Help with v=Ve^-t/CR

[mr2turbo]

Hi all,

I am currently doing a HNC in Electronics and i am struggling with the following equation, v=Ve^-t/CR. I'm not the greatest at maths and generally know enough to get through but this one has me stumped.

Our lecturer gave us the following question:

The voltage across the plates of a capacitor at any time in T seconds is give by the formula v=Ve^-t/CR. Given that V=200v C=0.1uF and R = 500k

A.find the initial rate of change of voltage (time=0)
B. Rate of change after 0.5 seconds.

I got as far as writing out the following: 200e^-0/50ms. The answer that was given was 10Mv per second but i cannot work out how he got there, even with assistance! There was some differentiation involved to, which totally lost me.

Any advice would be very much appreciated! Many thanks.

 

[chuckey]

The initial rate of change is easy, the time constant is .05S, so the initial rate of change is 200V in .05 S, or 4000 V/S . As the capacitor charges the voltage increases slower, so the rate of change is lower. In theory the capacitor voltage NEVER gets to the applied voltage, however for practical purposes it reaches 98% in 4 X the time constant. This is where the differentiation comes in, finding the change in the slope of the equation. But as the time is 10 times that of the time constant, I'll call it zero.
Frank

 

[mr2turbo]

Thanks for the reply Chuckey. I'm still a little lost though.

Your answer would apply to the charging of a capacitor and not discharging?

 

[LvW]

Te process of charging and discharging a capacitor through a resistor will happen in both cases with the same time constant T=RC.
However, in one case the exponenet is positive (e^t/RC) and in the other case it is negative (e^-t/RC)