(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
You measure the capacitance of a capacitor in an active circuit by observing the voltage across it and the current through it. That gives you, by Ohm's law, the impedance of the capacitor. Plug that in the the equation for capacitive reactance, and you get capacitance. Note: There is no such thing as a three phase capacitor. A capacitor is a two terminal device that resists a change in voltage inversely proportional to its capacitance. You connect one capacitor to one phase. If you have a "three phase capacitor", then you are talking about three capacitors. Deal with each one separately.
Capacitors are characterized by two values: their voltage, exceeding which will damage the capacitor (sometimes leading to a violent explosion), and their capacitance, as the name suggests. The voltage is expressed in volts. The capacitance is expressed in Farads. One (1) Farad is an amount of charge that makes the voltage across the capacitor terminals to rise by 1 Volt. If a 10mA current flows into the capacitor and it causes the capacitor's voltage to rise by 1V every second, the capacitor's capacitance is 10 milifarads. 1 Farad is a lot of charge, so for most applications, submultiples (microfarads and milifarads, mostly) are commonly used. A curious note: the more voltage a capacitor can handle, the (usually) bigger the size of it. At low voltages and low capacitance, the capacitance doesn't influence the size that much, though.
Q = CV, Q is charge, C is capacitance, V is voltage. C= Q/V = dQ/dV since it is linear function = 0.41F
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.
As capacitance is not affected by voltage, nothing will happen! Capacitance is affected only by the area of overlap of the plates, their distance apart, and the permittivity of the dielectric.
The reactance of a capacitor is a function of -- the capacitance of the capacitor -- the frequency of the voltage across the capacitor
You measure the capacitance of a capacitor in an active circuit by observing the voltage across it and the current through it. That gives you, by Ohm's law, the impedance of the capacitor. Plug that in the the equation for capacitive reactance, and you get capacitance. Note: There is no such thing as a three phase capacitor. A capacitor is a two terminal device that resists a change in voltage inversely proportional to its capacitance. You connect one capacitor to one phase. If you have a "three phase capacitor", then you are talking about three capacitors. Deal with each one separately.
Capacitors are characterized by two values: their voltage, exceeding which will damage the capacitor (sometimes leading to a violent explosion), and their capacitance, as the name suggests. The voltage is expressed in volts. The capacitance is expressed in Farads. One (1) Farad is an amount of charge that makes the voltage across the capacitor terminals to rise by 1 Volt. If a 10mA current flows into the capacitor and it causes the capacitor's voltage to rise by 1V every second, the capacitor's capacitance is 10 milifarads. 1 Farad is a lot of charge, so for most applications, submultiples (microfarads and milifarads, mostly) are commonly used. A curious note: the more voltage a capacitor can handle, the (usually) bigger the size of it. At low voltages and low capacitance, the capacitance doesn't influence the size that much, though.
Q = CV, Q is charge, C is capacitance, V is voltage. C= Q/V = dQ/dV since it is linear function = 0.41F
The reactance of a capacitor depends on its capacitanceand the frequency of the voltage across it.In general, the magnitude of capacitive reactance is . . .1 / (2pi x frequency x capacitance)At 100 Hz, that would be0.00159 / (capacitance) in Farads .
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.
As capacitance is not affected by voltage, nothing will happen! Capacitance is affected only by the area of overlap of the plates, their distance apart, and the permittivity of the dielectric.
I hear you saying that "joule" should be a unit of "ability"; that's pretty slippery."Joule" is a unit of energy. The capacitor does store energy.The energy stored in a capacitor is [ 1/2 C V2 ], measured in joules.' C ' is the capacitance, in farads.' V ' is the voltage across the capacitor, in volts.You can see that a larger capacitance (more farads), when charged to a smaller voltage,stores the same amount of energy. So the joules of energy stored in the capacitordepends on the capacitance and the voltage across it.The same energy can be stored in capacitors with different values of capacitance.In fact, the same capacitor can store different amountsof energy ... more or less joulesat different times.
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.
In order to double the voltage across a capacitor, you need to stuff twice as much charge into it.
Because the capacitor discharges. so voltage across the capacitor decreases.
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.