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Q: What current does a 175-ohm resistance draw on 220 Volts?
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A circuit whose resistance is 22 ohms and has an applied voltage of 440 volts will draw a current of?

20A By using V=IR


Which 12 volt bulb will draw the most current when connected to a 12 volt battery?

The bulb with the lowest resistance. Current = Volts / Resistance


How many watts in 20 volts?

You need to have the amperage to determine how many volts you get out of 20 watts.


What is the resistance of the heating element of an electric iron it the ampere draw is 8 amperes when 115 volts are applied?

Using Ohm's Law (E = I R) Voltage = Current x Resistance or switch around to get R = E / I: 115 volts / 8 Amperes = 14.375 Ohms The above is correct for DC current but is close enough to be used for AC current.


How does a 10 watts bulb work in a series connection?

A 10 watt bulb is defined by the voltage supply and the resulting current. So to make the math simple, suppose you have a 10 watt incandescent bulb designed to work at 20 volts. That means it will draw 1/2 amps. Watts = Volts x Amps. The resistance of the bulb is then Volts / Amps so in this case the resistance of the bulb would be 40 ohms. So our mythical bulb has a resistance of 40 ohms with 20 volts across the bulb in our example. Now if we put two of these bulbs in series with the same 20 volts we now have a total resistance of 80 ohms supplied by 20 volts and the circuit will draw 1/4 amp. This lower current will cause the bulbs to be dimmer.


How many watts is 20 amps?

Twenty amps is zero watts. You are missing one value. W = Amps x Volts. <<>> It depends on the resistance and the draw current in the electrical circuit.


How many amperes will the toaster draw if it has a resistance of 24 ohms?

It depends on how many volts it has.


What would the current draw be if there was six ten ohm resistors connected in series across a 120 volts?

The total current in a circuit consisting of six operating 100 watt lamps connected in parallel to a 120 volt source is 5 amperes. Since power is volts times amps, take 600 watts (100 times 6) and divide by 120 volts to get 5 amps.


How many amps does a 2x4' lay-in fixture draw at 277 volts?

Check the current draw that is on the label of the ballast.


How can you maintain a constant 13.5 volts DC at 20 amps even though the circuit will demand more amps as the temperature increases?

If the question asks how 13.5 volts can be supplied to a device that draws 20 amps (nominally), the supply responds to the setpoint selected (13.5 volts). The supply's voltage has the ability to actually change as the dynamic resistance of the device it supplies changes. That's weird because we want the voltage to stay the same. The supply is actually changing the amount of current it supplies as the resistance of the load changes, and this will keep the applied voltage fixed at 13.5 volts. How does that work? We know that for a given resistance, if we wish to supply a constant voltage, we will get a fixed amount of current draw. As the resistance changes (goes down) due to thermal effects, the supply will actually deliver more current to maintain the 13.5 volts. In this way, the supply can accomplish voltage regulation. It's classic Ohm's law. The volts equals the current times the resistance. If voltage is to remain constant, then the current times the resistance will have to remain constant. The onlyway this can happen is that as resistance goes down, current must go up. As the device heats up and its dynamic resistance decreases, the current it "demands" to keep the supplied voltage at the 13.5 volts goes up. The supply does all this automatically.


Two 12 volt batteries in series one has 4 amp's another has 4 amp's they are connected across a 2 ohm load the current flowing in the circuit is?

Volts = Current x Resistance. You have 24 Volts divided by 2 ohms and the draw will be 12 amps. Your batteries will fail quickly if not spectacularly.


Can you increase amps without having an increase in volts?

In general, increased resistance will lower current draw. See ohm's law (V = IR)