Q=CV, with Q=(5x10) = 50 Coulombs, and C=5x10^-6 Farads,
therefore: 50 = 5x10^-6 x V
V = 50 / (5x10^-6) = 10,000,000 Volts.
Seems like rather a lot! Maybe there's a typo in the question? Either way, use Q=CV to give the right answer.
is this right seems rather high to (50 volts?)
9200 volts my 1000 uF capacitor only holds 10 volts
A microfarad measures electrical capacitance. A farad can be defined as the charge in coulombs, which is 1 ampere per second, that a capacitor will accept for the potential across it to charge one volt. A microfarad is equal to one millionth of a farad.
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.
A capacitor is composed of metal plates. Voltage is applied to one, which causes electrons to build up on the other. This is reactive in nature, thus a capacitor is reactive. It stores a charge, and releases this charge when the voltage decreases.
A capacitor relies on the time it takes for the voltage across it to reach a certain value. When it is fully discharged, it acts like a piece of copper wire. ie: it conducts. The voltage across it now is zero. When a voltage is applied to it, it begins to charge, and so the voltage measured across it's terminals increases. This takes time. How much time depends on the voltage applied resistor connected to it and the value of the capacitor.This voltage is then used to energies a relay, turn a transistor on, fire an SCR, and lots of other things.
A5uf capacitor has 5*10-4 coulombs of charge stored on its plates
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.
9200 volts my 1000 uF capacitor only holds 10 volts
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.
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.
Voltage and current are two different things. Voltage is potential energy per charge, in joules per coulomb, while current is charge transfer rate, in coulombs per second. Its that same as saying that a battery has voltage but no current, because there is no load. Well, a capacitor resists a change in voltage by requiring a current to change the voltage. Once that voltage is achieved, there is infinite resistance to the voltage, and thus no current.
A capacitor can charge to its' maximum OR the voltage applied to it, whichever is LESS.
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)
calulate the voltage of a battery that provides 20 joules of energy to every 5 coulombs of charge