Circuit with capacitor question

[Timiop2008]

Hi Can anybody check the answers to this question?. I am not sure if they are correct. I would appreciate if anyone can help me with the last part, (c). Thanks

A Parallel Circuit contains three components attached in parallel a 4.5V Cell, a 2200uF Capacitor and a 5.6kΩ resistor. A switch in the circuit is moved into the on position so the capacitor discharges through the resistor.

a) What is the value of the Time Constant RC for this circuit?
Attempt:
T=RC
=5600X2.2
=12,320s

b) When the switch is turned on, show that initial discharge is about 0.8mA
Attempt:
1/RC
=1/12320
=8.17X10^-5 A
=0.817mA

c) Calculate the Current In the circuit after RC seconds
I do not really know how to do this question. Would you use I=Q/t ?

 

[Delphi51]

In (a), note that 2200uF = 2200*10^-6 F
In (b) you use 1/(RC) as if it is a current. Actually RC has units of seconds.
If the capacitor was initially charged to 4.5 Volts, then the initial current would be I = V/R = 4.5/5600.

If you just want the answer to (c), read the first sentence at
http://en.wikipedia.org/wiki/RC_time_constant
This is because the capacitor/resistor circuit voltage decays like
V = Vo*e^(-t/RC)

 

[Andrew Mason]

a) What is the value of the Time Constant RC for this circuit?
Attempt:
T=RC
=5600X2.2
=12,320s

No. The capacitance is in microfarads, not milli farads.

b) When the switch is turned on, show that initial discharge is about 0.8mA
Attempt:
1/RC
=1/12320
=8.17X10^-5 A
=0.817mA

V = 4.5 volts. R = 5.6kohms. Use Ohms law to find the current.

c) Calculate the Current In the circuit after RC seconds
I do not really know how to do this question. Would you use I=Q/t ?

Use:

V = V_0e^{-t/RC}

When t = RC seconds, what is V?

Use Ohm's law to work out the current from this voltage.

AM