Charge of circuit?

Question

Hi,

Consider the circuit in Figure P31.77 ( http://i137.photobucket.com/albums/q208/infinitbelt/p31-77alt.gif ), in which V = 90 V, R = 25 , and the switch has been closed for a very long time.


(a) What is the charge on the capacitor?

(b) The switch is opened at t = 0 s. At what time has the charge on the capacitor decreased to 10% of its initial value?


I tried finding the total resistance of the circuit, and I got 67.1. Then I used Q=VC to solve for Q but I am still not getting the right answer.



Thanks!

Answer 1

how bout i give you some pointers....
1. at t=0- (before switch opens) and we know the switch has been closed for a long time, this implies the capacitor is basically an open circuit to a DC voltage, so in essence, that branch of the parallel circuit is basically an open circuit.
2. according to 1) then we can calculate the V drop across R.
3. from 2., we know the V across the Cap when the switch opens, since the 60ohm R is now doing nothing in the cct.
4. use the exponential form of the equation for cap. charge to find the 10% region.

Answer 2

Under steady state conditions, there is no current in the capacitor leg, so the voltage is strictly dictated by the voltage divider with 60 ohms on top and 25 ohms on the bottom.

When the switch is opened, the 60 ohm resistor has no place for current to go, so is irrelevant. Therefore, the capacitor discharges though the series combination of the 25 ohm adn 10 ohm resistors.

Answer 3

For (a), Q = CV
For (b), Q(t) = (1 - e^(-t/RC)), where Q(t) is 10% of your answer from (a), R is the series resistance of discharge path(35 ohms), C is your capacitance(2.0uF), solve for t

It's been a while since I played with logs. Maybe someone else can take what I got and run with it. Good luck.