Capacitance, t=0 and t=infinity

  • Thread starter nomorenomore
  • Start date
  • Tags
    Capacitance
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
  • #1

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?
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. :H

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:
Physics news on Phys.org
  • #2
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:
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. :H

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.
 

1. What is capacitance?

Capacitance is the ability of a system to store an electric charge. It is measured in units of Farads (F) and is dependent on the geometry and materials of the system.

2. What happens at t=0 in relation to capacitance?

At t=0, the capacitance of a system is at its maximum value as there is no charge stored yet. This is also known as the initial or starting capacitance.

3. What happens at t=infinity in relation to capacitance?

At t=infinity, the capacitance of a system is at its minimum value as it has reached its maximum charge storage capacity. This is also known as the final or steady-state capacitance.

4. How does capacitance change over time?

In most systems, capacitance increases as time goes on until it reaches its maximum value at t=infinity. However, there are some exceptions where capacitance may decrease over time due to factors such as aging or degradation of materials.

5. How does temperature affect capacitance?

Temperature can affect capacitance in various ways. In some materials, capacitance may increase with temperature due to the decrease in resistance. In others, it may decrease due to changes in the dielectric properties. It is important to consider temperature when designing and using capacitive systems.

Suggested for: Capacitance, t=0 and t=infinity

Back
Top