It will take slightly less than one second (0.92 seconds) to charge a 1000uF capacitor to 12 volts through a 1000 ohm resistor if your power source is 20 volts.
The time constant of a 1000uF capacitor in series with a 1000 Ohm resistor is 1 second. (1x10-3 Farads times 1x10+3 Ohms = 1 second) It takes 1 time constant to reach 63% of a step change, 2 time constants to reach 86% of step change, and so forth using the equation VT = V0 (1 - e (-T/RC)). See notes below
12 is 60% of 20, so it will take about 0.92 seconds for the capacitor to reach 12 volts. In two seconds the capacitor will reach about 17 volts. In five seconds, five time constants, the capacitor will be considered to be fully charged, 99.3%, to 20 volts.
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Notes:
VT is voltage after a given number of seconds. V0 is the total initial voltage, in this case, 20 volts. -T/RC is the negative number of time constants for the exponential equation e-TC, which a charging capacitor exhibits. (More specifically, e-TC is the proportion of voltage across the resistor, while 1 - e-TC is the proportion of voltage across the capacitor.)
This equation is based on the fundamental equation of a capacitor...
dv/dt = i/c
... which states that the slope of the voltage is proportional to current and inversely proportional to capacitance. Plug this into an initial state differential equation, for the case of charging an initially discharged capacitor through a resistor, and you get
VT = V0 (1 - e (-T/RC))
(Derivation requires calculus, and that seems a little bit out of scope for this question.)
If a 10 microfarad capacitor is charged through a 10 ohm resistor, it will theoretically never reach full charge. Practically, however, it can be considered fully charged after 5 time constants. One time constant is farads times ohms, so the time constant for a 10 microfarad capacitor and a 10 ohm resistor is 100 microseconds. Full charge will be about 500 microseconds.
Just makes the capacitor charge at a slower rate, reduces input power.
because without using capacitor or resistor in a circuit,it cant be complete.Resistor is used to protect the circuit by giving a certain amount of voltage.Capacitor is used to charge and discharge purpose.
Yes, voltage matters when charging a capacitor. Capacitor charge rate is proportional to current and inversely proportional to capacitance. dv/dt = i/c So, voltage matters in terms of charge rate, if you are simply using a resistor to limit the current flow, because a larger voltage will attempt to charge faster, and sometimes there is a limit on the current through a capacitor. There is also a limit on voltage across a capacitor, so a larger voltage could potentially damage the capacitor.
Charge buildup between the plates of a capacitor stops when the current flow through the capacitor goes to zero.
when we replace the resistor with a capacitor ,the current will flow until the capacitor charge when capacitor will fully charged there is no current through the circuit because now capacitor will act like an open circuit. for more info plz E-mailt me at "zaib.zafar@yahoo.com"
If a 10 microfarad capacitor is charged through a 10 ohm resistor, it will theoretically never reach full charge. Practically, however, it can be considered fully charged after 5 time constants. One time constant is farads times ohms, so the time constant for a 10 microfarad capacitor and a 10 ohm resistor is 100 microseconds. Full charge will be about 500 microseconds.
The reason why resistor voltage decreases while a capacitor discharges is because the resistor acts like a source of electrical energy. As the capacitor discharges, it draws energy from the resistor, which causes the voltage across the resistor to decrease. This is because the capacitor is acting like a drain, and is taking energy out of the resistor, thus causing the voltage across the resistor to decrease. The resistor and capacitor work together in order to create a discharge circuit. This is done by connecting the capacitor to the resistor, and then to a voltage source. The voltage source supplies the energy to the resistor, and then the resistor transfers this energy to the capacitor. As the capacitor discharges, it takes energy from the resistor, which causes the voltage across the resistor to decrease. In order to understand this process better, it is important to understand the basics of Ohm's Law. Ohm's Law states that the voltage across a resistor is equal to the current through the resistor multiplied by the resistance. As the capacitor discharges, it takes energy from the resistor, which means that the current through the resistor decreases, and therefore the voltage across the resistor will also decrease.
Just makes the capacitor charge at a slower rate, reduces input power.
A: It is called discharging a capacitor. The charge will follow the rules of a time constant set up by the series resistor and the capacitor. 1 time constant 63% of the charge will be reached and continue at that rate.
because without using capacitor or resistor in a circuit,it cant be complete.Resistor is used to protect the circuit by giving a certain amount of voltage.Capacitor is used to charge and discharge purpose.
-- The quantity 'RC' has the physical dimensions of Time. -- If the capacitor is charging through a resistor, then 'RC' is the time it takes to charge up to (1 - 1/e) of the voltage it still has to go to become fully-charged. -- If the capacitor is discharging through a resistor, then 'RC' is the time it takes to discharge to 1/e of its present voltage. -- ' e ' is the base of natural logarithms, approximately 2.71828... -- 'RC' is called the 'time constant' of the resistor/capacitor combination.
If a resistor is connected in series with the capacitor forming an RC circuit, the capacitor will charge up gradually through the resistor until the voltage across the capacitor reaches that of the supply voltage. The time called the transient response, required for this to occur is equivalent to about5 time constantsor5T. This transient response timeT, is measured in terms ofτ= R x C, in seconds, whereRis the value of the resistor in ohms andCis the value of the capacitor in Farads. This then forms the basis of an RC charging circuit were5Tcan also be thought of as"5 x RC".
The resistor allows a slow charge to enter the capacitor. When this charge reaches a certain point the circuit activates and forces the capacitor to discharge. Once discharged the circuit reverses itself and starts the charge over again. The larger the cap and/or resistor the lower the frequency because it takes longer to charge the cap.
Yes, voltage matters when charging a capacitor. Capacitor charge rate is proportional to current and inversely proportional to capacitance. dv/dt = i/c So, voltage matters in terms of charge rate, if you are simply using a resistor to limit the current flow, because a larger voltage will attempt to charge faster, and sometimes there is a limit on the current through a capacitor. There is also a limit on voltage across a capacitor, so a larger voltage could potentially damage the capacitor.
Charge buildup between the plates of a capacitor stops when the current flow through the capacitor goes to zero.
Depends on the size of the battery and the capacitor. If both are small enough to fit in your hand, then some fraction of a second.