Can someone clarify whether or not a capacitor blocks current at DC?
simulate this circuit – Schematic created using CircuitLab
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\$\begingroup\$ You please go through the below link where you get to understand the basics of capacitors and analyse capacitor circuits en.wikipedia.org/wiki/Capacitor \$\endgroup\$– IamDpJul 23, 2013 at 13:38
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\$\begingroup\$ EEVBlog has a video on this topic: eevblog.com/2013/06/21/… \$\endgroup\$– JYeltonJul 23, 2013 at 15:47
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2\$\begingroup\$ insufficient preliminary research \$\endgroup\$– Nick AlexeevJul 23, 2013 at 16:42
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3 Answers
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The current in a capacitor is C dv/dt
It's as simple as that - if there is no change in voltage with respect to time (dv/dt) then there is no current flowing.
This is for a perfect capacitor with no leakage between its plates. If there is leakage there will be a small current that flows that is proportional to the applied voltage.
Polarized capacitors do not take kindly to reversing the working voltage and they will conduct current with a steady reverse voltage applied.
answered Jul 23, 2013 at 13:38
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Is current able to flow through a Capacitor at DC?
The answer depends on what you mean by "at DC".
If you mean "is current able to flow through a capacitor at DC steady state?", then the answer is no. Since the voltage across the capacitor is constant in DC steady state, the current through the capacitor is zero.
However, if you mean "is DC current able to flow through a capacitor?", the answer is, in the context of ideal circuit theory, yes.
In this (ideal) circuit, there is a 1A constant (DC) current through the capacitor. The voltage on the capacitor steadily increases at the rate of 1 volt per second.
Of course, such a circuit is not physical but it does serve to emphasize that, although it is often said that capacitors "block DC", this is, IMHO, misleading.
The fact is, as Andy aka points out, the current through an ideal capacitor is zero if there is a DC (unchanging) voltage across the capacitor.
But, it doesn't follow that there cannot be a "DC" current through a capacitor. If the current source bothers you in anyway, we could change the circuit to the following:
Here, the voltage source produces a voltage that increases at the rate of 1 volt per second. In this case, there is a constant (DC) current of 1A through the capacitor.
answered Jul 23, 2013 at 14:21
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3\$\begingroup\$ I'd argue that your 2nd circuit is not a DC circuit, since it has a source whose value varies with time. And your first has no DC solution, since not all of the circuit variables converge to a steady state value. But whether these are "DC currents through a capacitor" is more about semantics than engineering. \$\endgroup\$ Jul 23, 2013 at 14:58
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1\$\begingroup\$ @ThePhoton, neither my circuits nor the OP's are DC circuits since there is a capacitor present. As you point out, my 2nd circuit doesn't even have a DC source. Yes, the 1st circuit has no DC steady state solution. But it is a fact that the capacitor current is constant in both circuits, i.e., the capacitor does not "block" the constant (often called "DC") current. If you interpret the question quoted to be "is current able to flow through a capacitor in DC steady state", then the answer is almost trivially no. My answer serves to point out that the question is ambiguous. \$\endgroup\$ Jul 23, 2013 at 15:21
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1\$\begingroup\$ Perhaps OP should first decide on what is actually DC, because from what I can see, the question as it is now is only about semantics. For example back when I was first taught the definitions, "DC" meant that polarity of the voltage does not change and "AC" meant that the polarity does change. Neither implied anything about the actual waveform of the voltage and we even had pictures of AC and DC square waves in the book. \$\endgroup\$ Jul 23, 2013 at 16:28
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2\$\begingroup\$ @AndrejaKo, indeed, the terms DC and AC are ambiguous. For example, "DC circuit analysis" gives the solution when all voltages and currents are constant while "AC circuit analysis" gives the solution for sinusoidal steady state. I generally take "DC" to mean constant and "AC" to mean time varying with zero time average. The total signal is then the sum of the "DC" and "AC" components. \$\endgroup\$ Jul 23, 2013 at 16:40
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With an ideal capacitor, a sustained current cannot flow in this circuit. The key concept to understand here is that the capacitor will charge (which takes non-zero time, during which a current does flow). As the capacitor charges, the current approaches zero.
