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I apologize if the question is superfluous or already answered.

This is what I know: the charging of a capacitor from a dc source happens instantaneously. Since there is no series resistor to limit the current, then what actually prevents the current to become infinite and burn the capacitor at charging time?

On the opposite is the discharging. If a charged capacitor is shorted, it will burn. Otherwise said, the discharge can not happen as fast as the charging; why the capacitor can charge instantaneously but it can not discharge as fast?

What limits the charging current effectively preventing the burning of a capacitor, in the absence of any current-limiting resistor?

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The DC source has some internal resistance, aka it can't physically source infinite current (even if you directly short the terminals). Additionally Caps have ESR (Equivalent Series Resistance) that also limit current.

Do note that sometime it is prudent to put current limiting resistors at the gate of MOSFETS for example, to reduce current spikes.

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No, charging of the capacitor does not happen instantaneously...not in real life, and not in a properly constructed circuit analysis. Connecting a discharged capacitor directly across a voltage source violates the definition of "parallel" and KVL. Likewise for discharging. There must be resistance.

In essence, you are correct that if there is no resistance then the current, whether charging or discharging, must be infinite. So, if you want to ask a meaningful question you must assume that there is resistance in the circuit.

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This is what I know: the charging of a capacitor from a dc source happens instantaneously. Since there is no series resistor to limit the current, then what actually prevents the current to become infinite and burn the capacitor at charging time?

Alas, if a capacitor will really "burn" at high currents, it will burn regardless of which way the current is going.

On the opposite is the discharging. If a charged capacitor is shorted, it will burn. Otherwise said, the discharge can not happen as fast as the charging; why the capacitor can charge instantaneously but it can not discharge as fast?

I've never seen this happen. A single charge or discharge cycle will not realistically destroy a capacitor. I suppose there are might (might, I say) be caps which self-destruct in the event of a single event, but frankly I've never run across one in 50 years of messing around with electronics.

What can happen is that, if a capacitor is repeatedly charged and discharged at high frequency, some types can overheat and blow/burn up. In this case, you use a capacitor with a lower internal resistance (called ESR, short for "Equivalent Series Resistance"). For any capacitor, internal power is produced by the internal resistance of the cap, which produces a power related to the square of the current divided by the resistance, and if you make a thicker internal structure the resistance decreases. (This gets complicated at high frequencies due to phase shifts caused by both capacitive and inductive effects, but you don't need to worry about those just yet.)

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What limits the charging current effectively preventing the burning of a capacitor

That would be my question to you, because I don't know your power supply source or the wiring from the source to the capacitor. This impedance matters.

If your wire is 6" of 2/0 AWG wiring from a truck battery, that will have a very different impedance (effective resistance) than 10' of #26 wiring from a 1A wall-wart.

The load side of the capacitor will have impedance also. Both impedances factor into the actual current the capacitor will flow.

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  • \$\begingroup\$ At this point I’m not sure wether my questions are understood. If a cap burns at discharge (when shorted), why the same cap doesn’t burn at charge time (when it acts as a short for power supply)? This comes out of considering the capacitor a short when discharged which should raise the charge current high enough to burn through. And yet no capacitor burns when charging. Why is that? What limits the charging current through a capacitor? \$\endgroup\$ – WindSoul Feb 28 '20 at 17:15
  • \$\begingroup\$ @WindSoul I just told you. Here's the thing. Whenever drawing the circuit, always draw the wires. And draw the wires like they have a resistor in them, because wires do in fact have resistance/impedance. The power supply also has a built-in virtual resistance, though it might not be linear. \$\endgroup\$ – Harper - Reinstate Monica Feb 28 '20 at 17:34
  • \$\begingroup\$ The wire resistance is below milliohms. This can’t count as current limiting resistor at charge time. Not sure exactly what you mean by “I just told you”. \$\endgroup\$ – WindSoul Feb 28 '20 at 17:37
  • \$\begingroup\$ What is your charging source? It will have some resistance inherent in it. \$\endgroup\$ – SomeoneSomewhereSupportsMonica Feb 28 '20 at 20:19
  • \$\begingroup\$ 5VDC/2A is the source. Not sure of overload current, but I take it above 2A it will burn. Question is what size filtering capacitor is safe for the supply? Would a 1F completely discharged burn the supply at power on? \$\endgroup\$ – WindSoul Feb 28 '20 at 22:53

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