Changing the dielectric causes the capacitance to change.
An open circuit, by definition, has no continuity, therefore there is no current flow. A failed capacitor in an open circuit would have absolutely no effect.
The circuit becomes a pure resistance circuit where current and voltage are in phase with each others.
When the capacitor is removed from a circuit containing a lamp, the behavior of the lamp depends on the circuit configuration. In a basic circuit where the capacitor is used for filtering or smoothing, the lamp may flicker or turn off due to the loss of charge stabilization, leading to inconsistent voltage. If the capacitor is part of a timing or oscillation circuit, the lamp may stop functioning altogether. Overall, the removal of the capacitor typically disrupts the normal operation of the lamp.
It increases. The time constant of a simple RC circuit is RC, resistance times capacitance. That is the length of time it will take for the capacitor voltage to reach about 63% of a delta step change. Ratio-metrically, if you double the resistance, you will double the charge or discharge time.
The capacitance doesn't depend on the charge stored in it. The capacitor has the same capacitance whether it's charged by a DC and just holding it, or in an AC circuit where the charge on it keeps changing and reversing, or in a box on the shelf connected to nothing and not charged at all.
When a capacitor is connected to a circuit, the current flow through the capacitor initially increases and then decreases as the capacitor charges up.
Changing the dielectric causes the capacitance to change.
What happens to the current in a circuit as a capacitor charges depends on the circuit. As a capacitor charges, the voltage drop across it increases. In a typical circuit with a constant voltage source and a resistor charging the capacitor, then the current in the circuit will decrease logarithmically over time as the capacitor charges, with the end result that the current is zero, and the voltage across the capacitor is the same as the voltage source.
If voltage is increased, capacitance remains constant. Capacitance is determined by the physical properties of the capacitor, such as plate area, distance between plates, and permittivity of the material, and is not affected by changes in voltage applied across the capacitor.
The capacitance won't change, or it won't change significantly. The capacitance is simply the charge/voltage ratio - and if the charge doubles, the voltage will also double. Capacitance is determined by the physical properties of the capacitor (plate separation, plate area, and dielectric). The unit for capacitance (farad) is a coulomb per volt. So the capacitance is the amount of charge (coulombs) that the plates will hold at a given voltage.
An open circuit, by definition, has no continuity, therefore there is no current flow. A failed capacitor in an open circuit would have absolutely no effect.
A capacitor discharges when it releases the stored electrical energy it has accumulated. This typically happens when the capacitor is connected to a circuit or load that allows the energy to flow out of the capacitor.
When capacitors are connected in series, their total capacitance decreases. This is because the total capacitance is inversely proportional to the sum of the reciprocals of the individual capacitances. The voltage across each capacitor remains the same.
The circuit becomes a pure resistance circuit where current and voltage are in phase with each others.
When the capacitor is removed from a circuit containing a lamp, the behavior of the lamp depends on the circuit configuration. In a basic circuit where the capacitor is used for filtering or smoothing, the lamp may flicker or turn off due to the loss of charge stabilization, leading to inconsistent voltage. If the capacitor is part of a timing or oscillation circuit, the lamp may stop functioning altogether. Overall, the removal of the capacitor typically disrupts the normal operation of the lamp.
If by power supply you mean a voltage source, it really won't matter that the resistor is removed. The voltage source will provide infinite current, instantly charging the capacitor so that the capacitor's voltage is equal to the source.Alternative AnswerIf you are referring to an a.c. circuit, then a load current will continue to flow with its value being determined by the capacitive reactance of the circuit, and the resulting phase angle will lead the supply voltage be very close to 90 degrees.