Can I power a 240 volt (15 amp) stove on a 208 volt supply?
Good Lord...I'm not even going to say anything... Standard appliance ratings allow for a 10% +/- on voltage. This means, something that is rated to run on 240 volts can run on 10%, or 24 volts in this case, more or less. 240 - 24 is 216, obviously not within the 208 volts you say you have. 208 volts does come from a three phase system. However, just because that is the building supply doesn't mean you have three phase in your apartment. Oft times, three phase power is brought in in order to run mass quantities of motors more efficiently. You still only have 2 phases in your apartment which may give you 208-215 volts. Will the oven run on it? Yes. Will it slightly reduce the output of the burners? Only a little. Will you notice it? No. Hope this helps.
Two of the power outlets at home only read 104 volts?
What happened to me was an outlet in a house I used to live at never would work right and I used one of those free estimate coupons in the back of the phone book yellow pages.....and this big hunk of guy showed up (and he smelled really nice) and he had this tool belt that had every tool known to man that he kept on a belt (that was strapped on his 32 inch waist) and he found the breaker / fuse box and traced the wire to the outlet that was giving me the problem and he said the wrong size breaker was installed and he said that it needed a 5 amp breaker because the line was also being shared with the oven in the kitchen and I looked at him in the eyes and his big blue eyes were like laser beams on ice...and time froze...and before I knew it, he had one of those big hands on each of my shoulders saying, 'mam, are you alright?'....but anyway, he replaced the breaker and electrically speaking, things have been going happily ever after. He drops by every now and then and gives me a free safety checkup.
If an electron is accelerated through a potential difference of 45.5 volts, then what is the velocity acquired by it?
Whenever a charged particle q is accelerated through a potential difference of V, its kinetic energy increases by an amount of qV. This you can deduce simply by applying the total mechanical energy conservation principle on the system as follows:Now in the given problem q = e = charge of an electron, m = mass of an electron and potential difference V = 45.5 volts. On putting these values in the above expression we getv = 4 × 10^6 m/s.Regards!
What is the electrical energy stored initially by the charged capacitor? Find the maximum current in the induc?
E = 0.5 * (Q^2 / C) E = 0.5 * ((2.50E-6 C)^2 / 3.80E-6 F) E = 8.2237E-7 Joules E = 822.3684E-9 Joules E = 822.3684 nanoJoules http://en.wikipedia.org/wiki/Capacitor The maximum current is when the voltage is a maximum in the capacitor. E = 822.3684E-9 Joules = 0.5 * C * V2 V = 657.8947E-3 Volts V = 657.8947 milliVolts V = L * (di/dt) 657.8947 milliVolts = 80.0E-3 H * (di/dt) dt/dt = 657.8947 milliVolts / 80.0E-3 H di/dt = 8.2237 Amps http://en.wikipedia.org/wiki/Inductor
Electric field between two parallel plates?
electric field between such plates depends on their separation and the voltage between them... the equation is E = V/d E = 640 V/m V = ? d = 10.0 mm = 0.01 m or 10.0 x 10^-3 m(you have to work with the fundamental unit of length or distance which is meters(m)-that is why the unit of electric field is V/m) so 640 V/m =V/ 0.01 m transposing to make 'V' the subject of the formula: V = 640 V/m x 0.01m so V = 640 V/m x 1/100 m = 6.4 V answer is 6.4 V
What is the charge on this capacitor when the electrical energy stored in the capacitor is 7.78 J?
A capacitor consists of two closely spaced metal conductors of large area, separated by a thin insulating foil. It has an electrical capacity of 3000.0 μF and is charged to a potential difference of 69.0 V.
What electrical potential energy is stored in the 1.5-µF capacitor?
Two capacitors with capacitances of 1.5 µF and 0.25 µF, respectively, are connected in parallel. The system is connected to a 50-V battery. What electrical potential energy is stored in the 1.5-µF capacitor? The answer is: 1.9 x 10^ -3 J. How did they get this???
Calculate the charge on this capacitor when the electrical energy stored in the capacitor is 9.73 J.?
In order to estimate the way that C changes when the spacing is increased, we must make some assumptions about the geometry of the conducting electrodes, and their arrangement in space. We are not given any information about this, so strictly speaking, the question cannot be answered; many different geometries, each having different properties can be imagined. However, for the sake of simplicity, I will assume that this is a parallel plate capacitor. Further, I shall assume that when the plates are separated, the resulting space is filled with dielectric material of the same type as was there originally. If these assumptions are what is intended then the the capacitance will fall by a factor of 3 when the plates are separated. If the charge on the plates remains constant during the separation and in-filling process, we can calculate the increase in voltage using Q = CV Since the capacitance falls by a factor of 3, the voltage will rise by a factor of 3, C remaining constant. The energy stored in a capacitor can be calculated as 0.5*C*V^2 where C is the capacitance and V is the voltage. Since the voltage rises by a factor of 3, the energy stored will rise by a factor of 9 due to that cause. Since the capacitance falls by a factor of 3, the energy stored will fall by a factor of 3 due to that cause. therefore the overall change in energy will be an increase by a factor of 3.