Capacitor After A Long Time

What does it mean past charging and discharging a capacitor? What are the working principles of capacitor charging? What is the capacitor charging and discharging theory?

Charging a capacitor means the accumulation of charge over the plates of the capacitor, whereas discharging is the release of charges from the capacitor plates. The transient response of capacitor charging and discharging is governed past ohm'southward law, voltage law, and the basic definition of capacitance.

Capacitor Charging:

Suppose we have the circuit below, with capacitor C, voltage source Five and a toggle switch. Consider the capacitor is discharged initially and the switch is open. At some point in time, I move the switch to position 1, and let's say that time is t=0.

Charging Electric current of the Capacitor:

At fourth dimension t=0, both plates of the capacitor are neutral and can absorb or provide charge (electrons). Past closing the switch at time t=0, a plate connects to the positive final and another to the negative. The plate of the capacitor connected to the positive terminal provides electrons because the plate has comparatively more than electrons than the source positive last. And plate connected to the negative terminal absorbs electrons provided by the source negative terminal which has comparatively more electrons. This move of the electrons is the charging electric current during the charging phase.

The voltage beyond the Capacitor:

At fourth dimension t=0, the voltage beyond the capacitor plates is "absolutely nil". As electrons kickoff moving between source terminals and capacitor plates, the capacitor starts storing charge. The phenomenon causes a huge current at the moment when the switch is closed at time t=0. As charge stores, the voltage across the capacitor rises and the electric current between source and capacitor goes down. At some stage in the time, the capacitor voltage and source voltage become equal, and practically at that place is no current flowing. The elapsing required for that "no-electric current situation" is a 5-time abiding ($5\tau $). In this state, the capacitor is called a charged capacitor.

Capacitor Charging Equation

Current Equation:

The below diagram shows the current flowing through the capacitor on the fourth dimension plot. Current flowing at the time when the switch is closed, i.e. t=0 is:

$i(t=0)=i_{R}=i_{c}=\frac{Due east}{R}$

Where instantaneous current can be found using the post-obit formula:

$i_{c}=\frac{Eastward}{R}eastward^{-\frac{t}{RC}} $

Capacitor Charging current graph — The current passing through the capacitor during the charging phase

Voltage Equation:

The below diagram shows the voltage across the capacitor and resistor on the time plot. The post-obit formulas are for finding the voltage across the capacitor and resistor at the time when the switch is closed i.e. at t=0:

$v_{c}(t=0)=0$
$v_{R}(t=0)=Due east$

Capacitor Voltage Discharge Voltage graph — The voltage across the resistor during a charging phase

The formula for finding instantaneous capacitor and resistor voltage is:

$v_{c}=Eastward (1-e^{-\frac{t}{RC}})$
$v_{R}=Ee^{-\frac{t}{RC}}$

Voltage-across-capacitor-during-charging-phase — The voltage beyond the capacitor during the charging phase

RC Time Constant:

Here R and C are replaced with the Greek alphabetic character $\tau $ (Tau) and named every bit "RC time constant" measured in seconds. The capacitor takes $5\tau $ seconds to fully charge from an uncharged country to whatever the source voltage is.

RC Time Constant = $5\tau$

Electric current and Voltage Equation:

The current across the capacitor depends upon the change in voltage beyond the capacitor. If there is a irresolute voltage across it, will describe current but when a voltage is steady there will be no electric current through the capacitor. That's why it draws current for only a small corporeality of time during charging.

${ i }_{ c }=C\frac { d }{ dt } ({ V }_{ c })$

Capacitor Charging Summary :

At the instant when the switch was airtight, the capacitor draws a very large current that behaves similar a brusk circuit. At that moment almost zero voltage appears across the capacitor. Current in the circuit is only express by the resistance involved in the circuit.
When charging time ends, the capacitor behaves similar an open circuit and there is no current flowing through the capacitor and has a maximum voltage across it.

Capacitor Discharging:

Suppose the capacitor shown beneath is charged past a voltage source Due east, and then the voltage beyond the capacitor volition be raised to voltage Eastward. Now I move the switch to position ii in the following excursion, the capacitor is connected to resistive load instead of the voltage source. The capacitor will at present work equally a source for the resistor and voltage beyond the capacitor will start to lose its stored charge bypassing current. By losing the charge, the capacitor voltage will start to subtract. For a constant resistor, the current will also start to reduce as voltage decreases. Finally, the voltage beyond the capacitor will hitting the zero point at a 5-fourth dimension constant ($5\tau $). Similarly, the current will too go to zippo afterward the same time elapsing.

Capacitor Belch Equation:

To find the voltage and current of the capacitor at whatever instant, use the post-obit capacitor discharging equation:

$v_{c}=Ee^{-\frac{t}{RC}}$
$i_{c}=\frac{E}{R}east^{-\frac{t}{RC}}$

Capacitor-Discharging-current-graph — Current through the capacitor during discharging phase

Observe the above graph is beneath the zero lines because the direction of current menstruum during discharging phase is opposite to that of the charging phase.

As the resistor and capacitor are continued in series, and then the electric current is the same for both. Where voltage across the resistor is different and represented by the following formula:

$v_{R}=Ee^{-\frac{t}{RC}}$

Capacitor-Discharging-Resistor-voltage-graph — The voltage across the resistor during discharging phase

The discharging is too dependent upon resistance and capacitance and takes to completely discharge.

Initial Condition:

In all the above word, nosotros suppose an uncharged capacitor, yet, it may not e'er be the case. Here we are interested in charging a capacitor that has already some accuse stored on information technology. Because of the charge stored, the capacitor would have some voltage across it i.due east.

$v_{i}=\frac{Q_{i}}{C}$

$Q_{i}$ is the initial accuse stored on capacitor terminals which causes the initial voltage on its terminals $v_{i}$.

Now we are connecting the above capacitor to a circuit with source voltage East. There will be a deviation betwixt the source voltage and capacitor voltage, so the capacitor will start to charge and depict electric current according to the difference in voltage. The capacitor voltage will increment exponentially to the source voltage in 5-time contents.

Charging-Initially-charged-capacitor-graph — Voltage curve of the initially charged capacitor during the accuse phase

The graph above shows the voltage across the capacitor. The initial voltage is represented by the flat portion of the graph. And the charging phase is represented by the curve portion of the graph.

Find the transient voltage beyond the capacitor using the following formula:

$v_{f}=v_{i}+(v_{f}-v_{i})(one-east^{-(\frac{t}{\tau })})$

5$_{f}$ is the voltage of the source, and V$_{i}$ is the voltage of the charged capacitor before connecting to the circuit.

Determination :

No electric current flows through the dielectric during the charging and discharging stage except leakage current.
It takes 5 times constant to charge or discharge a capacitor fifty-fifty if information technology is already somewhat charged.
The capacitor voltage exponentially rises to source voltage where current exponentially decays down to zero in the charging phase.
As the switch closes, the charging current causes a loftier surge current which tin only be limited by the series
A discharged capacitor behaves like a curt circuit when initially connected to the circuit, which means causing a surge current initially.
A capacitor behaves similar an open circuit when it is fully charged, which ways not assuasive electric current through information technology.
In the discharging phase, the voltage and current both exponentially decay down to zero.