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How do you calculate the charge of a capacitor using random voltage and random time and random resistor?
MortenBrendefur
Battery, resistor, and capacitor are connected in series.
E = voltage of the battery, volts
R = resistance of the resistor, ohms
C = capacitance of the capacitor, farads
T = time since the circuit was completed, seconds
I = current in the circuit, amperes
Vc = voltage across the capacitor, volts
Q = charge on the capacitor, coulombs
e = base of natural logs = approx 2.7183
At any time 'T' after everything is connected up,
Vc = E x (1 - e-T/RC) volts
I = (E/R) e-T/RC amperes
or
I = (E - Vc) / R
Q = 1/2 C Vc2 coulombs
or
Q = 1/2 C E2 (1 - e-T/RC )2 coulombs
See ? Nothing to it.
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∙ 13y ago
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Why do resistor voltage decrease while capacitor discharges?
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.
Does voltage matter when charging a capacitor?
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.
Can a capacitor charge to its maximum by applying any voltage to it?
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A resistor R and a capacitor C are connected in series to a battery of terminal voltage V0 what equations relating the current you in the circuit and the charge Q on the capacitor de?
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How many time constants does it take to fully charge a capacitor?
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How do you charge a capacitor?
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What happens to the charge if you double the voltage across a capacitor?
In order to double the voltage across a capacitor, you need to stuff twice as much charge into it.
How much time constants to charge or discharge a capacitor?
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Why the capacitor time period formula is given as t equals RC?
-- 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.
The charge on a capacitor increases by 16 C when the voltage across it increases from 82 V to 121 V What is the capacitance of the capacitor?
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How long does it take for a battery to charge a capacitor?
Usually a tiny fraction of a second. Actually it will depend on the characteristics of the the capacitor, and of the remaining circuit (mainly, any resistor in series). The "time constant" of a capacitor with a resistor in series to charge from 0 to a fraction of (1 - 1/e), about 68%, of its final value. This time is the product of the resistance and the capacitance. After about 5 time constant, you can consider the capacitor completely loaded for all practical purposes - i.e., it will be at the same voltage as the battery.
