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What wattage will be generated if you apply 100 volts to a resistance of 50 ohms?
Wiki User
∙ 11y ago
The formula you are looking for is I = W/E.
Wiki User
∙ 13y ago
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What are Ohms Law and the Power Law?
Ohms Law is used for the measurement of resistance. You can find the resistance in a series circuit using this equation; amps=volts/resistance or volts=amps x resistance.The Power Law is used to find the wattage in the circuit. You can find the amount of watts using this equation: watts=volts x amps.CommentOhm's Law has nothing whatsoever to do with either resistance or power.The equation, R = E/I, is derived from the definition of the ohm, and not from Ohm's Law.Ohm's Law simply describes a linear relationship between the potential difference across a conductor and the current through it. It does NOT describe the relationship between potential difference, current, and resistance.
What is the amperage in an electric circuit when the voltage 120 volts and the resistance is 40 ohms?
Ohm's Law states Voltage = Current x Resistance. You rewrite the equation as Current = Volts / Resistance to solve for current.
What effect are there when resistance is increased in the size of the current?
Current is inversely proportional to resistance. If you double the resistance, you halve the current. Ohm's Law: Volts = Amps * Ohms Solve for Amps: Amps = Volts / Ohms
Why does an electric iron have a resistance of 15 ohms?
I never measured the resistance of an electric iron, but: the 15 ohm resistance is probably required to develop a required 806 watts of power. Using Ohm's law, I(current) = E(voltage)/R(resistance), therefore I(current) =110 volts/15 ohms = 7.33 Amps (current), and, Power (watts) = E(voltage) X I(current); 110 volts X 7.33 amps = 806 watts. The iron requires that wattage to develop enough heat in the sole of the iron. Hope this helps.
What is the difference between ohms and volts?
ohms is the unit of resistance while volts is the unit of voltage
How much heat is generated by 1 amp?
The heat generated by an electric current is in watts, Current x volts. So current alone is not enough to define it. By Ohms Law, Volts = Current x Resistance, so another way of expressing watts is (Current squared) x Resistance. If you know the current andthe resistance you can get the wattage, which will give you the heat generated.
How much power is consumed by a 6 ohm load with 12 volts applied?
Volts = Amps x Resistance Therefore Amps = Volts / Resistance
How do you measur ampere?
You use an "amp gauge" to measure amps in an actual circuit. It is hooked in series with the load. It can be placed anywhere in the circuit as long as it is hooked in series. Mathematically, you have to know the resistance, or wattage and voltage of a circuit. Volts=amps*resistance or amps=volts/resistance, or resistance=volts/amps. Ohms law!
How is the rating of a light bulb determined for instance 100W 60W is that how many watts an hour to run it?
No, the wattage is determined by the resistance of the filament in the light bulb. The formula to determine the wattage is Watts = Voltage (squared)/Resistance in Ohms. To find the resistance of a 120 volt light bulb use the formula, Resistance in Ohms = Voltage (squared)/Watts. So for a 100 watt bulb at 120 volts the resistance is 120 volts x 120 volts = 14400/100 = 144 ohms. For a 60 watt bulb at 120 volts the resistance is 120 volts x 120 volts = 14400/60 = 240 ohms. As you can see this holds true to Ohm's law, current is inversely proportional to the resistance of the circuit. The higher the resistance of a load, the harder it is for the current to flow. In this case less current results in less light being emitted from the filament in the light bulb.
What happen if you deliver 5v to a 12v device which is supposed to deliver 168watt with different amperage from 1A to 5A and not 15A?
Ohms law states current is directly proportional to the volts and inversely proportional to the resistance of the circuit. What will happen if you apply 5 volts to a 12 volt, device the amperage will drop proportionally. The wattage output will drop from 168 watts to 30 watts. The device fed with 12 volts and has an output of 168 watts will draw 14 amps. This calculates the resistance to be .85 ohms. The same device fed with 5 volts reduces the amperage to 6 amps resulting in an output wattage of 30 watts. Since the resistance remains the same no mater what, the only way to move the amperage value is to vary the voltage level.
What is cheaper to run a electric fireplace on a 120 volts or 240 volts outlet?
Voltage does not affect the cost to run. You pay for the wattage, the higher the wattage the more it will cost to run.
What is the Formula of volte convert into amp?
V/R = I meaning, Voltage (in volts) divided by Resistance (in Ohms) equals Current (in Amperes, a.k.a. Amps
Is using 110 volts cheaper than using 230 volts?
no its desame depending on wattage of loads
How can you convert amperes in to volts?
Amperes does not convert into volts they are two different values. Become familiar with the following formula and look at the relationships between the values. Volts = Amps x Resistance, Amps = Volts / Resistance and Resistance = Volts / Amps.
How much resistance should be on a 5.0 kw heat strip?
Depends on the voltage. Wattage is Volts x Amps. Resistance (ohms) is Volts divided by Amps. So on a 120V circuit, it would draw 41.66 amps. To do that, it would need a resistance of 2.88 ohms. But on a 240V circuit, it would draw 20.83 amps. That would require a resistance of 11.52 ohms. Determine the circuit voltage, then use that to figure the amps, then use that result to calculate the resistance necessary.
How much electricity is really used when one 220 voltage 150 watts bulb being plugged in 110 volts supply?
The current through a 220 volt 150 watt bulb is I = W/E = .68 amps. The resistance of that bulb is R = E/I = 324 ohms. The wattage used by the 220 volt bulb when 110 volts is applied W = E(sqd)/R = 37 watts. Half the voltage with the same resistance will quarter the wattage output.
What formula can be used to find wattage?
Watts= Volts x Amps
