I assume this is for an AC system. 110-120Vrms (typical wall socket voltage) = 155-170V sine wave peak to 0 volts. Similar for the current you gave - I assume it is in RMS?
In an AC system, 120Vrms * .5Arms = 60Watts
If the values you have are measured peak to 0 volts, Wattage = 1/2 * Voltage * Current
The power rating of the lamp can be calculated using the formula P = I * V, where P is power, I is current, and V is voltage. Plugging in the values gives P = 0.5 A * 120 V = 60 Watts. Therefore, the power rating of the lamp is 60 Watts.
100w V = IR I seem to remember from school days...
120 ohms of resistance, using ohm's law
*I=V/R
*Current (I) is equal to Voltage (V) divided by Resistance (R).
If it is incandescent the current will be 1 amp. I = W/E
The watts equal volts times amps, so 0.8 x 120 is 96 watts.
I assume you know this version of the formula for power.
Power = Potential difference x Current
So,
Power = 120 x 0.5 = 60 Watts
The answer is volts times amps, 120 x 0.5 so you have to use a calculator (except for those who have skills in mental arithmetic).
If it is a bulb designed to work at 120 v the current is 150 divided by 120 and the answer is in amps.
for a 12 v bulb, 12 volts x 0.5 amps = 6 Watts
for a 240 v bulb, 240 volts x 0.5 amps = 120 Watts.
The total current in the circuit would be 12 amps. When electrical loads are connected in parallel, the currents add up. So if each load draws 6 amps, the total current would be the sum of both loads, which is 6 + 6 = 12 amps.
A battery is rated to supply a certain number of volts. However, it actually supplies less, because they are "lost" as the current has to get out of the battery in the first place.(The battery has internal resistance)The amount of lost volts depends on the current being drawn:The less resistance a circuit has, the more current is drawn, because it's easier to flow.Example:If the circuit has little resistance, it draws a large current and the battery's internal resistance causes more lost volts.If the circuit has high resistance, it draws a small current and there are fewer lost volts.This is why when you short-circuit a battery (give it hardly any resistance to go through) it heats up and may explode. A large current is drawn and all the volts are used by the battery's internal resistance.
10V 600mA means that the device requires a voltage of 10 volts and draws a current of 600 milliamps (or 0.6 amps) to operate properly. It is important to match the voltage and current requirements when selecting a power supply or charger for the device to avoid damaging it.
The formula is Resistance= Voltage/ Amps(current) In your example: R=50/2.5, so the answer is 20 ohms.
Standby current is the current that a device draws when it is not actively performing its function. This current would be measured in amperage, and commonly amperes, milliamperes or microamperes would be the units of measurement. As an example of a device drawing standby current; a radio transmitter may not be actively transmitting, but the power supply is turned on, and the transmitter is ready to operate. In this case, the transmitter is drawing very little power. A computer can be in standby mode and drawing "standby current" , examples of which are also "hibernation" and "sleep" modes. The computer display and hard disk drives are turned off, and the CPU is throttled down to low power state. However, memory is kept active, which requires just a small amount of battery power.
A standard 3kW immersion heater will require a fuse rating of 13 A. This is because, it draws a current of 12 A.
All depends what voltage is applied across it.Assuming it's designed for operation at 120 VAC, then it draws 33-1/3 mA when it's turned on.(120 x 0.0333) = 4
Power is VI so 360 watts.
Power = Voltage * Current, so 120V x 9.5A = 1140 Watts. What the hell VCR draws 1140 Watts, though? Even early Sony U-Matic VCR's with mechanical everythings drew only 100 Watts.
The amperage output on an adapter is the rating applied by the design manufacturer. Connecting a load that draws more that the design limit of the adapter will damage the adapter. As long as your connected load stays under the adapters rating there is no problem.
An ups consists of battries as the main components .Batteries store charge (charge=current*time)... hence its units is so. ex. if its rating is 6AH and the load connected to in draws 1A, then the batter lasts for 1 hr.
To find the resistance necessary, one would need to know how much current the bulb draws. If one knows the current the bulb draws, then one would subtract the 14 volts from 120 volts then divide that by the current the bulb draws and one will find the resistance needed. Once this has been done, one would need to multiply the current drawn by the voltage drop to get the wattage rating necessary. Another important detail to note is that the power dissipated by the resistor will be much greater than the power consumed by the bulb itself. Finally if the bulb burns out the voltage across the contacts will be 120V. I would not recommend using this method to drop the voltage for the bulb.
The voltage across a DC device that draws 2A and consumes 12Wh/h is 12/2 or six volts.
1.7amp
Current equals power divided by voltage, so with 110 V across the load, a 900 W system draws about 8.18 A (if the voltage potential is double, 220 V, the current is half, 4.09 A).
Use a generator with a high enough rating to power the house, of course. Trying to power a house that draws 60A of current with a 10A generator is just never going to work.
Power = Voltage x Current P=V.I Power (in Watts) = 110V x 8.70A = 957W (Appx. 1kW) - Neeraj Sharma