It's time to Know about Charging and Discharging of Capacitor
Charging Of A Capacitor :
Consider a series RC network connected to a battery of voltage 'V' through a switch 'S' . Let us assume that the Capacitor initially uncharged. There is no current when switch'S' is opened.
If the Switch is closed at t=0 there will be a current through resistor and Capacitor will begin to charge . Note that during charging process the charge transfered from one place to other through the resistor , switch and battery untill the Capacitor is fully charged . The maximum Charge depends upon the EMF the battery . Once the maximum charge is reached the current in the circuit is zero. Suppose at any instant during charging the circuit current is 'I' and Charge on the Capacitor is 'q'
Applying Kirchoff's Voltage law
V-IR-q/C = 0
Initial Current :
At t=0 when the Switch is closed , the charge on the Capacitor is zero therefore the current is maximum (I0) and is given by
I0 = V/R -------------------at t=0
Charge on the Capacitor at any instant during charging :
The charge 'q' at any time during charging can be shown to be
q = Q(1-e^[-t/RC])
Q= Max Charge on capacitor = CV
Time constant て = RC
q= Q(1-[e^(-t/て)])
Voltage across capacitor at any instant during charging :
It can be shown mathematically that Voltage 'v' across capacitor at any time during charging is given by
v = V(1-e^[-t/て])
V = final Voltage
Discharging of a Capacitor :
Consider the circuit I) consisting of a capacitor with an initial Charge "Q" , a resistor R and a switch "S" when the Switch is open. There is a potential difference of Q/C across the Capacitor and zero potential difference across the resistor since I=0
If the Switch is closed at t=0 the Capacitor begins to discharge through the resistor . At the same time during Discharging . Let Circuit current be I and Charge on capacitor "q" . According to Kirchoff's Voltage Law potential drop across resistor = (IR) must be equal to potential difference across the Capacitor = (q/C)
IR = q/C
However the current in the circuit must be equal to the rate of decrease of charge on the Capacitor
I.e I = -dq/DT
Therefore
This we see that both the charge on the Capacitor and the circuit current delay exponentially at a rate determined by the time constant て=RC
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| Capacitor |
Consider a series RC network connected to a battery of voltage 'V' through a switch 'S' . Let us assume that the Capacitor initially uncharged. There is no current when switch'S' is opened.
If the Switch is closed at t=0 there will be a current through resistor and Capacitor will begin to charge . Note that during charging process the charge transfered from one place to other through the resistor , switch and battery untill the Capacitor is fully charged . The maximum Charge depends upon the EMF the battery . Once the maximum charge is reached the current in the circuit is zero. Suppose at any instant during charging the circuit current is 'I' and Charge on the Capacitor is 'q'
Applying Kirchoff's Voltage law
V-IR-q/C = 0
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| RC series Circuit |
At t=0 when the Switch is closed , the charge on the Capacitor is zero therefore the current is maximum (I0) and is given by
I0 = V/R -------------------at t=0
Charge on the Capacitor at any instant during charging :
The charge 'q' at any time during charging can be shown to be
q = Q(1-e^[-t/RC])
Q= Max Charge on capacitor = CV
Time constant て = RC
q= Q(1-[e^(-t/て)])
 |
| Charging of capacitor |
It can be shown mathematically that Voltage 'v' across capacitor at any time during charging is given by
v = V(1-e^[-t/て])
V = final Voltage
Discharging of a Capacitor :
Consider the circuit I) consisting of a capacitor with an initial Charge "Q" , a resistor R and a switch "S" when the Switch is open. There is a potential difference of Q/C across the Capacitor and zero potential difference across the resistor since I=0
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| Discharging of Capacitor |
If the Switch is closed at t=0 the Capacitor begins to discharge through the resistor . At the same time during Discharging . Let Circuit current be I and Charge on capacitor "q" . According to Kirchoff's Voltage Law potential drop across resistor = (IR) must be equal to potential difference across the Capacitor = (q/C)
IR = q/C
However the current in the circuit must be equal to the rate of decrease of charge on the Capacitor
I.e I = -dq/DT
Therefore
This we see that both the charge on the Capacitor and the circuit current delay exponentially at a rate determined by the time constant て=RC
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| Discharging of Capacitor |
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