Electronics - Capacitors - Discussion

Can anyone explain me?

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

Please explain in detail.

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

Thanks for giving brief explanation @Elek Kick.