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The formula you are looking for is W = E(squared)/R, W = Voltage (squared) divided by Resistance.
Just use Ohm's Law Voltage = Current x Resistance Amps = Voltage Divided By Resistance Amps = 120 / 260
Let R be the resistance, V the voltage, and I the current R = V / I R = 120 / 24 R = 5 ohm
V = I x R so current I = 1/2 amp. I bet the bulb is rated at 60 W because Watts = Current x Voltage. Where V = voltage (volt) I = current (ampere) R = resistance (ohm) Your question isn't easy to answer. A lamp has two 'resistances': a 'cold' resistance, and a 'hot' resistance. Before it is energised, it is cold, so its resistance is low; when it is energised, it becomes very hot, and its resistance increases significantly. So, the question is whether your '240 ohms' is the cold resistance or the hot resistance. If it is the cold resistance, then a current of 0.5 A will flow through it for a fraction of a second, then its resistance will increase significantly, and the current will fall to a very much smaller value.
Ohm's Law states Voltage = Current x Resistance. You rewrite the equation as Current = Volts / Resistance to solve for current.
Power (watts) = current (amperes) * voltage (volts) Current (amperes) = voltage (volts)/resistance (ohms) 120 watts = current * 120 volts current = 1 ampere 1 ampere = 120 volts/resistance resistance = 120 ohms
this is the formula R=E/I. 120/6=20 ohms of resistance.
I = E / R = 120 / 14 = 8.571 Amp. (rounded)
v/i=r so 120/0.25=480 480 ohms is the resistance
The formula you are looking for is W = E(squared)/R, W = Voltage (squared) divided by Resistance.
The equation that you are looking for is I = E/R. Amps = Volts/Resistance.
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.
Assuming DC and resistive loads, resistance equals voltage across the load, divided by the current through it. In this case 120/10 or 12 ohms.
If they are in series, the total resistance is 150 ohms so the current is 120/150 amps. If they are in parallel, the current is 120/100 plus 120/50 amps.
Just use Ohm's Law Voltage = Current x Resistance Amps = Voltage Divided By Resistance Amps = 120 / 260
Let R be the resistance, V the voltage, and I the current R = V / I R = 120 / 24 R = 5 ohm
V = I x R so current I = 1/2 amp. I bet the bulb is rated at 60 W because Watts = Current x Voltage. Where V = voltage (volt) I = current (ampere) R = resistance (ohm) Your question isn't easy to answer. A lamp has two 'resistances': a 'cold' resistance, and a 'hot' resistance. Before it is energised, it is cold, so its resistance is low; when it is energised, it becomes very hot, and its resistance increases significantly. So, the question is whether your '240 ohms' is the cold resistance or the hot resistance. If it is the cold resistance, then a current of 0.5 A will flow through it for a fraction of a second, then its resistance will increase significantly, and the current will fall to a very much smaller value.