Simple basic electrical question

Thread Starter

Nathan Hale

Joined Oct 28, 2011
159
Hello folks i have a rather basic question.
Let us say an electronic/ electrical device like a simple light bulb is rated at 120 volts at .25amps current. That would make the power consumption of the bulb 30 watts. will this same light bulb turn on and function normally if i were to give it 30 volts at 1 amps current?
my question in other words is , will an electronic / electrical device do what it is meant to do if you give it the right amount of power even while you are changing the ratios of the voltage and current.
thanks
 

kubeek

Joined Sep 20, 2005
5,748
...if i were to give it 30 volts at 1 amps current?
This is not possible. You can power it with either 30V or with 1A, but you cannot decide both at the same time. The bulb will have 120/0,25= 480ohms of resistance and this will dictate the relationship between voltage and current. So at 30V it will consume 62.5mA which is a little less than 2W. At 1A the voltage will be 480 volts and the power will be 480W.
 

Thread Starter

Nathan Hale

Joined Oct 28, 2011
159
This is not possible. You can power it with either 30V or with 1A, but you cannot decide both at the same time. The bulb will have 120/0,25= 480ohms of resistance and this will dictate the relationship between voltage and current. So at 30V it will consume 62.5mA which is a little less than 2W. At 1A the voltage will be 480 volts and the power will be 480W.
OK....Now... I have a pocket radio alright? it is supposed to have 2 AA ( 1.5 v each) batteries to run it. i am assuming that the radio will draw "x" amount of current from the 2 AA batteries. if i theoretically replace the 2 AA batteries with a single A battery ( of 1.5 v ) which also happens to have the capacity to give out "2x" the amount of current , the radio wont work....why wont the radio work??? it is still getting the same amount of power that it needs!! why wont the radio work??!!
 

BillB3857

Joined Feb 28, 2009
2,543
Ohm had the answer. Have you ever heard of him? He discovered the relationship between voltage, resistance and current. His discovery was so accurate that a law was even named after him; Ohm's Law.
 

WBahn

Joined Mar 31, 2012
26,398
This is not possible. You can power it with either 30V or with 1A, but you cannot decide both at the same time. The bulb will have 120/0,25= 480ohms of resistance and this will dictate the relationship between voltage and current. So at 30V it will consume 62.5mA which is a little less than 2W. At 1A the voltage will be 480 volts and the power will be 480W.
This is a reasonable answer to the gist of the question, but for completeness it should be noted that the cold resistance of an incandescent bulb (which is about the only kind that makes sense to even be discussing for a question like this) will be markedly lower at room temperature, probably by a factor of 12 or so. But as soon as you start applying power to it, it will start heating up and the resistance will increase. So at 30V it will be drawing quite a bit more than the 62 mA but no where near even the original 250 mA, let alone the 1000 mA that would be needed to get 30 W into it.
 

WBahn

Joined Mar 31, 2012
26,398
OK....Now... I have a pocket radio alright? it is supposed to have 2 AA ( 1.5 v each) batteries to run it. i am assuming that the radio will draw "x" amount of current from the 2 AA batteries. if i theoretically replace the 2 AA batteries with a single A battery ( of 1.5 v ) which also happens to have the capacity to give out "2x" the amount of current , the radio wont work....why wont the radio work??? it is still getting the same amount of power that it needs!! why wont the radio work??!!
For most circuits as you lower the voltage supply you lower the current draw. Consider a simply resistor to get an idea. If you apply V voltage across it then you will get a current I that is related to the voltage and the resistance, R, by Ohm's Law, namely V=IR. If you cut the voltage in half, then you will cut the current in half. If you try to increase the current this will only be possible if the voltage increases proportionately.
 

Thread Starter

Nathan Hale

Joined Oct 28, 2011
159
Ohm had the answer. Have you ever heard of him? He discovered the relationship between voltage, resistance and current. His discovery was so accurate that a law was even named after him; Ohm's Law.
but i am giving the device the power it wants!! why cant it just work? why does it matter to the device if i cut the voltage in half and doubled the current or cut the current in half and doubled the voltage?
 

WBahn

Joined Mar 31, 2012
26,398
but i am giving the device the power it wants!! why cant it just work? why does it matter to the device if i cut the voltage in half and doubled the current or cut the current in half and doubled the voltage?
HOW?!

You can't just SAY that you are going to give it half the voltage and twice the current and expect that to somehow magically happen! This is NOT government -- here, laws MUST be obeyed!

If I give you a 12 Ω resistor and hook it up to a 12 V battery, it will draw 1 A of current and dissipate 12 W of power. Now I give you a 6 V golf cart battery that is capable of delivering 100 A of current and you connect it to that same resistor. How much power will that resistor dissipate?
 

Thread Starter

Nathan Hale

Joined Oct 28, 2011
159
HOW?!

You can't just SAY that you are going to give it half the voltage and twice the current and expect that to somehow magically happen! This is NOT government -- here, laws MUST be obeyed!

If I give you a 12 Ω resistor and hook it up to a 12 V battery, it will draw 1 A of current and dissipate 12 W of power. Now I give you a 6 V golf cart battery that is capable of delivering 100 A of current and you connect it to that same resistor. How much power will that resistor dissipate?
well that 12 ohm resistor wont be sucking more than half an amp of current ...right? so that would make the power dissipation 3 watts. am i correct?
 

WBahn

Joined Mar 31, 2012
26,398
well that 12 ohm resistor wont be sucking more than half an amp of current ...right? so that would make the power dissipation 3 watts. am i correct?
That's correct. So now do you see why you can't expect to use just one AA battery instead of two with that radio and expect to get the same power? Or to power that light bulb with 1/4 of the expected voltage and get the same power?
 

Thread Starter

Nathan Hale

Joined Oct 28, 2011
159
That's correct. So now do you see why you can't expect to use just one AA battery instead of two with that radio and expect to get the same power? Or to power that light bulb with 1/4 of the expected voltage and get the same power?
While you are here can u please answer another nagging question i got???
i read recently that this power plant that had some sort of accident had a nameplate capacity of 852 MW. Does it mean this power station produces 852 MW in an hour or does it mean in a day or does it mean in an year. here is the site for the power plant i am talking about.https://en.wikipedia.org/wiki/Three_Mile_Island_Nuclear_Generating_Station
 

MrChips

Joined Oct 2, 2009
23,215
Power is a measure of physical, mechanical or electrical strength.

Energy is power exerted, delivered or consumed over time. Thus you multiply power by time to calculate energy.

Hence 852MW is a measure of power, not energy. If the power plant delivered this amount of power over one hour, that would represent energy of 852MWh.

Here is another example.

A 100W light bulb is the power rating of the light bulb.
If you run the bulb for 24 hours, the energy consumed is 100W x 24h = 2400Wh = 2.4kWh
Your electricity company will bill you for 2.4kWh consumed, not for 100W of power.
 

profbuxton

Joined Feb 21, 2014
419
Put simply, you will not get a 12ov rated light to light up with 30v applied. best you MAY get is a very dull glow if that(haven't tried it). Note that the light has a particular resistance as manufactured(for 120v). This resistance determines the current that will flow through the light(ignoring cold versus hot light for the moment).
So at 120v the light will "draw" .25A. What would you expect to draw at 60v( note no change in resistance). It will draw .125A. Half the current, half the watts(or lumens) output and so on.

Same applies to your radio. It requires 2 x 1.5v batteries(total 3v) for the circuits to function. It also is a fixed(ignoring output volume changes, which may increase current "draw") resistance. Replacing the 2 1.5v batteries with one 1.5v battery( of any capacity, could be 1000AH if it will fit) will not make the radio work because the internal circuits are designed by the designer to function with a 3v supply. Try putting a good AA battery in with a partly dead AA and also the radio will not work(not long at least).
Think about the light(radio) as being a fixed resistance(load). You cannot "force" current into the "load" just by increasing the supply capacity(ie bigger battery at same(or reduced) voltage as in your example. If you really need more current into your "load" you need to increase the applied voltage. But that may exceed the voltage rating of the "load" and result in a burned out light(due to excessive current) and damaged components in your radio(same reason).
 

WBahn

Joined Mar 31, 2012
26,398
While you are here can u please answer another nagging question i got???
i read recently that this power plant that had some sort of accident had a nameplate capacity of 852 MW. Does it mean this power station produces 852 MW in an hour or does it mean in a day or does it mean in an year. here is the site for the power plant i am talking about.https://en.wikipedia.org/wiki/Three_Mile_Island_Nuclear_Generating_Station
Power is like speed (not the drug kind). So you might have a car that can travel at a peak speed of 75 mph and that can go as much as 1200 miles in one day (due to things like needing to stop to get fuel, meal/bathroom breaks, rest requirements for the driver, and whatever else). It makes no sense to ask if that 75 mph is what it can do in an hour or in a day -- it is the instantaneous speed.

An analogy would be if fuel was being used at a rate of 10 gal/hr. Would it make sense to ask if that means that it's 10 gal/hr in an hour? No, that's a meaningless question. Another would be light-years, which is not a measure of speed but rather of distance, namely the distance that light would travel in a vacuum in one year. So if told that the nearest star is roughly four light-years away, it would be meaningless to ask if that was in a month, a year, or a century.

Joules is a unit of energy and one watt is a rate of producing or consuming energy, namely 1 joule/second. So 852 MW is a rate of 852 million joules per second. Just as you multiply an average speed by the time the car was traveling at that speed to get total distance, you multiply power by the amount of time operating at that power to get total energy produced. Hence at 852 MW steady output, the plant produces 852 MJ (megajoules) of energy in one second, a bit over 3 terajoules in one hour, and nearly 73 terajoules in one day. Because these are such huge values numerically, they are not convenient to work with, so we use a unit of watt-hours (W·h or W·hr, though that is not technically the right abbreviation) which is the amount of energy that a 1 W power level would produce in 1 hour, namely 3600 joules. So in one hour that plant would produce 852 MW·h and in one day 20.4 GW·h. If you followed this, then see if you can figure out the power output in kW·h/s (the answer is 237 kW·h/s). While that's somewhat a crazy unit, it is actually used sometimes because energy expressed in watt-hours is so common and things like how many watt-hours of energy there is in a ton of coal or a gallon of propane or a thousand cubic feet of natural gas or a gallon of diesel fuel is easily found and so if you want to compute your rate of consumption of fuel then in this way is actually quite convenient.
 

studiot

Joined Nov 9, 2007
4,998
OK....Now... I have a pocket radio alright? it is supposed to have 2 AA ( 1.5 v each) batteries to run it. i am assuming that the radio will draw "x" amount of current from the 2 AA batteries. if i theoretically replace the 2 AA batteries with a single A battery ( of 1.5 v ) which also happens to have the capacity to give out "2x" the amount of current , the radio wont work....why wont the radio work??? it is still getting the same amount of power that it needs!! why wont the radio work??!!
Despite what is being said the situation is actually more complicated.

Firstly you need to distinguish those devices that produce something even on wrong voltages.
This would include electric heaters, light bulbs,
I will call these electrical devices and you should read about Ohm's law.
Such devices will produce less on low voltage and more on high voltage, and the right amount on the correct voltage.
Obviously that cannot go on indefinitely, eventually if you keep increasing the voltage you will destroy the device.

Then there are more complicated devices, like your radio. This requires a minimum voltage to work at all. It is also more sensitive to overvoltage and would be destroyed if you connected a PP3 9volt battery instead.
These are usually electronic devices.

Notice I have only mentioned voltage so far.
Whilst is is possible to force a specific current into most devices, in this world we nearly always have a specified voltage (the mains, the battery voltage etc) and cannot specify the current.
The current is determined by the details of the device connected not by us.

The product of voltage and current is the power - which is the rate at which the device draws energy from the supply.
Again we cannot specify what power is drawn, since this depends upon the current (as well as the voltage) it is determined by the device as well as the supply voltage.

The longer we run the device the more energy we use hence the post which said energy = power times time.

You should study up on Voltage, current, power and energy and a good place to start is the AAC Textbook (the Ebook), accessible from the tabs at the top.
 

cmartinez

Joined Jan 17, 2007
7,310
It is my experience that for beginners in the field of electronics, a hands-on approach is normally the best way to teach basic concepts. I'd suggest you get a voltmeter and a few simple electronic components and find out for yourself all that has been said here.
Then again, not everyone has either the parts or the equipment (or even the space at home) readily available, in which case I'd recommend you log in to youtube and search for videos explaining ohm's law. There's plenty of them. And while these videos will mostly repeat what has been said here, for most of us the visual explanation of abstract concepts clarifies things in a way that language alone cannot.
The more senses you apply to any subject that you study, the easier it will be for you to understand it.
Kudos for your effort and enthusiasm.
 
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