The reactance of the capacitor is 0.339 ohms, therefore the total impedance is sqrt(4002+0.3392) = 400.0001 ohms. So the resistor drops very nearly 20 volts, very slightly less.
200WV is the working voltage of the capacitor. This is the value that should not be exceeded between the two terminals. 470uF is the capacitive rating of the capacitor. It means 470 micro Farads, or 0.47 Farads.
The time it takes to fully charge a capacitor depends on the capacitance and resistance of the circuit; the voltage is irrelevant. The equation you need is:t = 5RCwhere: t = time in seconds, R= resistance in ohms, and C =capacitance in farads.So you should now be able to calculate the time for yourself, but remember to convert the resistance into ohms and the capacitance into farads before you insert the figures into the equation.
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14 micro farads @ 1.4 seconds.. idk what it is at 2 seconds ;p
In the ac waveform of a capacitor the current waveform leads the voltage waveformcurrent is large to start until capacitor fills with it's voltage charge if that helpsAnswerThe terms 'leading' and 'lagging', used when describing power factor, are defined in terms of whether the load current is leading or lagging the supply voltage.In a capacitive circuit, the load current leads the supply voltage, so the power factor is leading.
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".
Because the timing is set by the time constant of a resistor and a capacitor. With R in ohms and C in Farads, the time-constant is RC in seconds. If the capacitor leaks the timing will be wrong.
200WV is the working voltage of the capacitor. This is the value that should not be exceeded between the two terminals. 470uF is the capacitive rating of the capacitor. It means 470 micro Farads, or 0.47 Farads.
Theoretically, forever, because as the voltage on the capacitor approaches the source voltage, the available current to charge the capacitor approaches zero.In practice, however, it simply depends on what you call "charged".In the simple example of a capacitor being charged from a voltage source in series with a resistance, the voltage is given by ...VT = Vs (1 - e -T/RC)... so, if your definition of "charged" is 99% of VT then T would have to be 5 RC's or 5 time constants.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".
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.
Capacitor is the name of the device and capacitance is a measure of farads in the capacitor. Capacitance is the capacity for storing charge in the capacitor as measured in farads, micro farads or millifarads.
The time it takes to fully charge a capacitor depends on the capacitance and resistance of the circuit; the voltage is irrelevant. The equation you need is:t = 5RCwhere: t = time in seconds, R= resistance in ohms, and C =capacitance in farads.So you should now be able to calculate the time for yourself, but remember to convert the resistance into ohms and the capacitance into farads before you insert the figures into the equation.
In an electronic circuit a capacitor can be used to block direct current. In general a capacitor stores electric charge. The charge in a capacitor is the voltage times the capacitance and that is also equal to the charging current times the time (all quantities in SI units - seconds, volts, amps, coulombs, farads)
A capacitor is like a 0.001 ohm resistor when the capacitive reactance is 0.001 ohms. Capacitive reactance, in ohms, is defined as -1 divided by 2 pi f C, where f is frequency in hertz, C is capacitance in farads, and the -1 means that current leads voltage. Plugging in 60 Hz, and solving for C, you get 2.65 farads. That is a very large capacitor. At 6000 Hz, you get 26.5 millifards, which is still very large. At 6 MHz, you get 26.5 microfards, but at 6 MHz, you need to consider parasitic inductance.
A: from a voltage source a capacitor will charge to 63 % of the voltage in one time constant which is define the voltage source Resistance from the source time capacitor in farads. it will continue to charge at this rate indefinitely however for practical usage 5 time constant is assume to be fully charged
A capacitor impedance is equivalent to 1/jwC, where j = i = imaginary number, w = frequency, and C = capacitance in Farads.
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