Fai current's (average) electromagnetic_force, & (average) heat_force Fah

If I were
 to convert Electricity
 into average_momentum
 e.g. & average_forces,
 then this is how I would (mathematically) proportion (it, algebraically).

To deflect its needle,
 a D’Arsenval meter’s
 average (e.g. steady state)
 electromagnetic_force

 F_ai=I*(R^{^0.5})*K_i, I=V/R & 1/R=1/(R^{^0.5})*(R^{^0.5})

 is proportional
 to the current I
 & rooted_resistance R^{^0.5}

 & the proportionality constant
 K_i=F_ai/(P_e^{^0.5})=1*[N/(A*(Ω^{^0.5})]
 which is 1*[Newton/(Ampere*rooted_ohm)]
 (&) that (can be determined
 by) the average electromagnetic_force F_ai
 divided by the rooted electrical_power
 P_e^{^0.5}=P_iv^{^0.5}=(I*V)^{^0.5} . 

That current’s electromagnetic force
 F_ai=F_av
 is the same
 as the voltage's V=I*R (conversion
 from current I
 using the proportionality_constant R=Resistance,
 has the same)
 average electromagnetic_force
 F_av=V*K_i/(R^{^0.5}), K_v=1/K_i=1*[V*(Ω^{^0.5})/N]

 F_av=V/((R^{^0.5})*K_v).

Either (voltage or current)
 average
 electromagnetic_forces (F_ai=F_av, are equal)

 F_ai=(Pe^{^0.5})*K_i

 F_av=(Pe^{^0.5})*K_i

 & is proportional
 to the rooted electrical_power P_e^{^0.5}=(I*V)^{^0.5}

 & (multiplied
 by) the proportionality constant
 K_i=F_ai/(P_e^{^0.5})=1*[N/(A*(Ω^{^0.5})].

 

Electricity’s
 average
 (random=)heating force
 F_ah=(P_e^{^0.5})*K_i

 is also proportional
 to the rooted electrical_power P_e^{^0.5}=(I*V)^{^0.5}

 & the proportionality constant
 K_i=F_ai/(P_e^{^0.5})=1*[N/(A*(Ω^{^0.5})],
 as (a random (scattering of the (repeated))) recoil, (Newton’s 3^rd law opposite & equal) force (equivalent).

Together
 both the electromagnetic_force F_ai

 & the random heating_force F_ah
 amounts
 are equivalent to an (arbitrary) average_force F_a=F_ai=F_ah

 & (are “multiplied” together
 to) make up the total electrical_power

 P_e=F_ai*F_ah/(K_i^{^2}), where the inverse squared proportionality_constant is

(K_i^{^2})=1*[(N^{^2}/((A^{^2})*Ω)].

So,
 the electrical_power
 P_e=(F_a/K_i)^{^2}
 is a squared average Force F_a^{^2}
 per squared proportionality_constant K_i^{^2}.

What does all that mean?

That means
 the average_force
 F_a=mom_a/t
 is equivalent
 to an mechanical average_force.

---

(Please note:
 James Watt’s mechanical Power P=P_m definition
 P_a=F_a*v_a
 is an (algebraic) “average”_Power P_a
 using an “average” (steady_state) Force F_a=m*v_a/t
 of mass m
 multiplied by average linear_acceleration a_a=v_a/t (my a_a2 2^nd order of motion);
 & an average_speed v_a=d/t
 of distance d
 per time t. 

My “electrical” average_Power formula P_e=F_ai*F_ah/(K_i^{^2})
 Pe#Pm
 looks very different
 from James Watt’s “mechanical” average_Power P_m=P_a=F_a*v_a.

James’ (average) mechanical_Power
 is an (average_)Force F_a
 multiplied by a(n average)_speed V_a;
 while my (average) electrical_Power
 is basically a “squared” (average_)Force F_a
 but divided by a squared constant.

E.g. A(n average_)speed
 is (definitely) NOT a(n average_)force!

Disclaimer:

1.
But that is the ONLY way
 I know how to do it (=the conversion formulas), yet.

2. ..(because)..
That is what the math
 & its (observed) proportions tells me.
)

---

& that (average_Force F_a)
 is the rate
 of the average_momentum
 mom_a=m*v_a
 per time t;

 where the average_speed v_a=d/t
 is the distance d
 per time t,
 for a mass m. 

& that (average_Force)
 F_a=F_ai
 is equal
 to the current’s
 average electromagnetic_force F_ai.

& if you know the average force
 F_a=F_ah
 then you also (already) know the average (random=)heating_force F_ah
 & the average electromagnetic_force F_ai=F_av
 for either the current (thru a Resistor, Resistance)
 F_a=F_ai
 or for voltage (thru a Resistor, Resistance)
 F_a=F_av
 or both (I*V) when rooted.

Note:

The units
 for the proportionality constant
 K_i=k_i*[N*s/C]
 work out to be
 a(n average_)momentum [N*s]
 per Coulomb [C],
 e.g.
 [N*s/C]=[kg*m/(s*C)]=[(kg/C)*(m/s)]
 where a mass to charge ratio [kg/C]
 is obvious(ly),
 multiplied with an average_speed [m/s].

E.g.1
Let be given
 the voltage V=1*[V]
 the current I=1*[A]
 the Resistance R=1*[Ω]

 the rooted_Resistance R^{^0.5}=1*[Ω^{^0.5}]
 the electrical_power P_e=I*V=1*[A]*1*[V]=1*[A*V]=1 [electrical_Watt]
 the rooted electrical_power P_e^{^0.5}=P_iv^{^0.5}=(I*V)^{^0.5}=(1*[A]*1*[V])^{^0.5}=1*[A*V]^{^0.5} .

The average electromagnetic_force is
 F_ai=I*(R^{^0.5})*K_i=1*[A]*((1*[Ω])^{^0.5})*1*[N/(A*(Ω^{^0.5})]=1*[N].

The average heating_force is (also)
 F_ah=I*(R^{^0.5})*K_i=1*[A]*((1*[Ω])^{^0.5})*1*[N/(A*(Ω^{^0.5})]=1*[N].

That (easy example) doesN’T tell much,
 so let’s try something else.

E.g.2
Let be given
 the voltage V=100*[V]
 the current I=5*[A]
 the Resistance R=V/I=100*[V]/(5*[A])=20*[Ω]

 the rooted_Resistance R^{^0.5}=(V/I)^0.5=(100*[V]/(5*[A]))^{^0,5}=(20*[Ω])^{^0.5}=~4.7*[Ω^{^0.5}]
 the electrical_power P_e=I*V=5*[A]*100*[V]=500*[A*V]=500 [electrical_Watt]
 the rooted electrical_power P_e^{^0.5}=P_iv^{^05}=(I*V)^{^0.5}=(5*[A]*100*[V])^{^0.5}=(500*[A*V])^{^0.5}=22.4*([A*V]^{^0.5}).

The average electromagnetic_force is
 F_ai=I*(R^{^0.5})*K_i=5*[A]*((4.7*[Ω])^{^0.5})*1*[N/(A*(Ω^{^0.5})]=22.4*[N].

The average heating_force is (also)
 F_ah=I*(R^{^0.5})*K_i=5*[A]*((4.7*[Ω])^{^0.5})*1*[N/(A*(Ω^{^0.5})]=22.4*[N].

Or else simply let

 F_a=F_ai=F_ah
 F_a=(P_e^{^0.5})*K_i, P_e^{^0.5}=P_iv^0.5=(I*V)^{^0.5}

 F_a=((I*V)^{^0.5})*K_i.

Such (average_momentum electrical heating) will deliver
 mom_a=F_ah*t, let time t= 1*[second]
 mom_a=22.4*[N]*1*[s]

 mom_a=22.4*[N*s]
 in 1 second (of time).
 

In other words,
 the 500*[W] electrical (power’s)
 average heating_force
 F_ah=22.4*[N]

 will deliver 22.4*[N*s]
 of average_momentum (heat)
 per second.

1000*[W] electrical
 will deliver 31.6*[N*s]
 per second;

 &
 10*[kW] electrical
 will deliver 100*[N*s]
 per second.

Edited November 27, 20232 yr by Capiert