Light Bulbs, Saws, Motors, Circuits and Basic Calculations

[W. Watson]

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.
--
Wayne Watson (Nevada City, CA)

Web Page: <speckledwithStars.net>

 

[John Popelish]

W. Watson wrote:

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.

First big mistake. The line voltage reading of 110 volts is
the RMS value.

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?

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.

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?

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.

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?

Dimmer.

Same question for a motor for a pump or drill.

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).

 

[Doug Miller]

In article <YKrOi.570$LD2.217@newssvr17.news.prodigy.net>, "W. Watson" <wolf_tracks@invalid.com> wrote:

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?

15A * 120V = 1800W

Hopefully, you don't need my help to compute the number of bulbs from that.

It would
seem 30 is the answer.

It would seem you need to try again.

I think a 100 watt bulb has a resistance of about 150 ohms.

About that, yes...

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.

Wrong. 110/120 **is** the RMS voltage.

The resistance of 30 100 watt light
bulbs in parallel would be 5 ohms. 75v/5ohms = 15A, so that's the limit.

As noted above, the voltage is 120V, not 0.7071 * 120V.

However, when did I put so much current into the bulbs, they started blowing
out?

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?

In other words, what's the practical limit here?

You should be able to deduce the practical limit from this simple equation:
120V * 15A = 1800W

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?

It doesn't change.

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.

Perhaps you should try some experiments...

--
Regards,
Doug Miller (alphageek at milmac dot com)

It's time to throw all their damned tea in the harbor again.

 

[Peter Bennett]

On Mon, 08 Oct 2007 08:09:47 -0700, "W. Watson"
<wolf_tracks@invalid.com> wrote:

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?

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.

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.

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

 

[W. Watson]

To All, thanks for the refresher and tips. It's been a very looong time
since I had to think about this stuff.

Peter Bennett wrote:

On Mon, 08 Oct 2007 08:09:47 -0700, "W. Watson"
wolf_tracks@invalid.com> wrote:

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?

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.

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.

When the saw draws more current, there will be more voltage drop in
the wiring, so the lamps will dim slightly.

--
Wayne Watson (Nevada City, CA)

Web Page: <speckledwithStars.net>

 

[Nobody]

On Mon, 08 Oct 2007 08:09:47 -0700, W. Watson wrote:

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?

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.

 

[Phil Allison]

"Nobody"

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.

** 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

 

[Doug Miller]

In article <zKadnW9IF6NqwZfanZ2dnUVZ_iydnZ2d@comcast.com>, John Popelish <jpopelish@rica.net> wrote:

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,

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.



--
Regards,
Doug Miller (alphageek at milmac dot com)

It's time to throw all their damned tea in the harbor again.

 

[W. Watson]

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.

When the saw draws more current, there will be more voltage drop in
the wiring, so the lamps will dim slightly.

--
Wayne Watson (Nevada City, CA)

Web Page: <speckledwithStars.net>

 

[redbelly]

On Oct 9, 10:31 am, "W. Watson" <wolf_tra...@invalid.com> wrote:

Back to the saw. So when a saw blade gets stuck or hampered, it's resistance
(impedance?) decreases?

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

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.

When the saw draws more current, there will be more voltage drop in
the wiring, so the lamps will dim slightly.

--
Wayne Watson (Nevada City, CA)

Web Page: <speckledwithStars.net

 

[W. Watson]

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.

redbelly wrote:

On Oct 9, 10:31 am, "W. Watson" <wolf_tra...@invalid.com> wrote:
Back to the saw. So when a saw blade gets stuck or hampered, it's resistance
(impedance?) decreases?

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

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.

When the saw draws more current, there will be more voltage drop in
the wiring, so the lamps will dim slightly.
--
Wayne Watson (Nevada City, CA)

Web Page: <speckledwithStars.net

--
Wayne Watson (Nevada City, CA)

Web Page: <speckledwithStars.net>

 

[Nobody]

On Wed, 10 Oct 2007 08:34:59 -0700, W. Watson wrote:

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.

"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.

 

[W. Watson]

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.

Nobody wrote:

On Wed, 10 Oct 2007 08:34:59 -0700, W. Watson wrote:

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.

"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.

--
Wayne Watson (Nevada City, CA)

Web Page: <speckledwithStars.net>

 

[redbelly]

On Oct 10, 11:34 am, "W. Watson" <wolf_tra...@invalid.com> wrote:

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,

Probably -- drawing more current would drop the line voltage, it's a
question of by how much and is it noticeable.

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.

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

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.



redbelly wrote:
On Oct 9, 10:31 am, "W. Watson" <wolf_tra...@invalid.com> wrote:
Back to the saw. So when a saw blade gets stuck or hampered, it's resistance
(impedance?) decreases?

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

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.

When the saw draws more current, there will be more voltage drop in
the wiring, so the lamps will dim slightly.
--
Wayne Watson (Nevada City, CA)

Web Page: <speckledwithStars.net

--
Wayne Watson (Nevada City, CA)

Web Page: <speckledwithStars.net

 

[Nobody]

On Wed, 10 Oct 2007 14:47:21 -0700, W. Watson wrote:

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.

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.

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).

 

[W. Watson]

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.

redbelly wrote:

On Oct 10, 11:34 am, "W. Watson" <wolf_tra...@invalid.com> wrote:
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,

Probably -- drawing more current would drop the line voltage, it's a
question of by how much and is it noticeable.

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.

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

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.



redbelly wrote:
On Oct 9, 10:31 am, "W. Watson" <wolf_tra...@invalid.com> wrote:
Back to the saw. So when a saw blade gets stuck or hampered, it's resistance
(impedance?) decreases?
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
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.
When the saw draws more current, there will be more voltage drop in
the wiring, so the lamps will dim slightly.
--
Wayne Watson (Nevada City, CA)
Web Page: <speckledwithStars.net
--
Wayne Watson (Nevada City, CA)

Web Page: <speckledwithStars.net

--
Wayne Watson (Nevada City, CA)

Web Page: <speckledwithStars.net>

 

[W. Watson]

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

Nobody wrote:

On Wed, 10 Oct 2007 14:47:21 -0700, W. Watson wrote:

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.
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.

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).

--
Wayne Watson (Nevada City, CA)

Web Page: <speckledwithStars.net>