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In this circuit you can find an RC circuit and 4A DC current source.

I thought that if we use a DC voltage source instead of a DC current source, this circuit would become open circuit after a while. Because the voltage on the capacitor is equal to the voltage source.

I've applied it the same way and thought about the case as DC current source. I think the result should be the same open circuit result.

But it's wrong and I don't know the reason.

This is my circuit and ? symbol is my question. Why this spot's current is not zero? I mean open circuit. This is simulink result and you can find this capacitor's current is still 4A (yellow line)!

Can you tell me why the current source in the RC circuit is not open circuit?

enter image description here enter image description here

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For a capacitor, Q = CV and, if we differentiate Q with respect to time we get current so: -

$$\dfrac{dQ}{dt} = I = C\dfrac{dV}{dt} $$

So, if \$I\$ is a constant current source then \$\dfrac{dV}{dt}\$ will be constant and that means that voltage will continue to rise linearly with time until your simulation runs out of numerical range or, in the real world, your "practical" current source is unable to generate the terminal voltage required to sustain a constant current.

I thought that if we use a DC voltage source instead of a DC current source, this circuit would become open circuit after a while. Because the voltage on the capacitor is equal to the voltage source.

Correct.

I've applied it the same way and thought about the case as DC current source. I think the result should be the same open circuit result.

Incorrect, for the reasons given above.

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  • \$\begingroup\$ thank u so much!! \$\endgroup\$
    – Chris Park
    Jan 13 '20 at 10:01
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Your current source pushes to the capacitor 4A. Its voltage grows 4 volts per second. The program does it because you have ordered the program do so. I guess that's why most people pay for that software.

I bet in a circuit which is made of real parts you sooner or later meet the max available voltage that the current source can produce. Or the insulation in the cap breaks. In simulations reaching the max available number is the roof.

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  • \$\begingroup\$ Can you explain a little more? Because I still don't know why the capacitor can't open and is short like wire. \$\endgroup\$
    – Chris Park
    Jan 13 '20 at 8:28
  • \$\begingroup\$ Its not short. It's a piece which takes new charge as long as someone has a device which can win the voltage in the capacitor. The voltage is the accumulated charge divided by the capacitance. An ideal current source pushes the selected current to anything, no matter how high voltage is needed. Leave it unwired, then you probably get an error message. \$\endgroup\$
    – user287001
    Jan 13 '20 at 8:33
  • \$\begingroup\$ You mean the current is pushing through the middle of the capacitor? So a few times later or if more complexity circuit r getting an error message? \$\endgroup\$
    – Chris Park
    Jan 13 '20 at 8:39
  • \$\begingroup\$ When one charges a capacitor the charge accumulates to the metal plates. No current goes through the space between the plates as long as it's a cap. The charging device pulls electrons from the plus side and pushes them to the minus side. The current goes thru the charging device, not in the middle between the plates of the cap. \$\endgroup\$
    – user287001
    Jan 13 '20 at 8:44
  • \$\begingroup\$ yes right, the plates as long as it's a cap, one side will be "+" charge, and the other one is will be "-" charge. so a few times later this cap has a same voltage as the source voltage and then become open circuit. But when it's an current source, it doesn't open. I'm confused. i'm so sorry... \$\endgroup\$
    – Chris Park
    Jan 13 '20 at 8:49

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