If I understood your question correctly,
'If you know the power rating of an appliance and the voltage of the line it is attached to, can you calculate the current used by the appliance?'
You are looking for this equation:
I = Current (Amps)
P = Power Rating (Watts) V = Voltage (Volts) I = P/V This equation is useful when calculating the fuse rating for an appliance.
For example, and appliance rated at 2400 watts, supplied with 240 volts has a 10 amp maximum.
I = 2400/240
however, when it comes to paying for electricity, the energy is measured in units, which can be found using the equation: P = Power (kW) t = Time (Hours) Energy Used (Units) = Pt for example, an appliance rated at 2 kW used for 3 hours, uses 6 units of energy.
Units = 2*3
current=watts(power)/voltage
V = I x R V = voltage, I = Current, R = Resistance or it can be calculate like this V = P / I V = Voltage, P = Electric Power, I = Current
KCL calculate current . KVL calculate voltage drop.
-- Apply a small, known voltage between the terminals of the device. -- Measure the current through the device with the voltage applied. -- Calculate the resistance of the device. It's (voltage) divided by (current).
Wattage, you mean power. Power = V * I. V - the voltage and I the current.
current=watts(power)/voltage
There are a number of ways to calculate that current, from which you're free to select the most convenient one. Here are a couple of them: Current = (the mains voltage)/(resistance of the appliance) Current = square root of (power consumed by the appliance/resistance of the appliance)
In its simplest form the equation to calculate the wattage of an electrical appliance is: Watts = voltage x current. If the appliance is in a AC supply use the Route mean square voltage (the stated AC voltage).
Yes and No. You have three types of adaptors: constant current with variable voltage output. constant voltage with variable current output. constant voltage with constant current output. What you are talking about is the latter. This means that the adaptor was created for a specific appliance requiring 12V/2A, which it will indeed use. Heavier appliance will not draw enough current/voltage and will malfunction. Lighter appliance will draw too much current/voltage and will overload/shortcircuit.
You just need the voltage and the current. Watts = Amps x Volts.
Wattage is unit of power which is the product of Voltage in V and Current in Amps. If you know the current drawn by the appliance with 1200 Watts then you can calculate the Voltage = Power/ Current. For eg. if the current drawn by the appliance is 100A then the voltage is 1200/100 i.e. 12 Volts.
How do you calculate voltage drop for starting motor current
formals to calculate exciation voltage of alternator
Not enough information. Power = current x voltage. Since voltage can be anything, there is no way to calculate power. Time is irrelevant; though once you have the power, it can help you calculate energy (energy = power x time).
to calculate the cable size of a run of 30 meters long you first will have to know the current of the appliance use the voltage drop formula V d = (mVxIxL)/1000 once the voltage drop is less than 2.5% of the nominal voltage, the cable should be upsize.
V = I x R V = voltage, I = Current, R = Resistance or it can be calculate like this V = P / I V = Voltage, P = Electric Power, I = Current
12 volts.