They told me P=IV so watts a volt amp?!

Discussion in 'Electricians' Talk' started by Bumbling Diyer, Jan 30, 2004.

  1. Bumbling Diyer

    Bumbling Diyer New Member

    Sorry... thought I'd get all the easy puns out of the way...

    Seriously though, could someone please clear up my ignorance about the difference between watts and VA? In particular, in relation to dimming low voltage lights...
     
  2. PPS

    PPS New Member

    Watts is the power output.

    VA is the current draw.

    So a 60 watt bulb draws 4 amps at 240 volts,

    Drop the voltage and to ge the same wattage you need to increase the current to get the same output.

    So a VA unit should be rated by the max it can handle.

    (It is 20 years since I did my City and Guilds, and I left the trade 16 years ago to go in to IT, so if I'm wrong don't hit me)
     
  3. andyspark

    andyspark Active Member

    If a 60 watt lamp draws 4 amps we're all in trouble.

    P = V x I

    60 = 240 x 0.25

    A 60 watt lamp draws 0.25 amp
     
  4. Fubar

    Fubar New Member

    VA is the current draw.

    So a 60 watt bulb draws 4 amps at 240 volts,

    P=IV therefore P/V = I.

    A 60 Watt bulb draws 0.25 amps at 240 Volts.

    --
    Michael
     
  5. Damocles

    Damocles New Member

    dunno mate, they told me to keep thinking about Ivy Watts, so i'd remember it, but that's teenagers for you.

    Alwyas thought VA was the same as Watts,just makes the pointita amps times volts
     
  6. Abe

    Abe New Member

    Seriously though, could someone please clear up my
    ignorance about the difference between watts and VA?
    In particular, in relation to dimming low voltage
    lights...

    In for a penny.............

    For a purely resistive load (a light bulb is close enough), they are the same.

    Things start to get hairy when driving highly reactive loads like electric motors and is due to the fact that we are considering alternating voltage and currents with mismatched phase. In this case the watts delivered to the load will be less than the VA applied, the ratio (W/VA) being referred to as the power factor. It is possible to apply power factor correction (PFC) to optimise the transfer of power by reducing the phase difference between voltage and current.

    VA is important in determining the correct current rating of cables and power controllers, as in the case of a low power factor, the current may be much higher than would be calculated from the wattage of the load.

    As far as dimming low voltage lighting is concerned, it is important that the VA rating of your dimmer is higher than the maximum VA input required by your "dimmable" transformer.
     
  7. Kipster

    Kipster New Member

    Alwyas thought VA was the same as Watts,just makes the pointita amps times volts


    Wattage is True Power.
    VA is Apparent Power.

    The difference between watts and VA is caused by the reactive portion of a circuit, i.e. the capacitive and/or inductive components.

    Power is only consumed by the resistive part of the load; current flowing in the reactive part does not result in power being dissipated.

    So in the case of low voltage lights the transformer needs to be sized to allow for the current flowing in the reactive part of the circuit, thats why there is a VA rating instead of watts.
     
  8. Damocles

    Damocles New Member

    Sounds like you are saying watts is the integral of instantaneaous current times voltage, wheras VA is rms current times rms voltage?

    'Power is only consumed by the resistive part of the load; current flowing in the reactive part does not result in power being dissipated.'

    Dont know about that bit. wouldnt you consider the inputs to a transformer driving an electric heater as an inductive load which dissipates power?

    surely this appears as an inductive load to the power source, i mean it is an inductor? while since i thought about this.
     
  9. The Trician

    The Trician New Member

    Alwyas thought VA was the same as Watts,just makes
    the pointita amps times volts


    Wattage is True Power.
    VA is Apparent Power.

    The difference between watts and VA is caused by the
    reactive portion of a circuit, i.e. the capacitive
    and/or inductive components.

    Power is only consumed by the resistive part of the
    load; current flowing in the reactive part does not
    result in power being dissipated.

    So in the case of low voltage lights the transformer
    needs to be sized to allow for the current flowing in
    the reactive part of the circuit, thats why there is
    a VA rating instead of watts.

    This is essentially correct, it is the inductance/capacitance (impedence) which has to be taken into account where transformers/chokes/Motors/PF correction circuits are concerned. For a purely resistive load, these factors acn be safely ignored.
     
  10. finjon

    finjon New Member

    I suppose the transformers should be rated in watts, rather than VA or kVA. Cos unless you know the P.F., you won't really know watt load you can connect.
     
  11. fusefinder

    fusefinder New Member

    "Mathematically, a watt is a scalar quantity resulting from the vector product of two vector quantities (volts and amps). It is not the simple algebraic product of the rms voltage times the rms current.

    "VA" on the other hand is the scalar quantity resulting from multiplying the rms magnitude of the vector quantities, volts and amps. This resulting quantity will never be smaller than the watts demanded by an instrument. Uninformed individuals incorrectly use VA to assess a product's overall efficiency and power demands. Correctly applied, VA is used to determine proper ac mains conductor gage and circuit breaker sizing as well as "power factor".

    "Power factor" is, for a given set of conditions, the watts consumed by a product divided by the VA necessary to deliver that power. The power factor will only equal 1.0 when the load is purely resistive; i.e., watts consumed equals the VA necessary. In real life this happens infrequently. More times than not, the power factor is a number less than 1.0.

    As the load becomes more reactive, a greater number of VA is required to deliver the same number of watts to the load. Assuming the rms mains voltage remains fixed, the decreased power factor necessitates a larger rms current be available to supply the same nominal number of watts to the load. The power utility and distribution grid must be capable of generating and distributing this disproportionately high level of current. This is why the utility companies are desirous of power factors near 1.0 for products.

    "Efficiency", is simply the work done by a system divided by the work supplied to the system. For an electrical load under a given set of conditions, this is usually the output power divided by the input power. The resulting number is always less than 1.0. The difference between the output power and the input power is the number of watts lost from the system in the form of heat. Lost watts can be converted into BTU's/Hr and equates to the amount of heat which is released into the ambient surroundings. "

    (Copied from some web page, URL too long to post...)
     
  12. The Trician

    The Trician New Member

    As you say, the most desireble and efficiant use of electrical energy is to have a power factor as close to '0' or Unity as you can achieve. This means equalising the inductive and capcacitive values so that they cancel each other out and you theoretically attain a value of'0' or 'unity'. This is attainable theoretically but not practically - a bit like a perpetual motion machine, where there is no energy used, and no losses incurred:(
     
  13. Damocles

    Damocles New Member

    So was that a yes or a no?
     
  14. finjon

    finjon New Member

    What was the question?
    Anyway, to continue in the show-off vein, banks of capacitors are used for Power Factor Correction, to cancel out the effect of inductive loads. The one leads, the other lags, they cancel the effects of one another. With resistive loads, the current and the voltage are in sync. Inductive, the current follows. Capacitive, the current leads.
    Electricity supply companies, (what's the British shorthand, in Ireland it's the ESB (Electricity Supply Board)), discourage poor PF's because it means their cables have to be bigger.
     
  15. The Trician

    The Trician New Member

    Yep, its all about phase angles etc!
     
  16. PPS

    PPS New Member

    OK,

    So in 15 years I forgot my maths (who hasn't)

    But rememberd Ohms Law, and remembered the basic's of VA?
     
  17. finjon

    finjon New Member

    It may be the other way round, I may have it **** about ***.
    Anyway, with an inductive load, eg a motor, when you connect an ammeter and a voltmeter, you have initially, a high current, and a low voltage, and vice verse with a capicitive load. In both cases,the ammeter and voltmeter go in opposite directions. Connect a motor , (for example), in the case of an inductive load. The ammeter peaks, drops, as the voltmeter rises. And vise verse with a capacitive load, where initially you have a voltage, and no current.
    (God bless the VEC College Of Technology, Kevin Street, Dublin. They taught us well!)
     
  18. finjon

    finjon New Member

    Why is ** censored? ** ** ** **. Stick that in your pipe and smoke it.
     
  19. Dr Who

    Dr Who New Member

    Anyone remember CIVIL ?
     
  20. fusefinder

    fusefinder New Member

    haven't heard that since '87! :) but a good way to remember what leads what!
     

Share This Page