An electric clothes dryer demands 22A from a 240V outlet at approximately 90% Power Factor. The power demand on the outlet should be about 240V x 22A x 0.9 = 4.75 kW. The active components in an electric clothes dryer are the heating element (100% PF) and the electric motor that turns the tumbler (70-80% PF).
The formula you are looking for is W = I x E. (W is watts, I is current in amps, E is volts)
Ohm's Law tells us that P=IV (Power, in Watts = Current, in Amperes X Voltage, in Volts).
Your clothes dryer uses P=22 x 240.
3,600
Yes. It draws less current.
Assume the supply as DC (Only resistance given) Voltage drop = 10X10X0.12 = 12V (approx)
375 kj
Likely one of the two electric elements is going out which creates very high resistance in the circuit and therefore draws very high current (amps) and causes the breaker to trip.
It will increase the current since the water heater is made of a heating element and which is resistive in nature. Ohms law states that V=IR where V is the voltage, I the current and R the resistance. Now the resistance will always remain constant. Thus, when the voltage is increased, the current will also increase.
4800 watts because watts = volts x amps.
watts = volts X amps, so: 15A X 240V = 3600 watts
1.7amp
We know that Voltage = Current x Resistance, so if E = I x R, then E = 20 x 12 = 240 volts, and the dryer must be plugged into a 240 volt outlet.
Yes. It draws less current.
The formula you are looking for is W = I x E.
By Ohm's Law Voltage = Current x Resistance R = V / I = 120 / 12 = 10 Ohms
11.5 x 240 = 2760W motor.
Power (Watts) = Current (Amps) * VoltagePower = 22Amps * 240 VoltsPower = 5,280 Watts5280
P=VI so I=P/V I= 60/230 I=0.261 A
Ohm's Law says Voltage = Current x Resistance V = 5 x 24 = 120 V.
Assume the supply as DC (Only resistance given) Voltage drop = 10X10X0.12 = 12V (approx)