When a capacitor is discharging, current is flowing out of the capacitor to other elements in the circuit, similar to a battery. Current flowing out of an element, by convention, is defined as negative current, while current flowing into an element, such as a resistor, is defined as positive current. Thus a discharging capacitor will always have a negative current.
Look closely at the circuit. If discharging is slower than charging, then there must be less current available for discharging. The equation of a capacitor is... dv/dt = i/c which means that the slope of the voltage is proportional to the current and inversely proportional to the capacitance. This does not matter if it is a charging or a discharging situation. For example, one amp of current into one farad of capacitance will be one volt per second, and negative one amp of current into (out of) that same one farad capacitance will be negative one volt per second. You don't give a lot of information in the question, but it sounds like you are analyzing the filter in a rectifier. The charge path is through the diode, which is a low impedance, high current, circuit, while the discharge path is only through the load. This would explain, in one case, why the discharge time is greater than the charge time.
The voltage goes to zero because a current path has been created between the positive and negative elements of the capacitor, discharging the stored charge and putting both the anode and cathode of the capactor at the same electrical potential. Thus, no voltage difference between them, which is why the voltmeter reads zero.
Because that is what a capacitor does, resist a change in voltage. It holds a certain amount of energy per charge (voltage), and to change that voltage requires current proportionally to the capacitance.
The product of resistance and capacitance is referred to as the time constant. It determines rate of charging and discharging of a capacitor.
The voltage across a capacitors given as a time constant t= 63% the resistor value multiply buy capacitor value. it doesn't matter if it goes more or less negative it will follow this function
When a capacitor discharges the discharge current flows in the opposite direction to the current used to charge it.
The 'conventional current' flows out of the positive side of the charged capacitor, and into the negative side. However, even though we never talk about it, we know that the things that actually carry the physical current around are the negatively charged electrons, and we know that when a capacitor is discharging, the electrons are flowing out of the negative side and into the positive side.
using CRO we can measure the rise time and fall time of the capacitor for further studies
Depending on the capacitor we are using it will have a cathode.For example if we take a unicapacitor(it will allow current on both sides) it will have a negative and a bi capacitor it will not have negative
It really depends on the experimental setup. If you have only a capacitor and a resistance in series, the current discharge from the capacitor will start high, then gradually go down. If you have a capacitor and an inductor in series, the current discharge will start being small, because the inductor will oppose any CHANGE in the current - that's how they work.
Because Alternet current has both positive as well as negative cycle capacitor does not conduct for negative cycle of the Alternet current and DC all are positive cycle thats why it capacitor conduct for DC not for AC
Explain how a discharging capacitor in an electronic divice produce complex waveform?
Look closely at the circuit. If discharging is slower than charging, then there must be less current available for discharging. The equation of a capacitor is... dv/dt = i/c which means that the slope of the voltage is proportional to the current and inversely proportional to the capacitance. This does not matter if it is a charging or a discharging situation. For example, one amp of current into one farad of capacitance will be one volt per second, and negative one amp of current into (out of) that same one farad capacitance will be negative one volt per second. You don't give a lot of information in the question, but it sounds like you are analyzing the filter in a rectifier. The charge path is through the diode, which is a low impedance, high current, circuit, while the discharge path is only through the load. This would explain, in one case, why the discharge time is greater than the charge time.
resistance does not produce currents . you need source (like voltage source , current source ,or , discharging capacitor) to generate current .
if the source is switched off there will be leakage slowly discharging the capacitor
It flows out of the capacitor into the external circuit
In capacitor we can observe that ac flows but dc doesnt. This flow of ac is due to the alternation of charging and discharging cycles of capacitor. Though there is not any actual flow of electricity we can observe that certain current flows in the relevant circuit. Displacement current defines that equivalent current. This was postulated by maxwell.