Tricky series circuit in a vintage wall light

Discussion in 'The Projects Forum' started by Emma Bannenberg, Jun 8, 2015.

  1. Emma Bannenberg

    Thread Starter New Member

    Jun 8, 2015
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    The vintage light in question is from the 1950s and hardwired to the household lighting circuit at 240v.

    My problem is finding the right wattage/voltage/ampage combination for two very different bulb sizes in the same circuit...please bear with me as I describe the light fitting and my experiments so far...

    The light is wired as a series circuit. The first bulb socket in the circuit is a tiny E10 (10mm screw in) intended to create a wash up the front of the light. The second socket in the circuit is a standard boyonet (B22) behind a metal plate, designed so light reflects back onto the wall. The light has been rewired by a qualified restorer, in the original wiring configuration - no changes. The light fitting is hardwired to a 240v household lighting supply. I've attached an image of the light.

    Here's the work I've done so far: I tried various bulbs to see if I could find a combination that would work. For example, I put a 240v 3w E10 bulb in the first socket in the series, it lights up fine (as long as a 240v bayonet bulb is in the second socket to complete the circuit), but the bayonet bulb doesn't light up. So I tried 25w, 40w, 60w and 100w bayonet bulbs but none work in the series circuit in combo with the 240v 3w E10.

    Then I put a 6.5v 300mA bulb in the E10 fixture, both the E10 and the B22 bulb light up as long as the bayonet bulb is over 40w. I am happy to use a 6.5v 300mA lamp in the series circuit but surely this is going to blow out constantly due to the many times higher voltage power supply?

    At this point I guessed that the problem was about combinations of wattage/voltate/ampage/resistance. I read up on Ohms Law and calculated out equations for power (wattage) output using the available variables for current, energy supply, resistance and voltage (using Kirchhoff's Law). I understand now that the problem is the voltage drop on the small lamp - high resistance in a low watt lamp = a big voltage drop and insufficient voltage to light the second, larger and higher watt lamp in the series.

    So...
    What combination voltage, wattage and ampage of bulbs will work together in this circuit, with each bulb working close to it's wattage capacity?
    Would 6.5v 300mA E10s burn out all the time due to 240v supply?
    If I use two 120v lamps on this 240v circuit does this solve my problem?

    Of course, in an ideal world I could use any combination of lamps for this light, but I am restricted to the voltage, wattage, ampage and resistance of available E10/MES 10mm Screw-in bulbs + B22/Bayonet Cap bulbs.

    Any answers greatly appreciated. If you hadn't guessed, I'm not electrical or engineering qualified or skilled in any way but I am very curious about how wattage, voltage, ampage and resistance are interacting in this circuit and I've learned a lot. It's interesting, but I also really want my light to work!

    Thanks!
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    Welcome aboard, Emma.

    First, an aside. There's no reason to post such a large file. This one is 21KB, or just over 1% of the 1.9MB you attached. Members that have slow connects will really appreciate the difference.

    _1016823.JPG

    It sounds like you've done a good job trying to get a handle on the basics -- I wish more of the students that post here were like you.

    So let's see what we can come up with.

    It's actually going to be impossible to get real accurate because light bulbs are highly non-linear, meaning that the resistance is not constant and varies with temperature, which varies with the voltage you put across them. So what we will come up with here assumes that the bulbs are actually running at their rated voltages. So let's call those voltages Va and Vb (Va can be the voltage rating of the wash bulb and Vb the rating of the main bulb).

    Since they are in series, we want Vs = 240Vac = Va + Vb

    Both bulbs will have the same current, I, so the wattage needed for each bulb will be Wa=I·Va and Wb=I·Vb.

    We want to get rid of I, so we solve for I in one equation and substitute it into the other.

    I = Wa/Va

    Wb=(Wa/Va)·Vb

    This can more conveniently be expressed as

    (Wb/Wa) = (Vb/Va)

    This says that the ratio of the wattages needs to be the same as the ratio of the voltages. You also know that the sum of the voltages needs to be 240V.

    So let's look at that 6.5V, 300 mA bulb and see what you would want for the other.

    Wa = 6.5V·300mA = 1.95W (let's call it two watts).

    Since 6.5V is so small compared to 240V, we can call Vb=240V, meaning that our voltage ratio is 37 so we would want a 74 W bulb. I would not go above this since that will put more voltage/current through the other bulb.

    There's nothing magical about that wattage and it should work for some range below this value. It's hard to determine how big that range might be because of those nonlinearities I was talking about (and also how much current the wash bulb needs to light up well enough).
     
  3. Emma Bannenberg

    Thread Starter New Member

    Jun 8, 2015
    4
    2
    Hey WBahn, Thanks so much for this solution, and sorry about the huge file size.

    I sensed there was a solution somewhere in my calculations but I didn't know which relationship to explore next to get to it. You've explained it perfectly clearly - and just undone the knot I've been trying to untie for a day or two. Thanks!

    I'll try out your calculations on a range of bulb voltages to ensure I understand it but just need to ask this: given that LED lights are preferable for the energy saving factor, will this calculation work the same for LEDs if I establish equal ratios of volts and watts or are LEDs a different beast in terms of current?
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,755
    4,799
    A problem that you'll probably encounter with LED lights is that they do not look anything like a resistor and the current through them is very non-sinusoidal. So I suspect that if you put them in series that they will not play well together. But that's a guess as I've never tried it and I'm not too familiar with the detailed characteristics of LED lighting. Perhaps someone else can shed some light (no pun intended) on things.

    Is there a reason not to wire the two bulbs in parallel? There's a reason that lights aren't wired in series any more.
     
  5. Emma Bannenberg

    Thread Starter New Member

    Jun 8, 2015
    4
    2
    I think they're wired in series due to limited space for wiring in the housing and tubular arm of the light, or perhaps it's just because back in 1950 when the light was manufactured they hadn't worked out this was not such a good idea. Thanks for all your help. I'll let you know how I go with it.
     
  6. Emma Bannenberg

    Thread Starter New Member

    Jun 8, 2015
    4
    2
    I found an original small bulb in another matching unrestored light fitting. 6.3v .33A and using the calculations above found a 75 watt bulb would be perfect. In Australia it's hard to buy old incandescent bulbs which have all been replaced with LED or low energy bulbs but I found a Halogen energy saving bulb rated 72 watts (providing equivalent to 100w) and this works perfectly, with both the small and the large bulb working at or near their capacity. Problem solved! Thanks so much for your help WBahn!
     
    ebeowulf17 and #12 like this.
  7. WBahn

    Moderator

    Mar 31, 2012
    17,755
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    It's so wonderful when theory and reality match. So often they don't because of the details that get simplified a way. Thanks for reporting back with the results and very glad I could help.
     
    ebeowulf17 likes this.
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