I have the exact same Unisaw. I connect my saw to my 30 amp dryer connector via a 30 amp twist lock converter and I have never experienced any issue.
I also have a 24" 5hp delta x5 planner that I also connect through my dryer outlet and also have never had an issue.

--
For full context, visit https://www.homeownershub.com/woodworking/amperage-for-5hp-table-saw-218022-.htm

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Mon, 8 Nov 2021 19:01:32 +0000, Ripsaw <9dd8514b1cf2e9993897409011a5bd99@example.com> wrote:

I have the exact same Unisaw. I connect my saw to my 30 amp dryer connector via a 30 amp twist lock converter and I have never experienced any issue.
I also have a 24" 5hp delta x5 planner that I also connect through my dryer outlet and also have never had an issue.

After 17 years he's probably figured it out or burned down his house.
Do you feel better after trolling?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Tuesday, March 2, 2004 at 6:20:32 AM UTC-6, Joe McDonald wrote:

In article <1078204646.761648@smirk>, <_firstname_@lr_dot_los-gatos_dot_ca.us> wrote:

What amperage breaker does one need for a 5HP table saw on a 240V
single phase circuit?

My logic: Typically, single phase 5HP 240V (or 230V) motors are rated
at 19.8 or 20 A. Given that a circuit is supposed to be loaded 80%
(look it up in the NEC sometime), this means a 25 A breaker would be borderline sufficient, and that a 30A breaker should be generous.
With a slow breaker (not a fuse), there should be no problems with the startup surge of the motor tripping the breaker.

Using Ohm's law this is a relatively simple question to answer...
1 horsepower = 745.7 watts
Power (watts) = I (amperage) x E (voltage)
When you have two values for the variables in the equation you can
figure out the third.
P 3728.5 (watts 1 HP x 5)
Therefore I = --- or I = ------
E 230 volts

230 volts = 16.21 amps
240 volts = 15.53 amps

Using the "80% rule", a 20 amp breaker should not exceed a continuous
load above 16 amps. It seems to me, that with the proper type of "slow" breaker, you should be OK with a 20 amp breaker.

Joe

PS
I am not an electrician or an electrical engineer. In fact, I have
never even played one on TV.

Ohm's law is for DC. It does not apply to AC.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Mon, 08 Nov 2021 20:30:47 -0500, Joe Gwinn <joegwinn@comcast.net>
wrote:

On Mon, 8 Nov 2021 16:07:24 -0800 (PST), Bob Davis
<wrobertdavis@gmail.com> wrote:

On Tuesday, March 2, 2004 at 6:20:32 AM UTC-6, Joe McDonald wrote:

In article <1078204646.761648@smirk>,
<_firstname_@lr_dot_los-gatos_dot_ca.us> wrote:

What amperage breaker does one need for a 5HP table saw on a 240V
single phase circuit?

My logic: Typically, single phase 5HP 240V (or 230V) motors are rated
at 19.8 or 20 A. Given that a circuit is supposed to be loaded 80%
(look it up in the NEC sometime), this means a 25 A breaker would be
borderline sufficient, and that a 30A breaker should be generous.
With a slow breaker (not a fuse), there should be no problems with the >>> > startup surge of the motor tripping the breaker.

Using Ohm's law this is a relatively simple question to answer...
1 horsepower = 745.7 watts
Power (watts) = I (amperage) x E (voltage)
When you have two values for the variables in the equation you can
figure out the third.
P 3728.5 (watts 1 HP x 5)
Therefore I = --- or I = ------
E 230 volts

230 volts = 16.21 amps
240 volts = 15.53 amps

Using the "80% rule", a 20 amp breaker should not exceed a continuous
load above 16 amps. It seems to me, that with the proper type of "slow"
breaker, you should be OK with a 20 amp breaker.

Joe

PS
I am not an electrician or an electrical engineer. In fact, I have
never even played one on TV.

Ohm's law is for DC. It does not apply to AC.

Sure it does, for RMS volts and amps, and pure resistances (like metal
wires and components).

Motors and inductors and capacitors are a different kettle of fish. An >ordinary ohmmeter measures DC resistance, which for an inductor is the >resistance of the copper wire winding, but AC current and voltage are >connected by AC reactance., which will be far larger than the DC
resistance at the usual operating frequency of that inductor.

A motor is a little different than an inductor. A motor is doing
"work". If the motor were 100% efficient, it would look like a pure
resistor. They aren't, of course, but the more efficient they are,
the closer to a resistor they appear.

80% efficiency for fully loaded small motors isn't too far off but
probably a little high. They get worse when lightly loaded but for
figuring circuit size, the fully loaded numbers are used. A safety
factor of another 20% is a good idea (if not code). Motors can be
overloaded. ;-)

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Mon, 8 Nov 2021 16:07:24 -0800 (PST), Bob Davis
<wrobertdavis@gmail.com> wrote:

On Tuesday, March 2, 2004 at 6:20:32 AM UTC-6, Joe McDonald wrote:

In article <1078204646.761648@smirk>,
<_firstname_@lr_dot_los-gatos_dot_ca.us> wrote:

What amperage breaker does one need for a 5HP table saw on a 240V
single phase circuit?

My logic: Typically, single phase 5HP 240V (or 230V) motors are rated
at 19.8 or 20 A. Given that a circuit is supposed to be loaded 80%
(look it up in the NEC sometime), this means a 25 A breaker would be
borderline sufficient, and that a 30A breaker should be generous.
With a slow breaker (not a fuse), there should be no problems with the
startup surge of the motor tripping the breaker.

Using Ohm's law this is a relatively simple question to answer...
1 horsepower = 745.7 watts
Power (watts) = I (amperage) x E (voltage)
When you have two values for the variables in the equation you can
figure out the third.
P 3728.5 (watts 1 HP x 5)
Therefore I = --- or I = ------
E 230 volts

230 volts = 16.21 amps
240 volts = 15.53 amps

Using the "80% rule", a 20 amp breaker should not exceed a continuous
load above 16 amps. It seems to me, that with the proper type of "slow"
breaker, you should be OK with a 20 amp breaker.

Joe

PS
I am not an electrician or an electrical engineer. In fact, I have
never even played one on TV.

Ohm's law is for DC. It does not apply to AC.

Sure it does, for RMS volts and amps, and pure resistances (like metal
wires and components).

Motors and inductors and capacitors are a different kettle of fish. An
ordinary ohmmeter measures DC resistance, which for an inductor is the resistance of the copper wire winding, but AC current and voltage are
connected by AC reactance., which will be far larger than the DC
resistance at the usual operating frequency of that inductor.

Joe Gwinn

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Mon, 8 Nov 2021 16:07:24 -0800 (PST), Bob Davis
<wrobertdavis@gmail.com> wrote:

On Tuesday, March 2, 2004 at 6:20:32 AM UTC-6, Joe McDonald wrote:

In article <1078204646.761648@smirk>,
<_firstname_@lr_dot_los-gatos_dot_ca.us> wrote:

What amperage breaker does one need for a 5HP table saw on a 240V
single phase circuit?

My logic: Typically, single phase 5HP 240V (or 230V) motors are rated
at 19.8 or 20 A. Given that a circuit is supposed to be loaded 80%
(look it up in the NEC sometime), this means a 25 A breaker would be
borderline sufficient, and that a 30A breaker should be generous.
With a slow breaker (not a fuse), there should be no problems with the
startup surge of the motor tripping the breaker.

Using Ohm's law this is a relatively simple question to answer...
1 horsepower = 745.7 watts
Power (watts) = I (amperage) x E (voltage)
When you have two values for the variables in the equation you can
figure out the third.
P 3728.5 (watts 1 HP x 5)
Therefore I = --- or I = ------
E 230 volts

230 volts = 16.21 amps
240 volts = 15.53 amps

Using the "80% rule", a 20 amp breaker should not exceed a continuous
load above 16 amps. It seems to me, that with the proper type of "slow"
breaker, you should be OK with a 20 amp breaker.

Joe

PS
I am not an electrician or an electrical engineer. In fact, I have
never even played one on TV.

Ohm's law is for DC. It does not apply to AC.

Yes it does - the R just means reactance instead of resistance -
wheather inductive or capacitive. With resistive loads it's the same
as with DC - read the resistance on your ohm-meter. With "inductive
loads" the reactance will always be higher than the resistance (draws
less current)

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Mon, 08 Nov 2021 20:30:47 -0500, Joe Gwinn <joegwinn@comcast.net>
wrote:

On Mon, 8 Nov 2021 16:07:24 -0800 (PST), Bob Davis
<wrobertdavis@gmail.com> wrote:

On Tuesday, March 2, 2004 at 6:20:32 AM UTC-6, Joe McDonald wrote:

In article <1078204646.761648@smirk>,
<_firstname_@lr_dot_los-gatos_dot_ca.us> wrote:

What amperage breaker does one need for a 5HP table saw on a 240V
single phase circuit?

My logic: Typically, single phase 5HP 240V (or 230V) motors are rated
at 19.8 or 20 A. Given that a circuit is supposed to be loaded 80%
(look it up in the NEC sometime), this means a 25 A breaker would be
borderline sufficient, and that a 30A breaker should be generous.
With a slow breaker (not a fuse), there should be no problems with the >>> > startup surge of the motor tripping the breaker.

Using Ohm's law this is a relatively simple question to answer...
1 horsepower = 745.7 watts
Power (watts) = I (amperage) x E (voltage)
When you have two values for the variables in the equation you can
figure out the third.
P 3728.5 (watts 1 HP x 5)
Therefore I = --- or I = ------
E 230 volts

230 volts = 16.21 amps
240 volts = 15.53 amps

Using the "80% rule", a 20 amp breaker should not exceed a continuous
load above 16 amps. It seems to me, that with the proper type of "slow"
breaker, you should be OK with a 20 amp breaker.

Joe

PS
I am not an electrician or an electrical engineer. In fact, I have
never even played one on TV.

Ohm's law is for DC. It does not apply to AC.

Sure it does, for RMS volts and amps, and pure resistances (like metal
wires and components).

Motors and inductors and capacitors are a different kettle of fish. An >ordinary ohmmeter measures DC resistance, which for an inductor is the >resistance of the copper wire winding, but AC current and voltage are >connected by AC reactance., which will be far larger than the DC
resistance at the usual operating frequency of that inductor.

Joe Gwinn

You will want a "high magnetic" breaker - which, fortunately for you
most "twinned" breakers are (at leat all Square D QO breakers are ). I
ended up using a twinned breaker for my central vac (120 volt 12 amp)
because it quite occaisionally tripped the single standard 15 amp
breaker on start-up and a single high magnetic breaker was not
readilly available locally and cost more to order in than my local
supplier charged for the twinned breaker. Hasn't tripped since (about
3 years) - and it tripped on 3 different single breakers - - -

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Monday, November 8, 2021 at 8:28:53 PM UTC-6, Clare Snyder wrote:

On Mon, 8 Nov 2021 16:07:24 -0800 (PST), Bob Davis
<wrober...@gmail.com> wrote:

On Tuesday, March 2, 2004 at 6:20:32 AM UTC-6, Joe McDonald wrote:

In article <1078204646.761648@smirk>,
<_firstname_@lr_dot_los-gatos_dot_ca.us> wrote:

What amperage breaker does one need for a 5HP table saw on a 240V
single phase circuit?

My logic: Typically, single phase 5HP 240V (or 230V) motors are rated
at 19.8 or 20 A. Given that a circuit is supposed to be loaded 80%
(look it up in the NEC sometime), this means a 25 A breaker would be
borderline sufficient, and that a 30A breaker should be generous.
With a slow breaker (not a fuse), there should be no problems with the >> > startup surge of the motor tripping the breaker.

Using Ohm's law this is a relatively simple question to answer...
1 horsepower = 745.7 watts
Power (watts) = I (amperage) x E (voltage)
When you have two values for the variables in the equation you can
figure out the third.
P 3728.5 (watts 1 HP x 5)
Therefore I = --- or I = ------
E 230 volts

230 volts = 16.21 amps
240 volts = 15.53 amps

Using the "80% rule", a 20 amp breaker should not exceed a continuous
load above 16 amps. It seems to me, that with the proper type of "slow"
breaker, you should be OK with a 20 amp breaker.

Joe

PS
I am not an electrician or an electrical engineer. In fact, I have
never even played one on TV.

Ohm's law is for DC. It does not apply to AC.

Yes it does - the R just means reactance instead of resistance -
wheather inductive or capacitive. With resistive loads it's the same
as with DC - read the resistance on your ohm-meter. With "inductive
loads" the reactance will always be higher than the resistance (draws
less current)

What is a real world example of purely resistive load supplied by AC?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Monday, November 8, 2021 at 7:30:56 PM UTC-6, Joe Gwinn wrote:

On Mon, 8 Nov 2021 16:07:24 -0800 (PST), Bob Davis
<wrober...@gmail.com> wrote:

On Tuesday, March 2, 2004 at 6:20:32 AM UTC-6, Joe McDonald wrote:

In article <1078204646.761648@smirk>,
<_firstname_@lr_dot_los-gatos_dot_ca.us> wrote:

What amperage breaker does one need for a 5HP table saw on a 240V
single phase circuit?

My logic: Typically, single phase 5HP 240V (or 230V) motors are rated
at 19.8 or 20 A. Given that a circuit is supposed to be loaded 80%
(look it up in the NEC sometime), this means a 25 A breaker would be
borderline sufficient, and that a 30A breaker should be generous.
With a slow breaker (not a fuse), there should be no problems with the >> > startup surge of the motor tripping the breaker.

Using Ohm's law this is a relatively simple question to answer...
1 horsepower = 745.7 watts
Power (watts) = I (amperage) x E (voltage)
When you have two values for the variables in the equation you can
figure out the third.
P 3728.5 (watts 1 HP x 5)
Therefore I = --- or I = ------
E 230 volts

230 volts = 16.21 amps
240 volts = 15.53 amps

Using the "80% rule", a 20 amp breaker should not exceed a continuous
load above 16 amps. It seems to me, that with the proper type of "slow"
breaker, you should be OK with a 20 amp breaker.

Joe

PS
I am not an electrician or an electrical engineer. In fact, I have
never even played one on TV.

Ohm's law is for DC. It does not apply to AC.

Sure it does, for RMS volts and amps, and pure resistances (like metal
wires and components).

Motors and inductors and capacitors are a different kettle of fish. An ordinary ohmmeter measures DC resistance, which for an inductor is the resistance of the copper wire winding, but AC current and voltage are connected by AC reactance., which will be far larger than the DC
resistance at the usual operating frequency of that inductor.

Joe Gwinn

You are right in the theory. In practice, doing calculations about motor current with simple ohm's law does not work because a motor is not a pure resistive load.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Bob Davis <wrobertdavis@gmail.com> writes:

What is a real world example of purely resistive load supplied by AC?

An incandescent light bulb.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Wed, 10 Nov 2021 05:22:37 -0800 (PST), Bob Davis
<wrobertdavis@gmail.com> wrote:

On Monday, November 8, 2021 at 8:28:53 PM UTC-6, Clare Snyder wrote:

On Mon, 8 Nov 2021 16:07:24 -0800 (PST), Bob Davis
<wrober...@gmail.com> wrote:

On Tuesday, March 2, 2004 at 6:20:32 AM UTC-6, Joe McDonald wrote:

In article <1078204646.761648@smirk>,
<_firstname_@lr_dot_los-gatos_dot_ca.us> wrote:

What amperage breaker does one need for a 5HP table saw on a 240V
single phase circuit?

My logic: Typically, single phase 5HP 240V (or 230V) motors are rated >> >> > at 19.8 or 20 A. Given that a circuit is supposed to be loaded 80%
(look it up in the NEC sometime), this means a 25 A breaker would be
borderline sufficient, and that a 30A breaker should be generous.
With a slow breaker (not a fuse), there should be no problems with the >> >> > startup surge of the motor tripping the breaker.

Using Ohm's law this is a relatively simple question to answer...
1 horsepower = 745.7 watts
Power (watts) = I (amperage) x E (voltage)
When you have two values for the variables in the equation you can
figure out the third.
P 3728.5 (watts 1 HP x 5)
Therefore I = --- or I = ------
E 230 volts

230 volts = 16.21 amps
240 volts = 15.53 amps

Using the "80% rule", a 20 amp breaker should not exceed a continuous
load above 16 amps. It seems to me, that with the proper type of "slow" >> >> breaker, you should be OK with a 20 amp breaker.

Joe

PS
I am not an electrician or an electrical engineer. In fact, I have
never even played one on TV.

Ohm's law is for DC. It does not apply to AC.

Yes it does - the R just means reactance instead of resistance -
wheather inductive or capacitive. With resistive loads it's the same
as with DC - read the resistance on your ohm-meter. With "inductive
loads" the reactance will always be higher than the resistance (draws
less current)

What is a real world example of purely resistive load supplied by AC?

Your oven. Water heaters, space heaters, light bulbs (sorta, though temperature dependent).

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Wed, 10 Nov 2021 05:22:37 -0800 (PST), Bob Davis
<wrobertdavis@gmail.com> wrote:

On Monday, November 8, 2021 at 8:28:53 PM UTC-6, Clare Snyder wrote:

On Mon, 8 Nov 2021 16:07:24 -0800 (PST), Bob Davis
<wrober...@gmail.com> wrote:

On Tuesday, March 2, 2004 at 6:20:32 AM UTC-6, Joe McDonald wrote:

In article <1078204646.761648@smirk>,
<_firstname_@lr_dot_los-gatos_dot_ca.us> wrote:

What amperage breaker does one need for a 5HP table saw on a 240V
single phase circuit?

My logic: Typically, single phase 5HP 240V (or 230V) motors are rated >> >> > at 19.8 or 20 A. Given that a circuit is supposed to be loaded 80%
(look it up in the NEC sometime), this means a 25 A breaker would be
borderline sufficient, and that a 30A breaker should be generous.
With a slow breaker (not a fuse), there should be no problems with the >> >> > startup surge of the motor tripping the breaker.

Using Ohm's law this is a relatively simple question to answer...
1 horsepower = 745.7 watts
Power (watts) = I (amperage) x E (voltage)
When you have two values for the variables in the equation you can
figure out the third.
P 3728.5 (watts 1 HP x 5)
Therefore I = --- or I = ------
E 230 volts

230 volts = 16.21 amps
240 volts = 15.53 amps

Using the "80% rule", a 20 amp breaker should not exceed a continuous
load above 16 amps. It seems to me, that with the proper type of "slow" >> >> breaker, you should be OK with a 20 amp breaker.

Joe

PS
I am not an electrician or an electrical engineer. In fact, I have
never even played one on TV.

Ohm's law is for DC. It does not apply to AC.

Yes it does - the R just means reactance instead of resistance -
wheather inductive or capacitive. With resistive loads it's the same
as with DC - read the resistance on your ohm-meter. With "inductive
loads" the reactance will always be higher than the resistance (draws
less current)

What is a real world example of purely resistive load supplied by AC?

I believe an oven or (non-induction) stove would come pretty close.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Wed, 10 Nov 2021 16:42:30 -0500, J. Clarke
<jclarke.873638@gmail.com> wrote:

On Wed, 10 Nov 2021 05:22:37 -0800 (PST), Bob Davis
<wrobertdavis@gmail.com> wrote:

On Monday, November 8, 2021 at 8:28:53 PM UTC-6, Clare Snyder wrote:

On Mon, 8 Nov 2021 16:07:24 -0800 (PST), Bob Davis
<wrober...@gmail.com> wrote:

On Tuesday, March 2, 2004 at 6:20:32 AM UTC-6, Joe McDonald wrote:

In article <1078204646.761648@smirk>,
<_firstname_@lr_dot_los-gatos_dot_ca.us> wrote:

What amperage breaker does one need for a 5HP table saw on a 240V
single phase circuit?

My logic: Typically, single phase 5HP 240V (or 230V) motors are rated >>> >> > at 19.8 or 20 A. Given that a circuit is supposed to be loaded 80%
(look it up in the NEC sometime), this means a 25 A breaker would be >>> >> > borderline sufficient, and that a 30A breaker should be generous.
With a slow breaker (not a fuse), there should be no problems with the >>> >> > startup surge of the motor tripping the breaker.

Using Ohm's law this is a relatively simple question to answer...
1 horsepower = 745.7 watts
Power (watts) = I (amperage) x E (voltage)
When you have two values for the variables in the equation you can
figure out the third.
P 3728.5 (watts 1 HP x 5)
Therefore I = --- or I = ------
E 230 volts

230 volts = 16.21 amps
240 volts = 15.53 amps

Using the "80% rule", a 20 amp breaker should not exceed a continuous >>> >> load above 16 amps. It seems to me, that with the proper type of "slow" >>> >> breaker, you should be OK with a 20 amp breaker.

Joe

PS
I am not an electrician or an electrical engineer. In fact, I have
never even played one on TV.

Ohm's law is for DC. It does not apply to AC.

Yes it does - the R just means reactance instead of resistance -
wheather inductive or capacitive. With resistive loads it's the same
as with DC - read the resistance on your ohm-meter. With "inductive
loads" the reactance will always be higher than the resistance (draws
less current)

What is a real world example of purely resistive load supplied by AC?

I believe an oven or (non-induction) stove would come pretty close.

Electric baseboards, electric resistor elements in a heat pump system
for when the temp is to cold.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Wednesday, November 10, 2021 at 7:22:39 AM UTC-6, Bob Davis wrote:

On Monday, November 8, 2021 at 8:28:53 PM UTC-6, Clare Snyder wrote:

On Mon, 8 Nov 2021 16:07:24 -0800 (PST), Bob Davis
<wrober...@gmail.com> wrote:

On Tuesday, March 2, 2004 at 6:20:32 AM UTC-6, Joe McDonald wrote:

In article <1078204646.761648@smirk>,
<_firstname_@lr_dot_los-gatos_dot_ca.us> wrote:

What amperage breaker does one need for a 5HP table saw on a 240V
single phase circuit?

My logic: Typically, single phase 5HP 240V (or 230V) motors are rated >> > at 19.8 or 20 A. Given that a circuit is supposed to be loaded 80%
(look it up in the NEC sometime), this means a 25 A breaker would be >> > borderline sufficient, and that a 30A breaker should be generous.
With a slow breaker (not a fuse), there should be no problems with the >> > startup surge of the motor tripping the breaker.

Using Ohm's law this is a relatively simple question to answer...
1 horsepower = 745.7 watts
Power (watts) = I (amperage) x E (voltage)
When you have two values for the variables in the equation you can
figure out the third.
P 3728.5 (watts 1 HP x 5)
Therefore I = --- or I = ------
E 230 volts

230 volts = 16.21 amps
240 volts = 15.53 amps

Using the "80% rule", a 20 amp breaker should not exceed a continuous
load above 16 amps. It seems to me, that with the proper type of "slow" >> breaker, you should be OK with a 20 amp breaker.

Joe

PS
I am not an electrician or an electrical engineer. In fact, I have
never even played one on TV.

Ohm's law is for DC. It does not apply to AC.

Yes it does - the R just means reactance instead of resistance -
wheather inductive or capacitive. With resistive loads it's the same
as with DC - read the resistance on your ohm-meter. With "inductive
loads" the reactance will always be higher than the resistance (draws
less current)

What is a real world example of purely resistive load supplied by AC?

Thanks to everyone who responded.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Wed, 10 Nov 2021 16:42:30 -0500, J. Clarke
<jclarke.873638@gmail.com> wrote:

On Wed, 10 Nov 2021 05:22:37 -0800 (PST), Bob Davis
<wrobertdavis@gmail.com> wrote:

On Monday, November 8, 2021 at 8:28:53 PM UTC-6, Clare Snyder wrote:

On Mon, 8 Nov 2021 16:07:24 -0800 (PST), Bob Davis
<wrober...@gmail.com> wrote:

On Tuesday, March 2, 2004 at 6:20:32 AM UTC-6, Joe McDonald wrote:

In article <1078204646.761648@smirk>,
<_firstname_@lr_dot_los-gatos_dot_ca.us> wrote:

What amperage breaker does one need for a 5HP table saw on a 240V
single phase circuit?

My logic: Typically, single phase 5HP 240V (or 230V) motors are rated >>> >> > at 19.8 or 20 A. Given that a circuit is supposed to be loaded 80%
(look it up in the NEC sometime), this means a 25 A breaker would be >>> >> > borderline sufficient, and that a 30A breaker should be generous.
With a slow breaker (not a fuse), there should be no problems with the >>> >> > startup surge of the motor tripping the breaker.

Using Ohm's law this is a relatively simple question to answer...
1 horsepower = 745.7 watts
Power (watts) = I (amperage) x E (voltage)
When you have two values for the variables in the equation you can
figure out the third.
P 3728.5 (watts 1 HP x 5)
Therefore I = --- or I = ------
E 230 volts

230 volts = 16.21 amps
240 volts = 15.53 amps

Using the "80% rule", a 20 amp breaker should not exceed a continuous >>> >> load above 16 amps. It seems to me, that with the proper type of "slow" >>> >> breaker, you should be OK with a 20 amp breaker.

Joe

PS
I am not an electrician or an electrical engineer. In fact, I have
never even played one on TV.

Ohm's law is for DC. It does not apply to AC.

Yes it does - the R just means reactance instead of resistance -
wheather inductive or capacitive. With resistive loads it's the same
as with DC - read the resistance on your ohm-meter. With "inductive
loads" the reactance will always be higher than the resistance (draws
less current)

What is a real world example of purely resistive load supplied by AC?

I believe an oven or (non-induction) stove would come pretty close.

99% close enough? A baseboard heater or any calrod heater, an
incandescent lamp. So that covers an electric water heater - and
heat-chaser cables - and electric in-floor heating cables - even an
electric blanket. (Yes there IS a small inductive element if it has a
coiled heat element - but virtually none if carbon fiber)

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