Capacitor in AC circuit plus DC offset

nik2009

Joined Jan 17, 2010
13
Hello!

This is not exactly a homework, but anyway.
I'm trying to figure out one thing 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.

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.

As I understand the voltage between point A's and the ground is
Ua = Ir2 * R2 + Ic * Xc
and
Ic = Ua / Xc

Where
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, the voltage drop between A and the ground depends on current through the capacitor but current through the capacitor depends on point A's voltage. 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

If the question is not clear enough I will gladly clarify.

Thank you!

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Jony130

Joined Feb 17, 2009
5,539
But Va is also equal to:
Va = (V1 + V2) - IR1 * R1
And Xc is constant because frequency of a AC signal has not change.

PS and don't forget about the phase shift. In capacitor current leads the voltage

Last edited:

nik2009

Joined Jan 17, 2010
13
Hi, Jony130!

I understand it isn't hard to find out Ic and Va values using math.

But how does the circuit "finds out" these values? It doesn't use the math. (I hope you understand what I mean . And sure this question isn't very practical, but still I'm curious to know)

I asked another person the same question and basically what he responded was that it is just the way circuits work - they obey Ohms law, Kirchhoff's laws, etc. And if I understand him correct, it isn't possible to explain why they work this way using some more "low" level concepts than these laws.

Do you think so too?

Jony130

Joined Feb 17, 2009
5,539
Yes I will give you the same answer. The "mother nature" has no problem with that circuit. Only the humans have a problem to understand the "mother nature" rules. And "mother nature" respect their own rules.

If you remove the capacitor from the circuit did you still have a problem with the voltage at point A. Maybe the problem lies in capacitor "action".

nik2009

Joined Jan 17, 2010
13
If you remove the capacitor from the circuit did you still have a problem with the voltage at point A. Maybe the problem lies in capacitor "action".
If I remove the capacitor I don't have the problem with Va.
Yes, the capacitor is puzzling me a little.

For example, V1 voltage decreases as a result Va decreases to a voltage below of the voltage capacitor holds (is charged to) at that moment and as a result the capacitor starts to discharge.
Now Ic (the discharge current through a capacitor) depends on Va.
Ic = Va / Xc.
But Va itself must satisfy Va = Ic * Xc.

So I can't understand when V1 decreases as a result what changes first Va or Ic. Va can't change first because it must "know" value of Ic. And Ic can't change first because it must "know" value of Va.

Do Va and Ic just change to appropriate values at the same time so neither of them comes first?

Jony130

Joined Feb 17, 2009
5,539
In this case the capacitor current will be first.

As you should know the capacitor oppose the change in voltage by supplying or drawing current.
Capacitor current is proportional the rate of voltage change across it (proportional to how quickly the voltage across capacitor is changing).
The faster the voltage change (frequency of a AC signal is high) the large the current flow through capacitor.

I = C * dV/dt

This means that to sustain current through a capacitor the applied voltage must change. The more rapidly voltage changes the larger the current. On the other hand if voltage is kept constant no current will flow no matter how large the voltage.
Likewise if the current through a capacitor is found to be zero, this means that the voltage across it must be constant, not necessarily zero.

In outer special case the voltage is a sine-wave.
So the rate of voltage change reach maximum when voltage signal crossing zero. And the rate of voltage change reach minimum when the voltage is at a peak value.

So in your circuit when V1 reach +5V the current in R1 reach his maximum value.
The capacitor current also will be at peak value and capacitor voltage will reach zero (2.5V).

nik2009

Joined Jan 17, 2010
13
Thanks Jony130.
I'm probably getting boring, but could you please answer one or two more questions

Let's imagine V1's voltage just changed to -5V so V1 + V2 = 10V. In this case, the Va's voltage is 2.5V and the current Ic is -10mA.

For example, when you are analyzing V1's voltage change, does your chain of thoughts looks more similar to A or B or C.

A. V1's voltage decreased consequently Va's voltage became 2.5V consequently after a tiniest period of time Ic became -10mA.
B. V1's voltage decreased consequently Ic became -10mA consequently after a very small period of time Va's voltage became 2.5V.
C. V1's voltage decreased consequently Va's became 2.5V and Ic became -10mA at the very same moment.

Thanks!

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Jony130

Joined Feb 17, 2009
5,539
My answer is C, when we dealing with sine-wave the current and voltage change simultaneously. So it is hart for as to analyze the circuit because we don't know the past.
Replace the sin-wave with square wave and know the circuit is much more easy to analyze.

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