Yes. The 125V is the rating of the bulb, which indicates the voltage it was manufactured to withstand. The voltage it is actually operated at does not have to be precisely 125V; As a matter of fact the voltage coming from your wall socket can vary from as low as 100VAC all the way to above 125 VAC.
So, screw that bulb in and light up your world!
A 300 watt bulb requires a circuit breaker of at least 2.5 amps. This can be determined by dividing the power (300 watts) by the voltage (120 volts). However, considering normal breaker sizes, a 5 amp breaker would be sufficient for a 300 watt bulb.
You can connect a maximum of 24 100-W bulbs in parallel on a 120-V home circuit without tripping the 20-A breaker. This is calculated by dividing the total circuit wattage (2400 W) by the wattage of each bulb (100 W) to find the number of bulbs that can be safely connected.
No, a 2 pole breaker is designed for a 240-volt circuit or for two separate 120-volt circuits that are out of phase with each other. For a 120-volt single-phase circuit, you would typically use a single-pole breaker.
Before energizing a breaker for a circuit, the circuit has to be clear of any short circuits or grounds. This can be accomplished by the use of a megger on the correct voltage setting.
A 60 watt light bulb, powered by a 120 volt circuit (obviously), will use .5 amps of electricity. Subtract a 25% safety margin from the breaker size (1.5 amps in this example) and you are left with 4.5 amps. .5 goes in to 4.5 eight times. That formula can be used for calculating all electrical loads (the PIE Formula). "P" for power in volts (120 or 240) times "I" (amps) equals "E" for energy used (in watts). Don't forget to leave the "margin of safety". Hope this helps.
A 300 watt bulb requires a circuit breaker of at least 2.5 amps. This can be determined by dividing the power (300 watts) by the voltage (120 volts). However, considering normal breaker sizes, a 5 amp breaker would be sufficient for a 300 watt bulb.
I = E/R = 120/25 = 4.8 amperes
A 60 W light bulb consumes more power (produces more light) than a 40 W light bulb in a given time interval when both are connected to a 120 V circuit. This is because power is directly proportional to the wattage rating of the light bulb.
Yes a 220 volt light bulb will run on a 120 volt circuit but at 1/4 of the wattage that the light bulb is rated at. A 100 watt light bulb on 220 would would be equal to a 25 watt light bult on 120 volt system.
Current or amperage, is wattage divided by voltage. 60 / 120 = .5 amp.
Current or amperage, is wattage divided by voltage. 60 / 120 = .5 amp.
Current or amperage, is wattage divided by voltage. 60 / 120 = .5 amp.
Power = (voltage) x (current) 60 = 120 x I I = 60/120 = 1/2 Amp.
you just have to do some math. if the can lights are the only thing on the circuit then times the wattage of each bulb by the total bulbs and only put 80% of the circuit breaker max on the line. ie 20 amp breaker and 120 volt power equals 2400 watts max. 80% of that is 1920 watts. 15 amp breaker and 120 volt power equal 1800 watts max. 80% of that is 1440 watts. if there are already other item on the circuit then you should test the line fully loaded with an amp clamp to see what its carrying already, and then deduct that from your totals
You can connect a maximum of 24 100-W bulbs in parallel on a 120-V home circuit without tripping the 20-A breaker. This is calculated by dividing the total circuit wattage (2400 W) by the wattage of each bulb (100 W) to find the number of bulbs that can be safely connected.
No, a 2 pole breaker is designed for a 240-volt circuit or for two separate 120-volt circuits that are out of phase with each other. For a 120-volt single-phase circuit, you would typically use a single-pole breaker.
Before energizing a breaker for a circuit, the circuit has to be clear of any short circuits or grounds. This can be accomplished by the use of a megger on the correct voltage setting.