The switch has been open for a vary long time, and is then at time T=0 closed. Determine the voltage V0(t) for t>0
voltage from a discharging capacitor is V*e^(-t/RC) Where V is the voltage stored in the capacitor, R is the resistance and C is the capacitance.
The Attempt at a Solution
1) My understanding is that the capacitor will charge up to 12 volts, however I have been told that it only charges up to 6 because the two 2kohm resisters make a voltage divider. I did not think the location in terms of other resisters made a difference for the voltage that a capacitor will charge to but I may be wrong.
2) Once the switch is closed I thought the voltage at the node where the switch is would be 4 volts because it would act as a voltage divider between the first 2kohm resister and the two 2kohm resisters in parallel. My friend says it will be 6 volts because the capacitor acts as an open circuit so it is only a voltage divider between two 2kohm resisters. Now if this is the case then would it matter as the only voltage at V0 is the decaying voltage from the capacitor?
3) For the resistance in the time constant will it just be the 2kohm resister at V0 plus the 2kohm resister in series at the switch, or will it be something else?
Thus my answer is (12e^(-t/.02))-4 While his answer is (6e^(-t/.02))-6 I have no idea who is right or if we are both wrong, so any help is appreciated.
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