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A circuit having resistor of 2 mega-ohm and capacitor of 1 muF is placed in series with a battery of 2 volt. Find the time after which the charge reaches 86.4% of its maximum value. |
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Answer» Solution :`C = 1 mu F = 10^(-6)` Farad, R= 2 mega ohm `=2 xx 10^6 ` ohm, E= 2 VOLT Now , `Cr= 10^(-6) xx 2 xx 10^6 = 2s` and `q/q_0 = (86.4)/(100) = 0.864` According to the relation `q= q_0 [1-e^(-1//CR)]` we get: `q/q_0 = 1- e^(-1//CR)` or , `0.864 = 1- e^(-1/2)` `e^(-1/2) = 1- 0.864 = 1.36 impliese^(1/2) = 1/(0.136)` `t= 2` In (7.352) = 4 sec |
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