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A `0.18 mu` F capacitor is first charged and then discharged through a high resistance. If it takes 0.5 sec for the chage to reduce to one forth of its initial value, find the value of the resistance. Given `log_(e) 4 = 1.386` |
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Answer» Here, `C = 1.08 mu F = 0.18 xx 10^(-6) F, t = 0.5 s, (q)/(q_(0)) = (1)/(4) = 0.25, R = ?` During dischanging, `q = q_(0) e^(-t//RC) :. e^(-t//RC) = (q)/(q_(0)) = (1)/(4)` `(-t)/(RC) log_(e) e = log_(e) 4 , (-t)/(RC) xx 1 = 0 - 1.386` `R = (t)/(1.386 C) = (0.5)/(1.368 xx 0.18 xx 10^(-6)) = 2 xx 10^(6) Omega = 2 M Omega` |
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