In order to double the voltage across a capacitor,
you need to stuff twice as much charge into it.
You charge a capacitor by placing DC voltage across its terminal leads. Make sure when using a polarized capacitor to place positive voltage across the positive lead (the longer lead) and negative voltage across the negative lead. Also make sure that the voltage you charge the capacitor to doesn't exceeds its voltage rating.
basically a capacitor will charge to the input DC level however it will mathematically never happen since capacitors charge at a certain rate the voltage drop across a capacitor will follow the R C time constant or 63% of the applied voltage for a unit time.AnswerIn the case of an a.c. supply, yes, there will be a voltage drop across a capacitor. In the case of an 'ideal' capacitor, this will be the product of the load current and the capacitive reactance of the capacitor.
(a) what is the total capacitance of this arrangement (B) the charge stored on each capacitor (C) the voltage across the 50 micro farad capacitor and the energy stored in it. 20v and 20+30+50 micro farad
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.
A5uf capacitor has 5*10-4 coulombs of charge stored on its plates
A capacitor can charge to its' maximum OR the voltage applied to it, whichever is LESS.
because resistance is restricting the current and voltage, so for it be accurate you need to know what the voltage and the amps are.AnswerCapacitance is quite independent of resistance and, therefore, it will NOT vary if resistance is changed.
Q = CV, Q is charge, C is capacitance, V is voltage. C= Q/V = dQ/dV since it is linear function = 0.41F
Charges may appear to flow through a capacitor, although in reality they don't.The degree to which charge appears to flow through a capacitor depends on therate at which the voltage across it changes.-- DC voltage doesn't change, so it doesn't appear to pass through a capacitor at all.-- AC voltage is always changing, and the higher its frequency, the more currentit appears to push through a capacitor.
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
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.
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.