### Discussion :: Capacitors - General Questions (Q.No.47)

Raghu said: (Oct 20, 2011) | |

Can anyone explain me? |

Priya said: (Nov 21, 2011) | |

Voltage across a capacitor for second time constant. will be 0.865 times of supply voltage. s0 6.92 volt. approximately equal to 7v |

Ajit Gupta said: (Mar 17, 2012) | |

Please explain in detail. |

Elek Kick said: (May 8, 2012) | |

We know that voltage across capacitor at any instant..... Vc = Vdc ( 1 - e^(-t/RC)) At second time constant i.e., 2t Vc = Vdc ( 1 - e^(-2t/t)) \\ RC = t Here Vdc = 8V Vc = 8V ( 1 - e^-2) Vc = 8V ( 1 - (1/e^2)) Vc = 8V ( 1 - (1/2.718^2)) \\ e = 2.718 exponential constant Vc = 8V ( 1 - (1/7.387)) Vc = 8V ((7.387 - 1) / 7.387) Vc = 8V ( 6.387 / 7.387 ) Vc = 8V ( 0.8646) Vc = 6.9168 or apporx. 7V t = RC , e = 2.718 and Vdc = X @ different time constant instant voltage will be.... 1t:-X*(1 - e^(-1t/t))=X*(1 - e^-1)=X*(1 -(1/e^1)) = X*0.632 2t:-X*(1 - e^(-2t/t))=X*(1 - e^-2)=X*(1 -(1/e^2)) = X*0.865 3t:-X*(1 - e^(-3t/t))=X*(1 - e^-3)=X*(1 -(1/e^3)) = X*0.950 4t:-X*(1 - e^(-4t/t))=X*(1 - e^-4)=X*(1 -(1/e^4)) = X*0.982 5t:-X*(1 - e^(-5t/t))=X*(1 - e^-5)=X*(1 -(1/e^5)) = X*0.993 |

Srikanth.P said: (Jan 12, 2017) | |

Thanks for giving brief explanation @Elek Kick. |

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