A capacitor of 2.5 mu F is charged through a resistor of 4 M Omega. In how much time will potential drop across capacitor will become 3 times that of resistor (ln 20.693)

A capacitor of 2.5 mu F is charged through a resistor of 4 M Omega. In

1.	A capacitor of 2.5 mu F is charged through a resistor of 4 M Omega. In how much time will potential drop across capacitor will become 3 times that of resistor (ln 20.693)
Answer» 13.86 s 6.93 s 1.386 s 69.3 s Solution :The growth of VOLTAGE in a CR circuit is given by `V=V_(0)[1-e^(-t//CR)]` where t is any TIME during the charging. For the given circuit, as per the conditions given `V_(c )(t)=(3)/(4)th` of the voltage applied so, `(3V_(0))/(4)=V_(0)[1-e^(-t//RC)]` or, `e^(-t//RC)=(1)/(4)` `RARR t=2RC log_(e )2` `rArr t=2xx4xx10^(6)xx2.5xx10^(-6)xx0.693=13.86 s`.

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A capacitor of 2.5 mu F is charged through a resistor of 4 M Omega. In how much time will potential drop across capacitor will become 3 times that of resistor (ln 20.693)

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