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A capacitor is being charged through a resistance of 3 mega ohm. If it reaches 75% of its final potential in 0.5 sec, find its capacitance. |
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Answer» SOLUTION :At any moment during the charging process, the potential difference across the plates of a capacitor is givenby `V= V_0 (1- E^(-t//CR))` where `V_0` is the final potential Give that `V/V_0 = 75% = 0.75 , t = 0.5 ` sec and `R= 3xx10^6 Omega` Hence , `0.75 = 1-e^(-0.5//(3xx10^6)C)` or ` e^(-0.5//(3xx10^6)C) = 1 0 0.75 = 0.25` or, `e^(-0.5//(3xx10^6)C) =1/(0.25) = 4` or , `(0.5)/(3xx10^6 C)= log_e 4 = 2.3026xx log_10^4` or , `C= (0.5)/((3xx10^6) xx 2.3026 xx 0.6021)` `= 0.12xx10^(-6) F = 0.1 12 muF` |
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