10a
144ka
The current draw in amps mulitiplied by the voltage.
A ammeter will tell you how much current draw the load is drawing
Most residential dryers run off a 30A 240V outlet. Indiviually their power ratings will vary, but they shouldn't draw more than 5.7kW. The typical dryer weighs something over 100 pounds to something under 150 pounds. Kind of a big window, but without specifics....
If a lightbulb has a resistance of 250 ohms, the voltage required for the bulb to draw a current of 0.5 A is 125 V. (Ohm's law: voltage equals current times resistance)Unfortunately, its more complicated than that...Is the resistance of 250 ohms the hot resistance or the cold resistance? It matters. It matters very much.Light bulbs have a dramatic positive resistance to temperature coefficient. It is not uncommon for the instantaneous on power to be 10 or 20 times the nominal value.So, if the 250 ohms is the measured resistance while operating at a current of 0.5 A, then 125 V is the correct answer. If the resistance is the cold resistance, you need to go back and find out the hot resistance at the desired operating point.
Ideal Voltmeter has an infinite resistance so it won't draw current from the circuit, but in real life ideal voltmeter doesn't exist.
The formula you are looking for is I = W/E.
It's 75/120 and the answer is in amps.
Power is measured in Watts, power (Watts) = E (volts) x I (current - amps) current is determined by the internal resistance (R) of the lightbulb, the lower the resistance the more current will flow. 120v x 0.5a = 60W 120V x 0.83a = 100W the 100W lightbulb will draw more current We also have Ohm's law: E(volts) = I (amps) x R (ohms) Household voltage stays the same at 120v we have for a 100w lamp: 120v = I x R R = 120v/0.83 amps R = 144.6 ohms for a 60w lamp: 120v = I x R R = 120v/0.5 amps R = 240 ohms The higher watt lamp has lower resistance.
Should be ok provided you do not draw too much current.
5500Watts/220V=25 Amps
Now this is yet another example of where a student obviously needs to pay more attention in class!! If you dare, ask your instructor. He or she isn't just there to draw a paycheck.
Assuming it is also rated for 120V., yes. The wattage doesn't change with an increase or decrease in voltage. However, the current draw does. When you double the voltage a load is hooked up to, the Amperage draw (current) drops in half. Example: if a 240 volt heater draws 6 amps, it will draw 12 amps if connected to 120V. If a 120V heater draws 15 amps, it will draw only 7.5 amps when connected to 240V. But power, or wattage stays the same, regardless. And this is what is used to calculate energy usage and therefore, cost. Please note the above answer says "if it is also rated for"
The only reason the dryer breaker will trip is it senses an overload or a short circuit on the circuit. To test this unplug the dryer and see if the breaker will stay latched. If it does then the wiring to the receptacle is not at fault. If you want to delve further into the problem, leave the dryer unplugged and remove the inspection panel at the back of the dryer and check the connections. Sometimes the screw terminals become loose and corroded and cause the dryer to draw more current. To compensate for the higher resistance at the faulty terminals the dryer will try to draw more current that the breaker will allow. If everything looks good after trying both of these things it is time for a repairman to come in and look at the dryer itself, as the fault is probably an internal problem within the dryer body.
Yes, a 130 watt fan can be plugged into a 120 volt receptacle. The current draw will be I = W/E, Amps = Watts/Volts = 130/120 = 1.08 amps.
1 amp
A 65 Watt incandescent light bulb should draw 65W/120V = 541.67mA
Treating this as an ideal resistance, voltage will be half, so current will be half. Power (which is V * I) is one fourth, so you will only draw 10W.