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Homework Help: Capacitor in AC circuit plus DC offset

  1. Oct 7, 2011 #1
    Hello guys!
    This is not quite a homework, but anyway.

    1. The problem statement, all variables and given/known data

    I'm trying to figure out some details about the circuit below.


    The circuit contains two voltage sources. V1 is 5V 100Hz AC source, V2 is 15V DC source.

    I know I can analyze this circuit simply treating it as if there are two currents. One is direct current, the other one is alternating current. But this is only a convenient way to analyze a circuit.
    And I'm trying to imagine what's happening on a bit "lower" level.

    3. The attempt at a solution

    So when AC source's voltage is increasing the current is increasing too and so more electrons are flowing through the wire. And bigger part of these electrons flow through the capacitor.

    When AC source is decreasing the current though the V1 and V2 doesn't change its direction but only decreases its value because the effective voltage across the circuit decreases but still is positive (E. g. 15V - 5V = 10V). But the current through the circuit branch with the capacitor changes its direction to the opposite.

    What I don't understand is how the voltage drop between the point A and the ground is formed.

    1. Is it like this: V1 voltage decreases -> the current decreases -> the capacitor "sees" that the current decreased and starts to discharge?
    If so, how exactly can the capacitor "see" that the current changed?

    2. Or it is like the following: V1 voltage decreases -> the current decreases -> the voltage drop across R2 decreases -> because the voltage drop decreased the capacitor starts to discharge?
    But in this case, the voltage drop across R2 is the same as the voltage drop between the point A and the ground. And the voltage drop between A and the ground depends on the currents through R2 and the capacitor, and the current through the capacitor depends on the voltage drop between A and the ground...

    If the question is not clear enough I would be glad to clarify.

    Thank you!

    Attached Files:

  2. jcsd
  3. Oct 7, 2011 #2


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    Staff: Mentor

    Take a look at what happens at point A if the capacitor is not there. The potential across the resistor stack will vary with time, and point A will be a certain fraction of that total voltage.

    Now when the capacitor is slapped on it acts like a sort of "voltage boat anchor" that tries to keep point A from changing potential too quickly. Current runs in and out of the capacitor as it tries to keep up with the ever changing "steady state" for the given circuit condition at any given instant. The rate at which current can flow in and out of the capacitor is determined by what the resistors will allow (hence the circuit has a time constant).

    When point A is being driven to a potential that is below what the capacitor currently holds, current must flow out of the capacitor, desiring to match that potential. Similarly, when point A is being driven towards a potential that is higher than that currently on the capacitor, current will flow into the capacitor to drive its potential higher.

    The capacitor's charging or discharging depends upon the potential it currently has on it and the potential that is trying to be established across it by the surrounding circuit. It doesn't "see" current, it "sees" potential.
  4. Oct 7, 2011 #3
    Hi gneill,
    Thank you for the answer.

    As I understand point A's potential is
    Ua = Ir2 * R2 + Ic * Xc
    Ic = Ua / Xc

    Ua - point A's potentiatel
    Ir2 - curren through R2
    R2 - R2 resistance in Ohms
    Ic - current through the capacitor
    Xc - the capacitor's impedance

    So to put this in words, potential of point A depends on current through the capacitor but current through the capacitor depends on point A's potential. And I can't understand what comes first here.

    E. g. if V1's voltage is changing then as a result Ua is changing but how the circuit knows the new value of Ua if it needs Ic to determine Ua. And to determine Ic the circuit needs Ua... And so on in circles :)
  5. Oct 7, 2011 #4


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    Staff: Mentor

    Well, that's the essential property of a physical system with mutually dependent and interacting quantities and qualities. It's also why we need differential equations to describe them!
  6. Oct 8, 2011 #5
    Thank you, gneill.
  7. Oct 9, 2011 #6
    Would it be correct to think that when V1's voltage changes then as a result Ua and Ic change to appropriate values at the same time so neither of them comes first/second?
  8. Oct 9, 2011 #7


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    Staff: Mentor

    Yes, that would be the macroscopic view of what happens.

    On the microscopic scale you would eventually be dealing with quantum effects, individual particles and averages of assemblies of particles (electrons), Bell curves and their outliers, and so on. The "which came first" argument becomes rather grey in that arena, but fortunately we're not dealing with things on that scale here! :smile:
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