ampere hour is the unit of electricity consumed by electrical load,for example you are consuming 5ampere per hour so unit of power will 5amp-hour,generally unit of power is kilowatthour,so 1000amp-hour =1KWH ampere =voltage/resistance, amp-hour the unit of power on which tariff is calculated for the consumption of electricity
Volt Ampere Hour indicate the capacity of a electric charge store able device like battery.
It consist three parts:
Volt: unit of electrical potential difference.
Ampere: Unit of Flow of Electric charge.
Hour : Its Big unit of time
Volt Ampere : Indicate the power of a dc system.(store able device Always be DC system)
Volt Ampere Hour : Indicate Power of a battery multiplied by time(in Hour) it will serve.
Let a battery having 'x' Voltage is connected with Load(Light, Fan, Electric Load). If in this system the flow of current is 'y' and if this system serve for 'z' Hour until finish all the energy stored in that battery then, we will say the capacity of the battery is xyz volt ampere Hour.
So, A battery of xyz volt ampere Hour will serve z Hour for a load taking y ampere at x volt.
Make xyz=800
Now Let x=12 Volt(Terminal potential difference of a battery)
then consider a load on x(=12) volt takes y=5 Ampear
then we got xyz=800
=> 60z=800
=>z=13.33
Means this system will continue for 13.33 Hour
Amperes are coulombs per second, so an ampere-hour is 3,600 coulombs, and that is a measure of charge. You are probably talking about watt-hour rating. Watts are joules per second, so a watt-hour is 3,600 joules, which is a measure of energy. Most often, energy measurements by the "electric meter" are in kilowatt-hours, which is the same as 3.6 megajoules.
In order to pass DC current through a transformer you would need some basic knowledge of what a transformer actually is. A transformer is simply put a coil of copper wire around metal plates of iron. When you put current onto the coil, the metal plates will build up a magnetic field. During the build up of this magnetic field, the current will be 'low' When the magnetic field is maximized, the coil will work as a shortcut on a DC power source. In order to use a transformer, ie transform voltage up or down, we need a fluctuating magnetic field. This can not be done without electronics of some variety, and the resulting voltage will be AC, not DC. The easiest way to use a transformer with a DC power source is to make a circuit that electronically change polarization of voltage back and forth. Depending on transformer, this can be as low as 50 times per second (EU standard 50 Hz AC 220 Volt, US standard 60 Hz AC 115 Volt) or higher like 100.000 times per second (100 Khz) as is in many high speed switch power supplies. It all depends on the transformer you use and what you want to use it for. If say you want to make a power source for an electric drill and you want to use a car battery as the source, then I would forget it. An electric drill need a lot of juice and in order to do so via a transformer, this transformer would need to be very very large. Every ampere of 12 Volt would only give you approx 0.07 Ampere of 115 Volt. A drill easily drain 800 to 1000 Watts. Approx 8 Ampere at 115 Volt. Following this, an ideal transformer with hardly any loss would eat away 100 Ampere every 8 ampere on output 115 Vac. With the loss factor (some of the transformed voltage will be given away as heat) and you would probably need approx 110 Ampere 12 Volt in order to generate 8 Ampere 115 Vac. A normal 60-65 Ampere car battery would then last a mere half hour fully charged. Another problem is that most batteries are not designed for such a load over any prolonged time (more than 30-60 seconds). Systems have been and are buildt that can withstand such a drain but they consist of many batteries connected in paralell in order to even the drain between all batteries.
These figures represent capacities of use for batteries. If a device uses 50 mA an hour on a 800 mAH battery, the device would operate for 16 hours before depleting the battery. If the same device was connected to a 700 mAH battery, the device would operate for 14 hours before depleting the battery.
Yes you can. All the 1300 mA rating means is that adaptor can supply devices up to 1300 mA. The old adaptor's limit was 800 mA. The new adapter has 500 mA more in reserve if it is ever needed.
800 cubic centimeters.
The first figure is the UTS of the bolt in hundreds of Mpa (so an 8.8 bolt fails at 800 Mpa) while the second figure is the percentage of UTS at which plastic deformation occurs, divided by 10. (so your 8.8 bolt stretches permanently at 80% of 800 Mpa, or 640 Mpa).
There are 1000 milliamperes in one ampere. Therefore, 800 milliamperes is equal to 800/1000 = 0.8 amperes.
No real comparison. In simplistic terms the D cell 1.5 volt flashlight battery will produce around 15 ampere hours. A 12 volt automobile battery will produce from 300 cold cranking amps (CCA) to 800 CCA or more.
800 kW in one hour is equal to 800 kWh (kilowatt-hours) of energy. This means that 800 kilowatts of power is being consumed or generated for one hour.
Yes
In the switch itself...
Yes, the 9 volt 800 milliamp adapter will work with the 9 volt 400 milliamp device. The device will only draw the amount of current it requires, so having a higher amperage rating on the adapter is fine. The voltage should match to ensure compatibility.
'http://wiki.answers.com/Q/If_breaker_CT_ratio_8001_when_current_of_any_phase_crossed_800_ampere_what_would_happen'
800 meters in 2 minutes is 14.9 miles per hour.
800 miles / 16 miles per hour = 50 hours
about 800 miles per hour
One hour
800/650 is about 1 hour 13 minutes 51 seconds, or 1.23 hours.