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Analysis of circuit with capacitors.

  1. Mar 10, 2014 #1
    1. The problem statement, all variables and given/known data

    Given the attached circuit, compute Tau, UC1 (t) and UC1 (∞).

    Then assume t = ∞ and C1 has the potential of Eth. Put the switch in the other position, so that C1 can discharge through R3. C2 will, of course, charge untill UC1 = UC2.

    Compute Tau, UC1(t), UC2(t) and IR3(t).

    From these calculations, it should be simple to graph it using the y axis as U or I and the c axis as Tau (time (s)).

    3. The attempt at a solution

    First things first. I attached the whole problem, since it's related to one circuit.

    Compute Tau, UC1 (t) and UC1 (∞).

    I started with transforming the circuit into a thevenin equivalent.

    RTH = R1||R2 = 2350Ω.
    ETH = ER2 = [itex]\frac{R2 * E}{R1 + R2}[/itex] = [itex]\frac{4700Ω * 8V}{4700Ω + 4700Ω}[/itex] = 4V.

    Tau = R*C = 2350Ω * 680*10-6 F = 1.598S

    UC1 (t) = ETH (1-e[itex]\frac{-t}{Tau}[/itex]) = 4V * (1-e[itex]\frac{-t}{1.598S}[/itex])

    UC1(∞) is my first problem, assuming the above is correct. By infinity, do they mean when C1 is fully charged? From earlier, I think 5 Tau used to be the definition of a fully charged capacitor, but this may be completely wrong, and knowing me, it is.

    Attached Files:

  2. jcsd
  3. Mar 10, 2014 #2


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    That would be my understanding.

    There is an easier way. The original circuit has two resistors forming a potential divider feeding the capacitor. So what's the maximum voltage the capacitor can reach? eg What voltage is it charging towards?
  4. Mar 10, 2014 #3
    I would assume 4V, as it's the same as R2 and the potential is divided equally between two resistors with the same value. And it's the same as I've set ETH to be.

    Is this wrong?

    Edit: does my calculations look ok by the way? Also, thanks a lot for taking your time to help me. Really appreciate it.

    Edit 2: When time is infinitively large, or as it approaches a huge number, e will approach 0, therefore:

    UC1(∞) = ETH(1-e[itex]\frac{-∞}{Tau}[/itex]) = 4V(1-0) = 4v.

    Does that look ok? Also, I can see I'm having some issues with e to the... power, but hopefully it's understandable.
    Last edited: Mar 10, 2014
  5. Mar 10, 2014 #4


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    Your calculations look fine for the switch in the first position, if you assume that V1 is being switched on at time t=0. Really I think they want you to just assume that the switch has been in that position for a very long time and you need to find the voltage on the capacitor before the switch is moved to the second position. That is, the steady state value of UC1 before the switching takes place.

    A simple approach to finding the steady state value for the voltage on a capacitor being driven by a fixed source and resistive network is to remove the capacitor in question and determine the potential across the open terminals. That avoids having to derive the exponential equations and taking limits (t → ∞).

    What I think they're interested in is the voltage on C1 as a function of time after the switch moves to the new position. What are your thoughts there?
  6. Mar 10, 2014 #5
    This would be ETH? I've found that to be 4V.

    So the calculation for that would be:

    UC1 (∞) = ETH = 4V?

    I've been reading a little in my book - which this course is based on (introductory circuit analysis by Boylestad) and from what I can find out, analysis of a circuit concerns(?) charging and discharging. Therefore, I'm thinking charging in the default position, and discharging in the other position (the switch position, that is).

    I'm thinking compute Tau, UC1(t) and UC1(∞) for the charging of C1.

    This is what I have so far (copied from opening post and my solution from the first quote in this post).

    I started with transforming the circuit into a thevenin equivalent.

    RTH = R1||R2 = 2350Ω.
    ETH = ER2 = [itex]\frac{R2 * E}{R1 + R2}[/itex] = [itex]\frac{4700Ω * 8V}{4700Ω + 4700Ω}[/itex] = 4V.

    Tau = R*C = 2350Ω * 680*10-6 F = 1.598S

    UC1 (t) = ETH (1-e[itex]\frac{-t}{Tau}[/itex]) = 4V * (1-e[itex]\frac{-t}{1.598S}[/itex])

    UC1 (∞) = ETH = 4V?

    Then, for the discharging of the circuit, t=∞ is achieved, and C1 has the voltage (potential) of ETH. The switch is set to the other position, and C1 will be discharged through R3. As follows, C2 will be charged untill UC1 = UC2. Here I'm supposed to graph the discharging of C1, the charging of C2 and the current graph for R1.

    I'm a bit confused here, but let's take the first thing first :p The charging with the switch set in the default position.
  7. Mar 10, 2014 #6


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    Looks like you've already handled that. Your method works fine. For the initial switch position UC1(∞) = 4 V as you found.
  8. Mar 10, 2014 #7
    Ah, great! :)

    When t=∞ is achieved and we put the switch in the other position, C1 starts to discharge through R3 and C2 is charged until UC1 = UC2.

    So here I'm supposed to graph the discharging of C1, charging of C2 and the current graph for R1.

    UC1 (0) = ETHe-t/1.598s = 4Ve-0/1.598s = 4V
    UC2 (0) = ETH(1-e-t/1.598s) = 4V(1-e-0/1.598s) = 0V

    UC1 (tau) = ETHe-1.598/1.598s = 2.94V
    UC2 (tau) = ETH(1-e-1.598/1.598s) = 2.53V

    UC1 (6tau) = ETHe-(1.598*6)/1.598s = 0.02V (0v)
    UC2 (6tau) = ETH(1-e-(1.598*6)/1.598s) = 3.99V (4v)

    These values are made to sketch the graphs for the charging and discharging of C1 and C2.

    Not sure how to compute IR3(0), IR3(tau) and IR3(6*tau). Also, there's a typo here. I says it's a good idea to find the values of IR3(0), IR3(tau) and IR3(6*tau) to sketch the graph for the current through R1 - I assume it's the current through R3 that's interesting?

    Here's my attempt anyway:

    IR3(0) = [itex]\frac{ETH}{R3}[/itex] * e -t/1.598s = 0.85mA * e-0/1.598s = 0.85mA

    IR3(tau) = 0.85mA * e-1.598/1.598s = 0.31mA

    IR3(6*tau) = 0.85mA * e-(1.598*6)/1.598s = 0.0021mA

    How does this look? :eek:

    I see now that C2 doesn't behave as it's supposed to. It charges to 4V and C1 goes down to 0v. So that's as if it were a basic charging/discharging of one capacitor. How do I get them to relate to each other in my calculations?
    Last edited: Mar 10, 2014
  9. Mar 10, 2014 #8


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    There are a few problems with your analysis, related to how the circuit finally ends up (a long time after the switching takes place). An excellent approach in these cases is to examine the initial and final conditions of the circuit and then simply "connect" the two states with the appropriate exponential curves.

    Okay, at the start C1 begins with a voltage of 4 V, while C2 begins at 0 V as you stated. At the end, both capacitors must end having the same voltage, whence the current stops flowing. You should have enough information there to determine the initial current and final current.

    Next, find the final voltage that will appear on the capacitors. You can look at this by way of the initially available charge that C1 brings to the game being divvied up between the two capacitors (since charge is conserved). You can then write an equation that connects the initial voltage on C1 to the final voltage on C1 via an exponential curve.
  10. Mar 10, 2014 #9
    The voltage over C1 initially is 4V, and finally it's equal to C2. The drop in potential across R3 factors in. Tau will also be different? Didn't think about that. How does two capacitors relate to each other when calculating the new RC? The book doesn't cover any circuit with two or more capacitors. Nor does the lecture notes.

    4V with 4700Ω = 0.85mA, but then I get 4V across the resistor as well. I'm a bit confused still.

    This is new to me. We've only worked with one capacitor. I'm trying though, believe me.

    I will go through what I have of notes, and see if I can find anything mentioned about your method here.

    Thanks a lot mate.
  11. Mar 10, 2014 #10


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    If you consider the circuit with the switch in the second position, a loop is formed. As far as the current is concerned, the capacitors are in series...

    Yeah, that's right. Initially capacitor C2 has no charge and thus 0 V across it. So to satisfy KVL around the loop the initial potential drop across the resistor must be the same magnitude as the initial voltage across the capacitor C1...

    This is where all the time you've spent looking at equivalent circuits comes in handy. If you have a series circuit, determine the net series capacitance and net series resistance to determine the tau.

    I will go through what I have of notes, and see if I can find anything mentioned about your method here.

    Thanks a lot mate.[/QUOTE]

    Glad to help!
  12. Mar 10, 2014 #11
    I will go through what I have of notes, and see if I can find anything mentioned about your method here.

    Thanks a lot mate.[/QUOTE]

    Glad to help![/QUOTE]

    I'm beyond tired right now, but I think I finally figured it out. Or at least made some progress.

    So, as you pointed out, the capacitors are actually in series. I've only studied basic electrical circuits for a few years, no wonder I still mix those up :cry:

    I've gone back to the Q = C * U formula (not going to write it all out unless I have to, because of the aforementioned tiredness). I came up with this formula eventually.

    UC1(∞) = UC1(0)[itex]\frac{CC1}{CC1 + CC2}[/itex] = 4V[itex]\frac{680uF}{680uF+680uF}[/itex] = 2V

    CRES = [itex]\frac{CC1 * CC2}{CC1 + CC2}[/itex] = [itex]\frac{680uF*680uF}{680uF+680uF}[/itex] = 340uF

    Tau=R3 * CRES = 4700Ω * (340*10-6F) = 1.598S

    So, the discharging of C1

    UC1(t) = UC1(∞) + (UC1(0) - UC1(∞)) e -t/Tau = 2V + 2Ve-t/1.598s

    Charging of C2

    UC2(t) = UC2(0) + (UC2(∞) - UC2(0))(1-e-t/Tau)
    UC2(t) = 0V + (2V - 0V)(1-e-t/1.598s) = 2V(1-e-t/1.598s)

    The current through R3.

    IR3(t) = [itex]\frac{UC1(t) - UC2(t)}{R3}[/itex] = [itex]\frac{(2V + 2Ve-t/1.598s) - (2V(1-e-t/1.598s)}{4700Ω}[/itex] = [itex]\frac{4v}{4700Ω}[/itex]e-t/1.598s

    How does this look? And sorry about the formatting. Had problems using the [ SUP][ /SUP] and [ SUB][ /SUB] functions withing the math fractions.
  13. Mar 10, 2014 #12


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    That looks excellent :smile: Well done.
  14. Mar 10, 2014 #13
    The weight that just got lifted off of my shoulders... I absolutely love this feeling!

    Thanks so much mate.
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