Electronics - Capacitors - Discussion

47. 

An 8 V power source is charging a capacitor. How many volts will be across the capacitor at the second time constant?

[A]. 1 volt
[B]. 3 volts
[C]. 5 volts
[D]. 7 volts

Answer: Option D

Explanation:

No answer description available for this question.

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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