P = I^2R
R = p(L/A)
P = power (watts)
I = current(Amps)
R = resistance(ohms)
L = length of wire(m)
A = cross sectional area of wire(m^2)
: They are directly related Either one increases power rating will increase. For an IC either one increases will dire-rate the component.
: They are directly related Either one increases power rating will increase. For an IC either one increases will dire-rate the component.
Wattage or power rating of a product can be calculated by multiplying voltage rating and current rating. (Power = Voltage x Current). e.g. if device is working at 12V and 2A is the current rating. It is 24Watt. Since Voltage = Current x Resistance , for a resistive load power can also be calculated by Power = Current x Current x Resistance = I^2 x R = I square R
What is the voltage and current of an integrated cicuit
The supply won't have to work as hard. It is perfectly acceptable, for example, to use a 1A, 12v supply to supply a 12v, .5A load. The current rating indicates the ability of the supply to dissipate heat caused by the current flowing. If the load current is above the power supply current rating, the power supply will overheat.
350
A zener diode with a rating of 500 mW will pass 50 mA at 10 V. (Power = voltage times current)Note: The question appears mis stated, in that it states a rating of 500 MW, not 500 mW. To my knowledge, there is no zener with a rating of 500 MW.
Power (watts) is voltage times current.
Ratio of voltage rating and current rating is called power factor in electricalAnswerPower factor can be defined in a number of ways -for example:cosine of the phase angleratio of true power to apparent powerIt has nothing to do with the ratio of voltage rating to current rating!
Each fuse has its own rating. It will be marked on the fuse somewhere.
It is generally not recommended to use a power supply that exceeds the current rating of the device you are powering. In this case, using a 24V 3A power supply instead of a 24V 2.3A could potentially overload the device and lead to damage or malfunction. It is safer to use a power supply that matches or slightly exceeds the current rating of the device.
3 amps