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Will 60watts 240volts blub glow if connected between phase and earth of a building not neutral wire when the building is properly earthed?
Wiki User
∙ 11y ago
If you are talking about a European residential supply, then the answer is yes -although you are using incorrect terminology. The correct term is 'line', not 'phase'. However, it is probably against the wiring regulations in your part of the world to wire a lamp between line and earth -however, this is often done, for test purposes, by electricians to confirm whether a line conductor is energised .
Wiki User
∙ 11y ago
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What is the current in an AC circuit with a 120-volts and a 60 watt light bulb?
Power = Volts x Amps ( P= V*A) Therefore Amps = P/V in this case .5 amps
How much energy does it take to light a 60-watt light bulb for an hour?
Actually, your wrong. Sure, the lighting ithat is set for proper illumination is 60watts, but if anyone is used to using a dimmer, you can dial down the power consumption for a typical incandescent bulb and it will still light. And as expected, the light level will decrease as you lower the dimmer setting as well.
What is the equation for finding amps?
I find amps by measuring with a 'clamp-on' ammeter. I use this information particularly to evaluate the performance of deep well pumps. If you are looking to determine amperage using known values such as Voltage, Resistance, or Power {Wattage} you will need to use "Ohm's Law". It's the base formula from which most all other formulas regarding electricity extrapolate it's functions from. Ohm's Law: E = I [X] R Where: E=Voltage or EMF [electro motive force] in Volts or "V" I=Current in Ampere as "A" or Amps, R=Resistance in Ohm's as "R" or the Omega symbol From thew known characteristics of E, I, and R another formula of Ohm's law can be used to determine power using two known values. Ohm's Law Power Formula: P = I [X] R Where: P=Power in Wattage/Watts as "W" I=Amp E=Volts Examples: You have a portable car heater {a heating element is just a large resistor for simplicity} and the specifications on its label says he device is a 12 volt appliance that uses 60 Watts of power You are asked to determine what size wire will be needed and the total resistance of the circuit.* Wire is sized for the amount of current ** it will be required to handle giving as little resistance {i.e. heat} as possible, but not over sized wasting the customers or your own money. Using the given values for E and P we can determine I. If, P = I X E then, 60Watts = I {Amps} [X] 12 Volts so, 60 = {I} [X] 12 or, {I} = 60/12 therefore, {I} = 5 Amps As a construction electrician we seldom are required to calculate for anything smaller than 14 guage wire {rated at 15 amps per code} so I am only guessing at 5 amps but I'm sure it is close to 22-24 AWG seeing 18 is good for 14 amps in fixture wiring. Then: If, E-I x R then, 12Volts = 5Amps {from above} [X] R so, 12 = 5 X {R} therefore, R = 12/5 therefore, R = 2.5Ohm's Thus: The device is a 12 Volt heater that uses 60 Watts of power that allows 5 amperes of current through the circuit that produces 2.5 Ohm's of resistance requiring a min. of 22 AWG wire for the network. .......or........... Go down to the hardware store and get a clamp on amp meter with multi function options to measure the resistance overall or individualy lol. [Note} I am an expert at justanswer.com and not familiar with other formats of similar venues. I apologize if my addition to this answer was added out of protocol or if I "stepped on" someones session answer. [*] (each entity of a circuit has it's own resistive value such as the wire, the heating element, and the devices used to control it which are all added together [Rt=R1 + R2 + R3 + etc....} to make the total resistance. [**] {current is the amount of electrons that passes through any given point of a circuit for one {1} second known as a coloomb and is measured in "Joules".
How do you convert 50watts to 60watts?
A: Easily invent a perpetual moving machine
Is 120volts more than 60watts?
They're different things. Watts are volts times current.
If you buy a 180 watt sterio and want four speakers do you need to buy 45 watt speakers?
Yes. That would be the ideal wattage. When if comes to watts you don't want to wire speakers to a system that will push more wattage than they can handle. Think of watts as an amount of power. If you push more power through to the speakers than they can handle than you'll blow them. Example: [Stereo]-----60watts------>[Speaker with 45 watt capacity] = Not good. You can wire a speaker with a higher wattage capacity than what the stero will put out. Example: [Stereo]-----60watts------>[Speaker with 60 watt capacity] = Good. Example: [Stereo]-----60watts------>[Speaker with 75 watt capacity] = Good too.
What is the current in an AC circuit with a 120-volts and a 60 watt light bulb?
Power = Volts x Amps ( P= V*A) Therefore Amps = P/V in this case .5 amps
How much power does my computer use while under load I have 8gb 1333mhz ddr3 RAM 1x AMD 6970hd and an i5 650 processor?
8watts for the memory, 20 watts for the hard drive, 80watts max for the I5 CPU, 10 watts for each fan, 25 watts average for the mainboard chipset, and up to 135watts for the 6970 video card, 8 watts for the DVDRW(writing). Those are all maximums. When the computer is idle and the video is not rendering new material, the whole system is probably 60watts +-.
How much energy does it take to light a 60-watt light bulb for an hour?
Actually, your wrong. Sure, the lighting ithat is set for proper illumination is 60watts, but if anyone is used to using a dimmer, you can dial down the power consumption for a typical incandescent bulb and it will still light. And as expected, the light level will decrease as you lower the dimmer setting as well.
What is 60watts to kwh?
We are talking about two different quantities, here. The watt is used to measure power, which is the rate at which your device is consuming energy. The kilowatt hour is used to measure how much energy has been consumed (or the work done) over a specified period of time. So you cannot simply convert one to the other. However, if you know the rate of consumption of energy, in watts, then you can determine how much energy, in kilowatt hours, your device consumes if you specify for how long that device is operating.For example, in one hour, your 60-W device will consume (60/1000) x 1 kilowatt hour. In 'n' hours, it will consume (60/1000) x n kilowatt hours. In other words, first convert the power in watts into power in kilowatts, then multiply that by the number of hours the device is operating.
Can you boil a pint of water on a 60watts peltier unit?
With enough insulation you can in theory boil any amount of water with any amount of energy, given enough time.The key is to supply more energy to the water than it looses.It takes 2260 joules to boil 1 cc of water, joules can be described at watt-seconds.One pint of water is approximately 473 cc, so that will requite 1,068,980 Joules to boil.Given perfect insulation; it would take 17816 seconds / 297 minutes / 5 hours to boil the water with a 60 watt heating element.Assuming that you do not use perfect insulation and given the fact that a peltier element won't give out all 60 watts on the one side, i'd have to say;No you can't boil a pint of water on a 60 watt peltier unit.
A lamp draws a current of 0.5 a when 120 V is applied across it What is the power rating of the lamp?
I assume this is for an AC system. 110-120Vrms (typical wall socket voltage) = 155-170V sine wave peak to 0 volts. Similar for the current you gave - I assume it is in RMS? In an AC system, 120Vrms * .5Arms = 60Watts If the values you have are measured peak to 0 volts, Wattage = 1/2 * Voltage * Current
How many kilowatts does it take to run a laptop?
Look at the power supply or back of any semi-sophisticated electronics device you want to know the kilowatt usage of. It should have one number on it with an "A" next to the number. This is the amperage. Multiply that by 120 (Alternating Current) and you have the wattage. Divide by 1000. After that, as you may be able to figure yourself already, if you use that device for one hour then you will be said to be using it for that many kilowatts. Example: I have a 1.5A laptop. (it may even say 1,5A) Multiply by 120. 180 watts Divide by 1000. 0.18 kilowatts If I use my laptop for one hour, it will consume at the most 0.18 kilowatts. It may be less because you do not use 100% of your CPU every second, and there may be power-saving features such as auto-off screen, modem or network card working. The best energy saving feature you can use is the one that turns off the screen, but keep in mind if you use anything like DSL/Cable that device usually has its own power supply and it can be more than a laptop. Also, if you want to do the math to find the cost of running your laptop for one straight month, multiply the kilowatts by 730.5 and then by your local power company's charge for each kilowatt-hour. Mine charges $0.1412, so the laptop would cost $18.56 if it was used at full capacity for approximately 30 days, and just multiplying again by 12 months gives me a yearly maximum of $222.80. You could probably reach that number if you turned off all power-saving features, ran a program that can hammer the processor like a math program using WiFi and/or networking simultaneously. If you have free time to burn, I recommend going to SETI@home and have your computer join the search for possible life beyond our world.
What savings can a CFL or Compact Fluorescent Lamp give?
Typical prices in the UK in 2008 are used in this answer only to illustrate the method of calculation needed to answer this question. To make a comparison for wherever you live, just substitute your own local currency and costs for the particular lamps you want to use (lamp purchase prices, cost of labour for lamp replacement and unit costs for electrical energy) and then you just have to do all the calculations! Any total money savings will depend on several things: - how much a lamp is used over a particular period of time - the purchase price of the lamp - the cost of installing/replacing the lamp. In a domestic "home" environment it is often assumed that the labour needed to replace a lamp costs nothing - unless, of course, you actually have to pay someone to help you to do it. But in any office, shop, factory or similar "business" environment there will definitely be a labour cost. For simplicity a very low labour cost to replace a lamp has been assumed in the calculations below, but this is only an estimate: it would have to be adjusted to take into account the actual costs of labour for the time needed to replace lamps at each site. Assumptions used in making the comparison: - an overall five year period of lamp use - a typical business use environment of (say) 5 days per week for (say) 50 weeks per year - a labour cost to make the replacement of (say) £2 per lamp - a 60 Watt Type R80 Reflector Conventional Incandescent Lamp (CIL) costs £1 and has a typical life of 1000 hours - a 15 Watt Type R80 Reflector Compact Fluorescent Lamp (CFL) having similar light output to the 60 Watt CIL, costs £15 and has a typical life of 5000 hours - mains electrical energy costs £0.10 per unit kiloWatt-hour (kWh) Scenario A - Average lamp usage 1 hour/day First compare the basic purchase price and labour costs associated with replacing the lamps over a period of 5 years. 60 Watt CIL: Total use over 5 years: 1hr/day x 5 days/week x 50 weeks/year x 5 years = 1250 hours. You would therefore need to buy 2 CILs, each lasting 1000 hours. Cost £1 each to buy plus, for each lamp, a £2 labour cost to fit it. Total CIL lamp purchase and installation cost over 5 years: £6. 15 Watt CFL Typical lamp life: 5000 hours over 5 years You would therefore need to buy 1 CFL. Cost £15 each to buy plus, for each lamp, a £2 labour cost to fit it. Total CFL lamp purchase and installation cost over 5 years: £17. Now compare energy usage over 5 years: 60 Watt CIL: (60Watts) x (1 hour/day) x (250 days/year) x (5 years) = 75,000 Watt-hours = 75kWh. At (say) 10 pence per kWh the CIL would use £7.50 worth of energy over the 5 year period. 15 Watt CFL: (15Watts) x (1 hours/day) x (250 days/year) x (5 years) = 18,750 Watt-hours = 18.75kWh. At (say) 10 pence per kWh the CFL would use £1.875 worth of energy over the 5 year period. Now compare the overall total costs: 60 Watt CIL: Buy-in price and labour costs £6 + energy £7.50 = £13.50 total cost. 15 Watt CFL: Buy-in price and labour costs £17 + energy £1.875 = £18.875 total cost So, with 1 hour/day average daily use, over a period of 5 years, using a CFL costs more money overall than using a CIL. The extra cost is: (£18.875 - £13.50) = £5.375 which is (£5.375/£13.50) = about a 40% extra money cost. Scenario B - Average lamp usage 4 hours/day First compare the basic buy-in price and labour costs associated with replacing the lamps over a period of 5 years. Total use over 5 years: 4hr/day x 5 days/week x 50 weeks/year x 5 years = 5000 hours You would therefore need to buy 5 CILs, each having a life of 1000 hours, or 1 CFL having a life of 5000 hours. 60 Watt CIL: 5 lamps at £1 each plus, for each lamp, a £2 labour cost to fit it. Total CIL lamp purchase and installation costs over 5 years: £15. 60 Watt CFL: Cost £15 each to buy plus, for each lamp, a £2 labour cost to fit it. Total CFL lamp purchase and installation costs over 5 years: £17. Now compare energy usage over 5 years: 60 Watt CIL: (60Watts) x (4 hours/day) x (250 days/year) x (5 years) = 300,000 Watt-hours = 300kWh. At (say) 10 pence per kWh the CIL would use £30 worth of energy over the 5 year period. 15 Watt CFL: (15Watts) x (4 hours/day) x (250 days/year) x (5 years) = 75,000 Watt-hours = 75kWh. At (say) 10 pence per kWh the CFL would use £7.50 worth of energy over the 5 year period. Now compare the overall total costs: 60 Watt CIL: Buy-in price and labour costs £15 + energy £30 = £45 total cost. 15 Watt CFL: Buy-in price and labour costs £17 + energy £7.50 = £24.50 total cost So, with 4 hours/day average daily use, the money saved by using CFLs instead of CILs over a period of 5 years is £45 - £24.50 = £20.50. This is a money saving of £20.50/£45 = about 45% money saving. Scenario C - Average lamp usage 8 hours/day First compare the basic buy-in price and labour costs associated with replacing the lamps over a period of 5 years. Total use over 5 years: 8hr/day x 5 days/week x 50 weeks/year x 5 years = 10,000 hours 60 Watt CIL: You would need to install 10 lamps lasting 1000 hours each at a cost of £1 each to buy plus, for each lamp, a £2 labour cost for fitting it. Total CIL lamp purchase and installation cost over 5 years: £30. 15 Watt CFL: You would need to use 2 lamps each lasting 5000 hours at a cost of £15 each to buy plus, for each lamp, a £2 labour cost for fitting it. Total CFL lamp purchase and installation cost over 5 years: £34. Now compare energy usage over 5 years: 60 Watt CIL: (60Watts) x (8 hours/day) x (250 days/year) x (5 years) = 600,000 Watt-hours = 600kWh. At (say) 10 pence per kWh the CIL would use £60 worth of energy over the 5 year period. 15 Watt CFL: (15Watts) x (8 hours/day) x (250 days/year) x (5 years) = 150,000 Watt-hours = 150kWh. At (say) 10 pence per kWh the CFL would use £15 worth of energy over the 5 year period. Now compare the overall total costs: 60 Watt conventional: Buy-in price and labour costs £30 + energy £60 = £90 total cost. 15 Watt CFL: Buy-in price and labour costs £34 + energy £15 = £49 total cost So, with 8 hours/day average daily use, the money saved by using a CFL instead of a CIL over 5 years is £90 - £49 = £41. This is a money saving of £41/£90 = about 45% money saving.
What is the equation for finding amps?
I find amps by measuring with a 'clamp-on' ammeter. I use this information particularly to evaluate the performance of deep well pumps. If you are looking to determine amperage using known values such as Voltage, Resistance, or Power {Wattage} you will need to use "Ohm's Law". It's the base formula from which most all other formulas regarding electricity extrapolate it's functions from. Ohm's Law: E = I [X] R Where: E=Voltage or EMF [electro motive force] in Volts or "V" I=Current in Ampere as "A" or Amps, R=Resistance in Ohm's as "R" or the Omega symbol From thew known characteristics of E, I, and R another formula of Ohm's law can be used to determine power using two known values. Ohm's Law Power Formula: P = I [X] R Where: P=Power in Wattage/Watts as "W" I=Amp E=Volts Examples: You have a portable car heater {a heating element is just a large resistor for simplicity} and the specifications on its label says he device is a 12 volt appliance that uses 60 Watts of power You are asked to determine what size wire will be needed and the total resistance of the circuit.* Wire is sized for the amount of current ** it will be required to handle giving as little resistance {i.e. heat} as possible, but not over sized wasting the customers or your own money. Using the given values for E and P we can determine I. If, P = I X E then, 60Watts = I {Amps} [X] 12 Volts so, 60 = {I} [X] 12 or, {I} = 60/12 therefore, {I} = 5 Amps As a construction electrician we seldom are required to calculate for anything smaller than 14 guage wire {rated at 15 amps per code} so I am only guessing at 5 amps but I'm sure it is close to 22-24 AWG seeing 18 is good for 14 amps in fixture wiring. Then: If, E-I x R then, 12Volts = 5Amps {from above} [X] R so, 12 = 5 X {R} therefore, R = 12/5 therefore, R = 2.5Ohm's Thus: The device is a 12 Volt heater that uses 60 Watts of power that allows 5 amperes of current through the circuit that produces 2.5 Ohm's of resistance requiring a min. of 22 AWG wire for the network. .......or........... Go down to the hardware store and get a clamp on amp meter with multi function options to measure the resistance overall or individualy lol. [Note} I am an expert at justanswer.com and not familiar with other formats of similar venues. I apologize if my addition to this answer was added out of protocol or if I "stepped on" someones session answer. [*] (each entity of a circuit has it's own resistive value such as the wire, the heating element, and the devices used to control it which are all added together [Rt=R1 + R2 + R3 + etc....} to make the total resistance. [**] {current is the amount of electrons that passes through any given point of a circuit for one {1} second known as a coloomb and is measured in "Joules".
