Electronics & Software – Tips & Guide


When a voltage source is connected to a capacitor is it an open circuit

voltage source is connected to a capacitor

if your source is alternative, then the capacitor will charge and discharge, charge and discharge. source is continuous, then it will charge until it has the same voltage as the source, after which there will be no more potential difference and therefore no current, it will behave like an open circuit.

if the voltage source is DC, and once the capacitor is charged to the source voltage, it will no longer draw power, in this stationary state it will act as if there were none and the source will see an open circuit.

if your voltage source is DC, there is will have an initial overvoltage when the capacitor is charged.After the capacitor has reached At the source voltage, the current in the capacitor will stop, and you will have practically an open circuit.

However, if your voltage source is AC, where the voltage goes to zero volts from time to time, the capacitor will charge and discharge itself constantly. no open circuit here !!!!

no! it is a reactive component. if the voltage is constant, any additional load placed on the source would be shared by the capacitor, which would increase the capacity of the source to resist this load. if the source voltage changes, the capacitor resists this change until it is balanced at the new level. like a balloon in a water pipe, if the pressure is constant, the ball does not do anything, but if you open or close a valve, the balloon will grow or shrink depending on the pressure, if you apply too much pressure, the balloon burst.

If you connect the capacitor without series resistor, the capacitor will charge very quickly. At the moment when you connect the power source, a potentially very large current flows for a very short time, which is to say that, for a moment, from the point of view of the power supply, the capacitor will appear as a short circuit. During this short period of time, the potential difference across the capacitor changes from a value discharged from 0v to a charged value equal to but opposite to the supply voltage, in which case an open circuit to the supply.

the capacitor is charging during which the current flows. after that, the current stops. it’s with the DC circuit. when the alternating current is applied, the capacitor which charges and discharges makes the alternating current seem to pass through it. A line to remember is a capacitor that blocks the DC current and passes AC while the inductor goes from DC and blocks the AC current.

The values must be in the range for the alternative frequencies to be compatible. it’s not an open circuit

when a voltage source is connected to a capacitor, is it an open circuit?

no. Assuming the capacitor is not charged, there will be a surge of current when the instantaneous voltage is applied first. Normally, there is a resistance in the circuit to limit the inrush current. a time constant is calculated with r x c. we generally accept that it takes 5 time constants for a capacitor to be fully charged (5 x rc). when fully charged, the capacitor offers a very high resistance to the DC power supply, but a very high resistance does not mean an open circuit. an open circuit would remove the capacitor.

Here is the strange thing about capacitors; when a discharged capacitor is connected to a DC power supply (some are sensitive to polarity), it starts charging immediately (giving the impression of a closed circuit), the fact is that, although the current is flowing both towards and outside the capacitor, no current actually passes through it (think of the plates as two car parks, one filling up like the other empty) this current will circulate only for a short time, once that the capacitor will be fully charged, it will behave as an open circuit (which, in one sense, it is)

if an AC power source is connected to a capacitor, the current will flow constantly. no, when a voltage source is connected to a capacitor, it behaves like a closed circuit (unless it was already charged)