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Light Bulbs, Saws, Motors, Circuits and Basic Calculations

Discussion in 'Electronic Basics' started by W. Watson, Oct 8, 2007.

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  1. W. Watson

    W. Watson Guest

    The application here is basic home electrical circuitry. It's been awhile
    since I've handled simple calculation, so let's see how this flies.

    How many 100 watt light bulbs can I get into a basic 15A circuit? It would
    seem 30 is the answer. I think a 100 watt bulb has a resistance of about 150
    ohms. AC voltage is 110, and the RMS voltage is about 0.7 of that, so let's
    call it 75 volts to make things easy. The resistance of 30 100 watt light
    bulbs in parallel would be 5 ohms. 75v/5ohms = 15A, so that's the limit.
    However, when did I put so much current into the bulbs, they started blowing
    out? In other words, what's the practical limit here?

    Now suppose I throw into a common household circuit a circular saw with a
    few lights plugged in. What happens to the current through the bulbs? Do
    they dim or brighten? Suppose I'm cutting a board, and the blade gets stuck
    or the saw begins to work harder through a dense piece of wood. Do the bulbs
    dim or get brighter? Same question for a motor for a pump or drill.
     
  2. First big mistake. The line voltage reading of 110 volts is
    the RMS value.
    Nominal line voltage most places n the U.S. is now 120 volts
    RMS. So a 100 watt lamp draws about (from Watts=current
    times voltage) .83 amperes. You are not supposed to load
    any branch circuit more than 75% of its trip rating, so you
    are allowed to load a 15 amp branch circuit to .75*15=11.25
    amps. 11.25amps/0.83amps per lamp =13.55 lamps. Lets call
    that 13 to not exceed the continuous limit.
    The extra load current slightly sags the line voltage
    because of the I*R drop of the wiring, where I is the total
    current and R is the resistance of he wiring.
    Same answer.

    If you see lights that get brighter, they are not on the
    same branch as the varying load, but probably on the other
    half of the 240 volt source. Most residential power sources
    consist of a 240 volt transformer secondary (in that pole
    pig hanging on the nearest telephone pole). This secondary
    is center tapped, and that center tap is grounded. All 120
    volt branch circuits connect loads from one end of the
    secondary or the other to the center tap. 240 volt loads
    are connected between the ends of the secondary.

    If you load a branch circuit to one end of the secondary,
    the voltage drop that load produced in the conductor back to
    the center tap, adds to the voltage between load end of that
    conductor and the opposite end of the secondary. So loads
    across that other half of the secondary that share this
    center tap conductor, will see a voltage rise when loads are
    increased on the other half of the secondary. If this
    effect is severe, it indicates that shared center tap
    conductor has excessive resistance (there is a lose
    connection somewhere along it).
     
  3. Doug Miller

    Doug Miller Guest

    15A * 120V = 1800W

    Hopefully, you don't need my help to compute the number of bulbs from that.
    It would seem you need to try again.
    About that, yes...
    Wrong. 110/120 **is** the RMS voltage.
    As noted above, the voltage is 120V, not 0.7071 * 120V.
    Hard to see how you blew out the bulbs at 120V. Maybe you meant to say you
    tripped the circuit breaker, or blew a 15A fuse -- which would be no surprise.
    How did you determine how much current you put into the bulbs, anyway?
    You should be able to deduce the practical limit from this simple equation:
    120V * 15A = 1800W
    It doesn't change.
    Perhaps you should try some experiments...
     
  4. AC voltage is normally specified as the RMS value. The normal
    household outlet in North America delivers 120 V RMS. You are
    complicating things by calculating the resistance of the lights - you
    just need to look at the power that can be delivered by the 15A
    circuit.

    A 15 amp circuit can deliver 15A * 120V = 1800 watts to a resistive
    load such as light bulbs, so you could theoretically operate 18, 100
    watt bulbs on a 15 amp circuit. In practice, you should limit the
    load to about 80% of the breaker rating.
    When the saw draws more current, there will be more voltage drop in
    the wiring, so the lamps will dim slightly.

    --
    Peter Bennett, VE7CEI
    peterbb4 (at) interchange.ubc.ca
    new newsgroup users info : http://vancouver-webpages.com/nnq
    GPS and NMEA info: http://vancouver-webpages.com/peter
    Vancouver Power Squadron: http://vancouver.powersquadron.ca
     
  5. W. Watson

    W. Watson Guest

    To All, thanks for the refresher and tips. It's been a very looong time
    since I had to think about this stuff.
     
  6. Nobody

    Nobody Guest

    As others have pointed out, 120V * 15A ~= 1800W = 18 * 100W bulbs.

    But don't turn them all on simultaneously. When you first apply power,
    the filament is cold, and has a much lower resistance than when it's hot,
    so you get a substantial current surge at turn-on.
     
  7. Phil Allison

    Phil Allison Guest

    "Nobody"

    ** Nothing to worry about though.

    A 100 watt lamp comes on very quickly & the current surge duration is no
    longer with 18 lamps than it is with one. It will be easily handled by
    typical domestic breakers.

    A single 1800 watt rated lamp is a different matter - the larger filament
    takes much longer to heat and it will likely trip a 15 amp breaker.

    A transformer rated at 1800 watts has a truly monstrous switch on surge, so
    much so that a special "soft start" circuit is essential for use on domestic
    power circuits.



    ....... Phil
     
  8. Doug Miller

    Doug Miller Guest

    Sorry, this is incorrect.

    The figure is 80%, not 75%, and it applies only to "continuous loads" as
    defined in the National Electrical Code: a load where the maximum current is
    expected to continue for three hours or more. This probably does *not* apply
    to most residential lighting circuits.
     
  9. W. Watson

    W. Watson Guest

    Back to the saw. So when a saw blade gets stuck or hampered, it's resistance
    (impedance?) decreases? I would imagine that if it continues to "be stuck"
    then, that it eventually draws more current than can be handled by a 15A fuse.
     
  10. redbelly

    redbelly Guest

    As I understand it, when a motor is stopped the Back-EMF (the reverse-
    voltage induced in the coils due to a changing magnetic field) goes to
    zero. This increase the net voltage, and hence current, of the motor.

    Not sure if I'm 100% accurate on the description, but that's
    essentially what's going on.

    Mark
     
  11. W. Watson

    W. Watson Guest

    I'll buy that, but it's been a very long time since I've heard the the term
    back-EMF, which shows how long I've been away from this stuff. If this is
    correct, then wouldn't that imply the bulbs in the other part of the
    parallel circuit dim, and, if the condition remains that the 15 amp fuse
    finally blows? I suppose the saw might effectively become an open circuit
    under certain conditions, and maybe just die without any effect on the fuse.

    Just for kicks I looked back-emf up in Wikipedia.

    Back electromotive force is a voltage that occurs in electric motors where
    there is relative motion between the armature of the motor and the external
    magnetic field. That didn't help me. :)
     
  12. Nobody

    Nobody Guest

    "relative motion between the armature of the motor and the external
    magnetic field" is a long-winded way of saying that the motor is turning.

    A motor is also a dynamo/alternator (a dynamo is an alternator commutated
    to produce a fixed polarity). If you apply voltage across the terminals,
    it spins. If you spin it, it generates a voltage across the terminals.
    The latter is true whether you spin it using some external force or by
    applying a voltage across the terminals.

    For a given spin direction, the polarity of the generated voltage is the
    same as is required to spin it in that direction. The applied voltage
    tries to drive current into the motor from +ve to -ve. The generated
    voltage tries to drive current out of the motor from +ve to -ve (i.e. in
    the opposite direction), reducing the amount of current flowing in (a
    perfect motor spinning freely would draw zero current).

    If the motor stalls, it stops behaving as a generator and produces no
    back-EMF. In essence, the armature is just a coil of wire. For a DC motor,
    that's a short circuit; a small DC power drill might draw a couple of
    amps while running but could draw 50 amps if stalled.
     
  13. W. Watson

    W. Watson Guest

    A question here if on its way (if the user doesn't release the off switch)
    to 50A will it likely burn out the windings or whatever and produce an open
    circuit? It sounds like chances are small it will go open.
     
  14. redbelly

    redbelly Guest

    Probably -- drawing more current would drop the line voltage, it's a
    question of by how much and is it noticeable.
    The times that I have had a motor stop (on a power drill), and I
    applied power for several seconds before realizing I had to stop and
    manually un-jam the drill bit, there was a faint burnt odor coming
    from the drill. But it continued to work just fine (after clearing
    the drill bit).

    However, applying power to a stuck motor for a longer time, it could
    be a different story.

    Just out of curiosity, what are you trying to do? Setting up a
    workshop? Or just refamiliarizing yourself with electronics?

    Regards,

    Mark
     
  15. Nobody

    Nobody Guest

    If the supply can actually provide the motor's stall current, and there's
    nothing else to interrupt or limit the current (e.g. a fuse), the windings
    will burn out. The windings will be rated based upon normal operating
    current; for a DC motor, stall current can be much higher.

    If the motor is being driven from a PSU, it may (read: should) have
    current limiting, or a fuse on the DC (output) side, or a fuse on the
    mains side. Failing that, some part of the PSU may burn out before the
    motor does.

    If you're driving a DC motor from a 12V car battery, and it stalls, the
    windings will probably be the first thing to go (the motor might have a
    thermal fuse, but this typically won't be replacable).

    With an AC motor, the inductance of the windings will limit the current in
    the event of a stall, so the stall current isn't as extreme as for a DC
    motor. Not that it will survive being stalled for extended periods, but
    you'll probably get chance to switch it off (whereas a DC motor could
    realistically be on fire in the time it takes to register that it's
    stalled).
     
  16. W. Watson

    W. Watson Guest

    Just re-familiarizing myself. After pondering resistive circuits, I started
    thinking about motors and other devices and how they effect outages and
    dimming. I think I've gotten as much as I need. If I go deeper, I'd need to
    go backing to computing impedances, and lots of other stuff that has long
    since disappeared from my brain. In school, they never really talked much
    about blowing fuses or overloading circuits in general classes on
    electricity. Some of that I suppose you learn the hard way, experience.
     
  17. W. Watson

    W. Watson Guest

    Interesting. I think that'll about satisfy my curiosity on the practical
    side of this.
     
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