R = U / I = 240 / 20 = 12 Ohms.
Since Ohm's law states that current is voltage divided by resistance, if you place 240 volts across 10 ohms, you will get 24 amperes. This is true for AC and DC. It is also true for inductive or capacitive circuits, except that the calculation of apparent resistance (reactance) and phase angle of current to voltage becomes more complex.
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 a 10 amp fuse can be used to protect a 240 volt circuit. The amperage rating of a fuse is based on the given amperage load of the circuit. The voltage rating on a fuse must match or be higher than the voltage that is applied to the fuse. In other words you can not use a 240 volt fuse on a 277, 347, 480 or 600 volt circuit but it can be used on a 120 volt, Manufactures of switching equipment today make it impossible to interchange different voltage fuses to be installed in higher voltage switches.
106 amps
I am assuming that its a 240 Volt AC circuit supplying an inductive load with a fault loop impedance of 1.9 ohms at the time of the short circuit. The power factor is assumed to be 0.8 The instantaneous earth fault current value would be; Current = (Voltage x Power Factor) / Impedance (240 x 0.8) / 1.9 192 / 1.9 = 101 Amps. However this may be a trick question as it doesn't ask for an instantaneous value, the fuse will limit the fault current to 15 amps and should disconnect the circuit within 0.4 seconds.
The resistance of the conductor can be calculated using Ohm's Law: resistance (R) = voltage (V) / current (I). Plugging in the values gives: R = 240 V / 120 A = 2 ohms.
To provide 240 ohms of resistance. What those 240 ohms do in an actual circuit depends on the intention of the designer.
An ideal lossless transformer with a primary voltage of 1440 and a secondary voltage of 240 will deliver 4 amperes to a 60 ohm load by ohms law 240 volts divided by 60 ohms causes 4 amperes to flow through the 60 ohm load. The power delivered to the 60 ohm load will be 4 amperes times the 240 volts or 960 watts. The current in the primary is divided by the primary to secondary ratio of 6. Thus, the primary current will be 4 amperes divided by 6 or 2/3 ampere. The power going into the primary will be 1440 volts times 2/3 amperes or 960 watts. In a real transformer there is losses that make the numbers change. The output voltage will be slightly less and the resistance and reactance of the windings lower the output.
Power (Watts) = Current (Amps) * VoltagePower = 22Amps * 240 VoltsPower = 5,280 Watts5280
Ohm's law: Volts = Amps * Ohms, or Amps = Volts / Ohms 240 volts / 8500 ohms = 28 milliamps
240 amps AC
Using Ohms Law, the answer is 120/0.5 = 240 Ohms.
Power = Amperes x Voltage
87 to 90 YJ fuel guage sender is: 0 ohms = empty; 44 ohms = 1/2; 88 ohms = full
1200
Power = E2 / RResistance = E2 / power = (240)2 / 3,000 = 19.2 ohmsCurrent = E / R = 240 / 19.2 = 12.5 amperes
If 3 identical 45-ohm resistors are connected in parallel, the net effective resistance of the bunch ...and the load seen by the battery ... is 15 ohms. The current supplied by the battery is60/15 = 4 Amperes.(This assumes that the battery is capable of supplying 4 amps at 60 volts, or 240 watts !)