A 16 V battery, a resistor with R = 10^6Ω and a 6-μF capacitor are connected in series with a switch. Find the time taken for the capacitor;
a)To charge to a voltage value of 12V
b)Time to discharge to a value to 7V, after it has fully charged to 16V

To start this problem I figure I need to find the Time constant T=RC, which is 10^6Ω*6μF =6e^6s.

Now I am stuck on how to find the charging and discharging times. If someone could point me in the right direction I would be greatly appreciated. I think I have to use the formula Vc = V(1 - e-t/RC) , but Im not sure how to do this. I assume for part a I plug in the 12V value somewhere, but Im not sure where.

Any suggestions?

 

http://www.electronics-tutorials.ws/rc/rc_1.html

Probably worth doing a quick charge and discharge drawing with full charge values plotted. To make sure it looks right too.

 

For the charging part of the question you would use ....

Vc = V(1 - e-t/RC)

assuming the initial voltage is 0V.

Here you would substitute Vc=12 and V=16 then solve for t.

The capacitor voltage discharge through R would take the form ....

Vc=V*e-t/RC

Where Vc=7 and V=16 and again solving for t

 

For the charging part of the question you would use ....

Vc = V(1 - e-t/RC)

assuming the initial voltage is 0V.

Here you would substitute Vc=12 and V=16 then solve for t.

The capacitor voltage discharge through R would take the form ....

Vc=V*e-t/RC

Where Vc=7 and V=16 and again solving for t

So was my time constant correct? According to the other post I had it wrong, but I think that I might be confused on what a charging circuit is. I will do the calculations using mine, but if you could clarify that I would be very grateful.

 

Hi ehelms, dcarson7 has time constant correct at 60sec.

You just have to keep powers of 10 in mind. A quick mental shortcut is

[X uF*Y MΩ]==[XY seconds]