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# Is it plausible for voltage across capacitors to be more than supplied voltage?

Topic: Resistor case size
July 18, 2019 / By Hanny
Question: For example if the voltage supplied is 1000v would it be plausible to have the voltage across the two capacitors to be 6000v?

## Best Answers: Is it plausible for voltage across capacitors to be more than supplied voltage?

Dolly | 8 days ago
If you have only capacitors, no. Capacitors in series will act like a voltage divider, similar to the way resistors make a voltage divider. If you also have either an inductor, or nonlinear elements such as diodes, then yes. In the case of AC in and AC out, an inductor (any inductive reactance) can resonate with the capacitance if the values are sized properly, causing voltages and currents to build well above the supplied voltage. Look up LC resonance for more information. In the case of a circuit with AC in and DC out, capacitors and diodes can be combined in such a way to form voltage doublers, triplers, etc., meaning the output DC voltage can be 2x, 3x, 4x, etc. what you would expect from a normal rectified output. A capacitor placed across the output will have a voltage much higher than the peak value of the applied AC voltage.
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We found more questions related to the topic: Resistor case size

Ok.... There are a couple of names/analogies for voltage. Voltage is sometimes called "electromotive force" and this makes the most sense in terms of water pressure. Current is actually a measure of moving charge. For example 1 Amp (the unit of current is an Amp) = 1 Coulomb/Second travelling past a point (actually a datum - but whatever). An electron has a charge of −1.6×10^−19 Coloumbs so 1A of current indicates the number of electrons flowing past a point. The Voltage, or electromotive force represents the force behind each electron. Therefore it should make sense that power=voltage x current... 1A @ 100V has the same power as 2A @ 50V. The 2A @50V has twice as many electrons at half the force so they end up delivering the same power. In terms of water pressure think about the voltage as the how much force (pressure) there is in the pipe and the current representing how much water (or how big the pipe is). It is called potential because the electromotive force does not necessarily have to be referenced to zero, it can be negative, and zero is arbitrary anyway. You can have current flow from a 100V point to a 50V point. Since electrical circuits can be made and destroyed very easily (as opposed to water pipes) it makes sense to take a more general approach than referencing everything to zero. For example if you had a 6V battery and a 24V battery referenced to the same point you would have 18V of potential difference between the 2 batteries and if something where to short circuit it would act as though it had 18V across it - not 6 or 24. Water pressure can use the potential model if you think about a river. Before the Hoover Dam there is all sorts of pressure and "potential energy" but after the turbines much of the "potential energy" has been sucked out of the water and after each irrigation tap the water loses more pressure and the river gets tamer and tamer - with less potential energy. The Colorado River doesn't even reach the Ocean all the time anymore because it has water removed for irrigation and the river flows so slowly (due to lack of pressure or potential) that near the end it evaporates before it reaches the ocean. As for voltmeters... It depends. Analog meters use the voltage being measured to induce a magnetic field that moves a needle - but that uses a whole different Physics concept (Ampere's law). In the case of digital meters it utilizes Ohm's law to reduce the voltage to something the meter can measure, and then it uses Analog-to-Digital converters which then feed into a display. There are many types of A/D converters that all work differently internally. Honestly - you would need to be at the Junior year in an Electrical Engineering program to fully understand how volt meters work. At your level it is best to just accept them. As for high voltage boxes... Most importantly whatever is in the box can probably kill you. Your body has a lot of internal resistance and the high voltage will easily overcome it, cause your hear to stop, cause electrical burns, etc. To make matters worse human muscles naturally tense up with getting a strong electrical shock so if you accidentally touch something your hand clamps around it. It is basically a danger sign - just like a cliff. The cliff doesn't suck you in but you might want to be warned. Air has a resistance to it, and it can also "break down". Lightening is an example of where there is enough voltage to "break down" the air, ionize it, and turn it into a plasma at which point the broken down area conducts. If for example it take 1 million volts to break down 10" of air, then 100,000 volts will break down in 1", 1,000 volts in .1", etc. The electricity could "arc" just like lightening if you get close enough. That is why high voltage wires are purposely separated from each other by more than plastic.

Carolyn
I can think of one way to do it. Charge up capacitors to 1000v, then disconnect it and bring the capacitors plates 6 times further then they were. This will increase the voltage six times, because C * V = Q, and capacitance of the plate capacitor is inversely dependent on the distance between the plates. In a standard circuit it won't work. Also when you connect the capacitor back to the same circuit the capacitor will start discharging back to 1000v (if caps are in parallel, or less if they are in series). If there is a potential difference - it means that electrons will be flowing, one way or another.
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Angelina
that truthfully relies upon on the voltage fee of the cacapacitors jointly as a ccapacitor blocks dc voltage that's going to bypass ac voltage. that's going to save dc voltage yet I truthfully have not heard a capacitor to exceed the rated fee.Is that 6000 vac or dc?
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Will
They would have to be larger capacitors.... 1000v supplied doesnt mean total so i mean u would need different capictors.
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Originally Answered: Max voltage at the base of a transistor?
VBE is usually about 0.6 to 0.7 volts. You do NOT apply a fixed voltage to the base (I'm assuming common emitter). The base is a current device, like a forward biased diode, and should be driven by a current, not a voltage. In your case, you have 12 volts and 60 ohms, which is a current of about 180 mA, a bit high, but OK for a power device. Checking, this part has a built in 50 ohm resistor B to E, which changes things. It limits the gain severely. You can email me if you want more help. .

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