show that the rate of change when the capacitor is discharging from the supply voltage value will b the same as when charging from the same value at any time given. vc=vs(1-e-t/cr) . VS= 10, c=10uf, R=20ohms.

Question

anonymous · Answer

VS is the initial charge?

anonymous · Answer

No Vs = d.c. Source voltage.. 
rate of change of what exactly?

anonymous · Answer

are you talking about rate of change of voltage across capacitor?

anonymous · Answer

rate at which charge is accumulating in first case ,,and rate at which it is discharging ,,,in 2nd case maybe

anonymous · Answer

yes voltage across capacitor

radar · Answer

The time constant (RC) is the same for both charge and discharge for this case.  It is the time constant which establishes the rate of change.  The time constant for this situation is 2 10^-4

anonymous · Answer

how do i show that in an equation?

radar · Answer

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