Voltage of RC circuit?

Question

http://i292.photobucket.com/albums/mm37/NoM_2008/Untitlercd.jpg

What is the capacitor voltage at time 5ms after the switch is closed? Can you explain too pls.. thanks..

Answer 1

http://en.wikipedia.org/wiki/RC_circuit

look at the VC(s)=... formula.

Answer 2

This is a standard RC circuit question the Capacitor is at 10 Volts so the charge in coulombs = Qmax = C/Vmax

When the switch is closed the capacitor discharges at an exponential rate. The time constant t for the circuit is RC = 10000 X 0.00001 = 0.1

The charge remaining on the capacitor after 5mS (0.005) is Q(T)

The formula is Q(T) = Qmax x e^(-T/t)

Where T = 5mS or (0.005) and t = RC time constant = 0.1

Solve for Q after time T this is Q(T) =  Qmax x e^ (-0.005/.1) = 0.95

in other words the charge on the capacitor is now 95 % of the original

Now multiply Vmax (10) by 95 % = 9.5 Volts

The voltage on the capacitor after 5mS is 9.5 Volts

This looks rather complex so I hope you have study notes to look at.

It assumes an understanding of exponential curves and the RC time constant. The website below may help

Answer 3

Sorry U show no battery in the circuit.