• Time evolution of RC series circuits.

Pre-requisite skills:  Knowledge of the exponential and logarithm functions, basic differential equations (optional).

Approximate completion time:  Under an hour.

Provide sufficient detail to verify that the assignment was completed in a meaningful manner.

MERLOT site:   RC Circuits, by Fu-Kwun Hwang.  (direct link). 

Applet by Fu-Kwun Hwang

(a) Starting at the positive terminal of the EMF source, sum the voltage drops around the circuit.  Denote the EMF with the variable V.    Do not use numerical values for any physical properties at this point.

2. For those having an understanding of basic differential equations, answer the following.  (For others, skip to Question 3).

(a) From your answer to the previous question, and the fact that the capacitor is initially uncharged, show that the charge Q on the capacitor evolves with time according to the relationship:

Q(t) = CV-  D Exp[-t/(RC)],

where D is a constant to be determined and Exp[-t/(RC)] means "the exponential of -t/RC."

(b) Using the fact that the capacitor is uncharged, show that this equation reduces to

Q(t) = CV{1 - Exp[-t/(RC)]}.

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3.   The left-hand side of the above equation is the charge separation of the capacitor, which can also be written as Q(t) = CV(t), where V(t) is the voltage drop across the capacitor as a function of time.   In other words

CV(t) = CV{1 - Exp[-t/(RC)]}.

(Yes, we could cancel the capacitance values on both sides, but we want to leave the C on both sides of the equation since the applet plots the time evolution of the product CV, and not just the voltage V.) We will now test this relationship and see what it has to say theoretically about the time evolution of the voltage drop across the capacitor plates.

(a)  From this relationship, verify that at t = 0 the voltage drop across the plates of the capacitor will be 0.

(b) What does this fact say about the charge separation across the plates at t = 0?

(c) What is the maximum value the product CV(t) will achieve, no matter how long the circuit is connected?  (Hint:  Let t become infinite.)

(d) How long will it take the charge to build to half the maximum value obtained in previous question [Question 3(c)]?   This time is called the time constant of the circuit.

4.  Now let's verify the charge-time evolution relationship numerically.  Notice that in the upper half of the applet is a plot of CV versus t.

(a)  Press the "Start" button and describe the behavior of the plot.  On a sheet of graph paper, reconstruct the plot, complete with labeled axes and numbered tic marks.  (Note:  The plot line needs to start at 0 on the vertical CV axis since the capacitor is initially uncharged.  If it does not, press "Reset" and start over.)

(b)  Notice that CV tend towards a maximum value.  What is this value?  

(c) Is this maximum value of CV what you expected theoretically?

(d) On your plot, indicate the time at which CV reaches half its maximum value.  Write down on the plot this value of time.   (Remember, we call this time the time constant).

(e)  Is this value close to what you expected theoretically?  How far off is it?  (Use a percent error or percent difference.)

5.  Now I want you to answer some conceptual questions.

(a) When the applet is running, why does the current drop in the circuit?  (In other words, why does the charge slow down?)  Answer in terms of the charge polarity on the plates of the capacitor.

(b) The height of the three light-blue vertical bars represent the voltage drop values across each component of the circuit.  Furthermore, these values are also shown numerically.   Explain how these three values are related.  As the applet runs, explain what happens to these values and why.

(c) Notice that charge flows through the circuit, even though the capacitor is, in essence, a break in the circuit.  Explain in terms of the flow of electrons and the charge separation on the capacitor plates.

(d) When the switch is flipped back such that the battery is taken out of the circuit, why does the capacitor discharge?

6.  Now it is time to discharge the circuit.

(a)  Once the current slows to a creep, use the mouse to flip the switch such that the battery is shorted.  (Note: Do not press "Stop" during this process.)   What happens?  Explain why in terms of the charge separation on the capacitor and the flow of charge in the circuit. 

(b) Ultimately, what will happen to the charge separation on the capacitor?

(c) Draw the resulting plot on your graph paper.    Compare the two lines (charging and discharging).

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