Power is volts times amperes, so 120 V and 50 A would be 6000 watts.
However, it also depends on phase angle and power factor, something that is related to reactive loads such as motors and power supplies. As a result, the power may actually be less, so it is more correct to say 6000 volt-amps. The case of 6000 watts being the same as 6000 volt-amps is only true for a purely resistive load such as a toaster.
You know if current is flowing in a bulb circuit because, if there is enough power (voltage times current), the bulb will illuminate. If there is current, but not enough power to illuminate the bulb, you will need to measure the current with an ammeter to see if there is any current.
Power = (current) times (voltage)Current = (Power) divided by (voltage)Voltage = (Power) divided by (current)
Since power is current times voltage, doubling current while keeping voltage the same will double the power. Ignoring slight non-linearity, if the power doubles, the heat will double.
The unit of power is watts, the unit of current is amps, and the unit of voltage it volts. Power = Voltage X Current Voltage = Power / Current Current = Power / Voltage In electricity, power is symbolized with a P, current with an I, and voltage with a V. The real formula looks like: P = V x I V = P / I I = P / V
Voltage and current are actually inversely proportional to one another. The formula P=IV is what you need to look at here, where P is Power, V is voltage, and I is current. Rearranging the equation you will see that V = P/I. You can see that if you increase voltage, while holding power constant, current is reduced. Now, to your question. The losses on a transmission line are proportional to the current flowing on the line, so transmitting at high voltage (and hence low current) is beneficial as it reduces the amount of power that is lost due to resistance in the line itself.
POWER=VI. V=voltage I= current
Assume you are saying that the current and voltage are in phase and you want to know how power is affected. When Voltage and Current are in phase the Power Factor is 1 and you have maximum power being applied. When Voltage and Current are not in phase, Power Factor decreases from 1 toward zero.
Power = Voltage * CurrentIsolating current, we getCurrent = Power/Voltage, I = 800W/100V = 8 amps
You know if current is flowing in a bulb circuit because, if there is enough power (voltage times current), the bulb will illuminate. If there is current, but not enough power to illuminate the bulb, you will need to measure the current with an ammeter to see if there is any current.
Power = (current) times (voltage)Current = (Power) divided by (voltage)Voltage = (Power) divided by (current)
Since power is current times voltage, doubling current while keeping voltage the same will double the power. Ignoring slight non-linearity, if the power doubles, the heat will double.
The unit of power is watts, the unit of current is amps, and the unit of voltage it volts. Power = Voltage X Current Voltage = Power / Current Current = Power / Voltage In electricity, power is symbolized with a P, current with an I, and voltage with a V. The real formula looks like: P = V x I V = P / I I = P / V
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You do not need ohm's law to relate power to current and voltage. Power is current times voltage. If you know current and voltage, you do not need to know resistance.
Power flowing into a transformer must match the power flowing out (minus losses which are minimal). If this is not the case, there's something wrong. Differential protection monitors current only; Current flowing into one side of the transformer will be equal to current flowing out the other side scaled by the turns ratio of the transformer. Since the turns ratio is equivalent to the voltage ratio, this is easily set.
Voltage and current are actually inversely proportional to one another. The formula P=IV is what you need to look at here, where P is Power, V is voltage, and I is current. Rearranging the equation you will see that V = P/I. You can see that if you increase voltage, while holding power constant, current is reduced. Now, to your question. The losses on a transmission line are proportional to the current flowing on the line, so transmitting at high voltage (and hence low current) is beneficial as it reduces the amount of power that is lost due to resistance in the line itself.
Voltage x current = power (watts)