(t//0.082)]` or, `e^(-t//0.082) = 1-(70.7)/(100) = 0.293` or, `e^(t//0.082) = 3.413` `therefore t= 0.082 log_e = 3.413 = 0 ` SEC. |
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