Capacitance, t=0 and t=infinity

In summary: The current is proportional to the time rate of change of the charge (the derivative) i=dQ/dt. The current is 0 at t=0 and increases to a maximum value at t=∞. At t=∞ the voltage on the capacitor is equal to the electromotive force ε. In summary, when switch S1 is closed at t=0, there is no current or charge on the capacitor. As time passes, the current increases and the charge on the capacitor increases, causing the voltage on the capacitor to increase. At t=∞, the voltage on the capacitor is equal to the electromotive force ε and the current is at its maximum value. The resistance of a capacitor is infinite, and its

nomorenomore

Homework Statement


1. Referring to the picture, switch S1 is closed at t=0.
a) What is the current in the circuit loop at t=0 and t=∞?
b) What is the voltage on C at t=0 and t=∞? (4M)

physexam1q7.png


(Picture's Link: http://s29.postimg.org/bfsbw2hc5/physexam1q7.png)

Homework Equations

When charging,
1.) ε - iR - q/c = 0
2.) max. current = -ε/R
3.) max. charge = C*ε
4.) q = Qfinal(1 - e^(-t/(RC))
5.) i = (Qfinal/RC)*e^(-t/(RC))

The Attempt at a Solution

:H

a) In fact I don't understand what t = 0 means. So when t = 0, is there any current yet?
For t = ∞, I tried to sub t = ∞ to equation 5) and use equation 3) to find the Qfinal. But it doesn't seem correct. Should I do it this way?
b) When t = 0, voltage on C = ε - iR? I assume the current has already thrown through the circuit when t = 0?
For t = ∞, should I sub t = ∞ into equation 5)? But it seems i = ∞ then? doesn't seem correct.

Very confused.:( Please help.

(Plus, how do we calculate the resistance of a capacitor? When I google it, I found someone said it's assumed to be zero. So in calculation, I assumed its resistance to be zero?)

Last edited: Dec 18, 2014 Homework Helper 15,542 1,917 nomorenomore said:

Homework Statement


1. Referring to the picture, switch S1 is closed at t=0.
a) What is the current in the circuit loop at t=0 and t=∞?
b) What is the voltage on C at t=0 and t=∞? (4M)

physexam1q7.png


(Picture's Link: http://s29.postimg.org/bfsbw2hc5/physexam1q7.png)

Homework Equations

When charging,
1.) ε - iR - q/c = 0
2.) max. current = -ε/R
3.) max. charge = C*ε
4.) q = Qfinal(1 - e^(-t/(RC))
5.) i = (Qfinal/RC)*e^(-t/(RC))

The Attempt at a Solution


a) In fact I don't understand what t = 0 means. So when t = 0, is there any current yet?

Initially, the switch is open. No current flows and there is no charge on the capacitor. You start measuring time at the instant when you close the switch. Current starts to flow, but there is no charge on the capacitor yet.

nomorenomore said:

For t = ∞, I tried to sub t = ∞ to equation 5) and use equation 3) to find the Qfinal. But it doesn't seem correct. Should I do it this way?

What did you get for the current at t = ∞?

nomorenomore said:

:H

b) When t = 0, voltage on C = ε - iR? I assume the current has already thrown through the circuit when t = 0?
For t = ∞, should I sub t = ∞ into equation 5)? But it seems i = ∞ then? doesn't seem correct.

You should know how the voltage on the capacitor is related to the charge on it, and that the charge is increased in the rate of current flow..
Current flows onto the capacitor till the voltage across the capacitor is less than the electromotive force ε. At t=0, the charge is 0. What is the current then according to equation 1.)?
As the charge accumulates on the capacitor, the capacitor voltage Uc increases and the voltage across the resistor Ur decreases. As i=Ur/R , the current decreases. After very long time the capacitor voltage become very close to the emf and the currents tends to zero.

Try to sketch how q and i changes with time according to 4.) and 5.).

nomorenomore said:

Very confused.:( Please help.

(Plus, how do we calculate the resistance of a capacitor? When I google it, I found someone said it's assumed to be zero. So in calculation, I assumed its resistance to be zero?)

The capacitor has infinite resistance. It has capacitance. The voltage is proportional to the charge Uc=Q/C.