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Voltage and light bulb

Discussion in 'LEDs and Optoelectronics' started by pidja105, Nov 23, 2015.

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  1. pidja105

    pidja105

    106
    1
    Oct 16, 2015
    Hello,
    I have a simple question:
    If i have 220v 40 watts, and i put into the 24v battery, it gives me small light, if i put it to 220v(i don't know what is power of the battery and inverter), it gives me full light, what is the problem, voltage or power(voltage*amperage)?
     
  2. shrtrnd

    shrtrnd

    3,712
    468
    Jan 15, 2010
    It's difficult to give you a specific answer, because we're not sure physically what you mean.
    My first thought is that your inverter may not be able to produce 40 watts, the information on the name-plate should tell you that.
    Your set-up has 220VAC, an inverter, a light bulb, and a battery, ... but we don't know from your description what the physical configuration of your set-up is.
    Can you give us more detail about exactly how you have everything wired together?
     
    Last edited: Nov 23, 2015
  3. Martaine2005

    Martaine2005

    2,649
    718
    May 12, 2015
    I think he is asking whether the voltage or amps alters the brightness of the bulb.
    If I am correct, it will be the voltage.
    It's the voltage and Current that give the power (Watts).

    Martin
     
  4. BobK

    BobK

    7,680
    1,685
    Jan 5, 2010
    Lower voltage will result in lower current.

    Power is voltage times current, so lowering the voltage will lower the power by the square of the voltage ratio. I.e. if you use 1/2 the voltage you will get 1/4 the power. This is only true for resistors, though. A light bulb (incandescent) has a non-linear voltage to current response because it's resistance increases with increased temperature, which depends on the power it is consuming.

    So, in your example, putting 24V on a 220V light bulb will result in lower current.
    If it is 40W at 220V the resistance can be determined by:

    P = V * V / R
    R = V * V / P = 1210Ω

    At 24V, if the resistance stayed the same, we would have:

    P = V * V / R = 24 * 24 / 1210 = 0.47W

    The actual resistance will be lower at that power level, so it might be more like 1W.

    When you use an inverter, you up the voltage to 220V and the bulb will use it's full power.

    Bob
     
    davenn and Martaine2005 like this.
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