Want this question answered?
Current (amps) = power (watts) / voltage = 100/240 = 0.42 amps
There's more to it than just adding some larger wire. You must upgrade your meter, your weatherhead, conduit, etc. Your power company will handle the cable from the pole to your meter. After that...its up to you. 4/0 aluminum or 2/0 copper SE or USE from the meter to panel. This change has to go all the way back to the main power transformer. It's illegal to do this yourself and the power company will not turn your power back on of you do. Call an electrician.
1.5kva has enough electrical power to supply 100 volts of electricity that is at 15 amps. You can also supply 300 volts at 5 amps and 1000 volts at 1.5 amps.
The formula you are looking for is I = W/E. Amps = Watts/Volts.
Given: Power P = 100 W. Reference power Po = 10^−3 W = 1 mW. Reference power level LPo = 0 dB. Get power P when entering power level LP: P = Po×10^(LI/10) W = 10^−3×10^(LP/10) W. Get power level LI in dB when entering power P in W. LP = 10×log (P / Po) dB = 10×log (P / 10^−3) = 50 dB. The reference power may be different, then the power level will be different.
Current (amps) = power (watts) / voltage = 100/240 = 0.42 amps
what is the fault in the transformer, it trips when it is charged.it is charged through the 100 amps MCCB.
a 1.5 kVa source of electrical power has the capacity to supply 100 volts at 15 amps, 300 volts at 5 amps, or 1000 volts at 1.5 amps.
There's more to it than just adding some larger wire. You must upgrade your meter, your weatherhead, conduit, etc. Your power company will handle the cable from the pole to your meter. After that...its up to you. 4/0 aluminum or 2/0 copper SE or USE from the meter to panel. This change has to go all the way back to the main power transformer. It's illegal to do this yourself and the power company will not turn your power back on of you do. Call an electrician.
A 24 volt DC power supply provides DC amps, not AC amps. You cannot draw 1.8 amps AC from a DC power supply, without some kind of inverter stage.That is the answer to the specific wording of the question. Now the answer to the question I think was originally intended...If 1.8 amps AC is being supplied to a 24 volt DC power supply, what would the current supplied by the power supply be?Power is volts times amps, so power supplied to the power supply is 120 VAC (assumed) times 1.8 amps, or 216 watts. If the power supply is 100% efficient, then the power input equals the power output, so use the some equation to take 216 watts and divide by 24 volts, and you get 9 amps.Keep in mind, this is ideal state, assuming 100% efficiency, and no real power supply will be that.
The equation for power factor is PF = True power in watts/Apparent power in Volt Amps.
1.5kva has enough electrical power to supply 100 volts of electricity that is at 15 amps. You can also supply 300 volts at 5 amps and 1000 volts at 1.5 amps.
The formula you are looking for is I = W/E. Amps = Watts/Volts.
Ohms is a unit of resistance. Amps is a measure of current. They are related by Ohm's Law where Voltage = Current x Resistance. In your example Resistance = 100 x 10 to the 6th power So V in Micro volts over 100 would yield current expressed as micro amps. So if you had 100 volts you would have 1 micro amp.
P = V x I x pf
USB 3.1 (C) can deliver a power output of up to 100 watts (20 volts and 5 amps).
Given: Power P = 100 W. Reference power Po = 10^−3 W = 1 mW. Reference power level LPo = 0 dB. Get power P when entering power level LP: P = Po×10^(LI/10) W = 10^−3×10^(LP/10) W. Get power level LI in dB when entering power P in W. LP = 10×log (P / Po) dB = 10×log (P / 10^−3) = 50 dB. The reference power may be different, then the power level will be different.