# graph of time constant rc network

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#### bhuvaneshnick

##### Member
in the below image u see that time constant is 0.37 of initial value and 63% of final value.if that so see the second graph in that image the point should be on 0.37 (from initial value) but actually the time constant point is on 0.63.
this is messing with my i dont know what i misunderstood.

#### dougy83

##### Well-Known Member
What it's saying, in a round about kind of way, is that one time constant is 63% of the way from the initial value to the final value. So, if discharging from 1V to 0V, 63% from 1V to 0V is 0.37V. If charging from 0V to 1V, 63% from 0V to 1V is 0.63%.

#### bhuvaneshnick

##### Member
If charging from 0V to 1V, 63% from 0V to 1V is 0.63%.
from charging 0v to 1v.its 63 percent of final value .so its at 0.63.but other statement says it is 0.37 of initial value.its initial value starting from 0.so it would be on o.37 rather that 0.63.Thats what i mean.

#### dougy83

##### Well-Known Member
The 0.37 relates to when it's discharging from 1V to 0V. You realise that 63% + 37% = 100%, right?

#### bhuvaneshnick

##### Member
you mean 0.37 relates discharging and 63% relates to charging
if that right for charging 1-0.37
and discharging we say 0.37
is that so

#### dougy83

##### Well-Known Member
It looks like that is what they teach in your course. Do you know where these numbers come from (i.e. which equation)?

#### bhuvaneshnick

##### Member
ya its something form e power tou
the we get pwr -1 .that is 0.37
no no i declare i dont know where it come from

#### dougy83

##### Well-Known Member
The formulae for the charge curve you showed above is [LATEX]V_c = 1 - e^{-t/\tau}[/LATEX] and the discharge curve is modelled by [LATEX]V_c = e^{-t/\tau}[/LATEX].

After one time constant, [LATEX]\tau[/LATEX], the charge curve [LATEX]V_c = 1 - e^{-t/\tau}; t \to \tau[/LATEX] becomes [LATEX]V_c = 1 - e^{-\tau/\tau} = 1 - e^{-1} = 1 - 0.3679 = 0.6321[/LATEX]

Likewise, After one time constant, [LATEX]\tau[/LATEX], the discharge curve [LATEX]V_c = e^{-t/\tau}; t \to \tau[/LATEX] becomes [LATEX]V_c = e^{-\tau/\tau} = e^{-1} = 0.3679[/LATEX]

#### bhuvaneshnick

##### Member
really a nice explanation thanks for that.say whether the my following thinking are correct with image file i added
1) in the picture they added what u said above that is both, charge and discharge formula for t>=0 because after switch is open its charging and while on the moment before open its on discharging. and so the curve is exponentially gros.is that right ?
2)if i make t<0 have equation of Vc=y(0)e^-at.because i have only discharging. graph is exponentially decay.right?
3)if i make t>0 i have equation only of Vc=Vs(1-e^-at)

say whether i am right or wrong.thanks

#### dougy83

##### Well-Known Member
1) The moment before the switch is opened, i.e. t<0, all current is flowing through the switch and none into the capacitor. Charge/discharge equations don't apply here and the voltage across the capacitor is 0V.
2) For t < 0, Vc = 0V
3) For t >= 0, Vc = Vs.[1 - exp(-t/tau)], where tau = RC

#### bhuvaneshnick

##### Member
okay.does fully discharged capacitor behaves like an open circuit

#### dougy83

##### Well-Known Member
okay.does fully discharged capacitor behaves like an open circuit
No, short circuit

• bhuvaneshnick

#### bhuvaneshnick

##### Member
see the below image
1)remove the capacitor and close the switch from the figure it act as potential divider circuit so R2 resistance have 3v
2) now consider the figure i uploaded with capacitor and switch open.the capacitor get charge and after closing switch, capacitor discharges the voltage and at steady state our circuit looks like as i mentioned in first line (1).that it act as potential divider with 3v on R2.so it is open circuit not an short circuit.

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