Want this question answered?
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.
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 counter acts the inductivity (decreases it) without impacting the resistivivity, thus increasing the power factor, or resistivity / inductivity ratio.
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.
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.
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 total capacitance from capacitors that are connected in series are added up inversely; 1/Ctotal = 1/C1 + 1/C2 + ... + 1/Cn, where Cn is the capacitance of the nth capacitor.
The circuit becomes a pure resistance circuit where current and voltage are in phase with each others.
There's no effect since the capacitor was already faulty i.e it was like not in the circuit. Install a healthy capacitor because it will improve the power factor of the fluorescent lamp circuit thus reducing energy wasted.
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.
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.
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.
When the terminals of a capacitor are connected together, the capacitor will discharge, returning to a zero potential state. Capacitors resist voltage change, meaning that if the capacitor is in a circuit that has zero voltage potential, the capacitor will eventually achieve zero potential. If the capacitor is in a circuit that has a 5 volt potential, the capacitor will seek and attempt to maintain that 5 volt potential (provided that the capacitor is rated at 5 volts or more). In an AC circuit, the capacitor will tend to smooth out the sin wave of the current, resisting change in both directions. In a DC power supply circuit, a capacitor will tend to reduce the voltage "ripple", and if the circuit is designed properly, will provide a smooth DC voltage. Shorting the terminals of a capacitor is effectively what often happens in many circuits; it's not a problem.
The capacitance counter acts the inductivity (decreases it) without impacting the resistivivity, thus increasing the power factor, or resistivity / inductivity ratio.