Be it AC or DC any device who's nameplate reads 10 volts.
10 amps 250 volts or 16 amps 250 volts.
The power can be calculated using the formula P = V x I, where P is power in watts, V is voltage in volts, and I is current in amps. Plugging in 240 volts and 10 amps, the power would be 2400 watts.
Watts = Volts x Amps x Power Factor Maximum Power Factor is 1 for resistive load.
Using the formula Power(P) = Voltage(V) * Current(I) 5 = 10 * I I = .5 amps Current is .5 amps
There is no translation between volts and watts, they measure different things. Power (in watts) equals potential difference (in volts) times current (in amps) P=IV=(I^2)R=(V^2)/R
10 amps 250 volts or 16 amps 250 volts.
The power can be calculated using the formula P = V x I, where P is power in watts, V is voltage in volts, and I is current in amps. Plugging in 240 volts and 10 amps, the power would be 2400 watts.
It means that a)the the power source gives outputs between 10 and 32 volts DC current OR b)The device needs beteen 10 and 32 volts DC current
Power = Volts * current [p = E*I]; 120volts*10amps = 1200watts
Watts = Volts x Amps x Power Factor Maximum Power Factor is 1 for resistive load.
Basically if you know the Voltage supply and the power used by an appliance then you use the formula for power which is Power = Volts x Amps. Rearrange so Amps (current) = Power / Volts If power was 2400 Watts and Volts was 240 the Current would be 2400 / 240 = 10 Amps
15 KOhms times 10 mA = 150 volts. 150 volts times 10 mA = 1.5 watts.
Using the formula Power(P) = Voltage(V) * Current(I) 5 = 10 * I I = .5 amps Current is .5 amps
Power = volts times amps, so an appliance drawing 10 amps at a line voltage of 110 volts is consuming 1,100 watts. Keep in mind, however, that in a non purely resistive load, the phase angle of amps to volts might not be zero degrees, so the calculation becomes more complex, and depends on power factor, or phase angle.
Divide Watts by Volts ; this gives you Amps.
230 volts
Power = voltage times current, and the power loss is the loss in the line, I^2 * R. At 11,000 volts, the current will be (11,000 / 415 = ) 3.77% of what it is at 415 volts. So the power loss in the line at 11,000 volts will be (3.77% ^2 = ) .14% of what it is at 415 volts.