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.
Ripple voltage in a capacitor-input filter primarily arises from the charging and discharging cycles of the capacitor. When the rectifier conducts, the capacitor charges to the peak voltage of the input signal. As the load draws current, the capacitor discharges, causing the voltage to drop until the rectifier conducts again, resulting in a voltage ripple. The magnitude of this ripple depends on factors such as the load current, capacitance value, and input frequency.
The product of resistance and capacitance is referred to as the time constant. It determines rate of charging and discharging of a capacitor.
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.
The charging time of a capacitor is usually lower than the discharging time because during charging, the voltage across the capacitor is increasing from zero to its maximum value, which initially allows a higher current to flow. During discharging, the voltage across the capacitor is decreasing from its maximum value to zero, resulting in a lower current flow. This difference in current flow affects the time it takes for the capacitor to charge and discharge.
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.
Explain how a discharging capacitor in an electronic divice produce complex waveform?
In a circuit with a capacitor, resistance and capacitance are related in how they affect the charging and discharging process of the capacitor. Resistance limits the flow of current in the circuit, which affects how quickly the capacitor charges and discharges. Higher resistance slows down the charging and discharging process, while lower resistance speeds it up. Capacitance, on the other hand, determines how much charge the capacitor can store. Together, resistance and capacitance impact the overall behavior of the circuit with a capacitor.
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