The voltage across a DC device that draws 2A and consumes 12Wh/h is 12/2 or six volts.
Ohm's law: Voltage = Current times Resistance Solve: Resistance = Voltage divided by Current So, a device drawing 50ma with 150V has a resistance of 150 / 0.05, or 3000 ohms. p.s. Since power is volts times amps, that device is dissipating 7.5 watts.
Exactly...you answered your own question. Each DC bulb will drop voltage according to its resistance and the amount of current it draws.
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.
thanks to ohm,,who invented the answer.according ohms law,V voltage=R(element)xI amps.then.....answer is....100 volts.
The voltage drop across the wiring in the wall is calculated using Ohm's Law: V = I * R, where V is the voltage drop, I is the current, and R is the resistance. Plugging in the values, V = 10 A * 0.12 ohms = 1.2 V. Therefore, the voltage drop across the wiring in the wall is 1.2 V.
A device that wants to be supplied with 4v, will be destroyed if connected to 12v. Do not confuse voltage with Amp/Hour capacity, which is what I think you are looking for. Make sure the voltage matches the device. Then determine how much current the device draws. The power supply will be rated as to how much current it can deliver and for how long.
Wire sizing of a feed conductor is based on the amperage that a device draws. To calculate amperage from KVA a voltage of the supply has to be stated. Without this voltage and whether the transformer is single or three phase an answer can not be given.
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).
CMOS is better than single MOSFETs because the complementary MOSFETs in CMOS always have one off and the other on, reducing the idle current to only leakage current and the output voltage exactly equal that of either the power or ground as there is no voltage drop across the MOSFET that is on. With just one MOSFET the device draws current anytime it is in the on state, even if idle.
The specification "8.4V 600mA" indicates an electrical device's voltage and current requirements. The "8.4V" refers to the voltage the device operates at, which is 8.4 volts, while "600mA" denotes the current it draws, measured in milliamps (mA), equivalent to 0.6 amps. This means the device needs a power supply that can provide 8.4 volts and can deliver up to 600 milliamps of current for optimal performance.
To calculate the wattage of a device, you need to know both the current (in amperes) and the voltage (in volts). If the voltage is not provided, you cannot accurately calculate the wattage. In this case, with only the current (4 A) given, you cannot determine the wattage without knowing the voltage as well.
The supply device must have an output voltage that matches that of the load, and a current rating that exceeds that of the load. So you cannot use a load that draws 2 A from a supply device that is rated at only 1 A.