Conductance
Equation
En analogía a la conductividad que es el valor inverso de la resistividad con conductivity $1/Ohm m$ and resistivity $Ohm m$
$ \rho_e =\displaystyle\frac{1}{ \kappa_e } $ |
se puede definir la conductancia como el inverso de la resistencia con conductivity $1/Ohm m$ and resistivity $Ohm m$:
$ G =\displaystyle\frac{1}{ R }$ |
ID:(3847, 0)
Conductivity
Equation
Una vez se ha calculado la conductividad total sumando sobre las contribuciones de cada uno de los iones con concentration of ions i $mol/m^3$, conductivity $1/Ohm m$ and molar conductivity ions of type i $m^2/Ohm mol$
$ \kappa_e =\displaystyle\sum_i \Lambda_i c_i $ |
se puede calcular la resistividad con concentration of ions i $mol/m^3$, conductivity $1/Ohm m$ and molar conductivity ions of type i $m^2/Ohm mol$
$ \rho_e =\displaystyle\frac{1}{ \kappa_e } $ |
y con ello calcular la resistencia con conductor length $m$, resistance $Ohm$, resistivity $Ohm m$ and section of Conductors $m^2$
$ R = \rho_e \displaystyle\frac{ L }{ S }$ |
y aplicar la ley de Ohm tradicional con conductor length $m$, resistance $Ohm$, resistivity $Ohm m$ and section of Conductors $m^2$
$ \Delta\varphi_R = R I_R $ |
ID:(3848, 0)
Conductivity with an ion
Equation
Dado que con concentration of ions i $mol/m^3$, conductivity $1/Ohm m$ and molar conductivity ions of type i $m^2/Ohm mol$
$ \kappa_e =\displaystyle\sum_i \Lambda_i c_i $ |
se tiene que para el caso de un ion es con concentration of ions i $mol/m^3$, conductivity $1/Ohm m$ and molar conductivity ions of type i $m^2/Ohm mol$:
$ \kappa = \Lambda_1 c_1 $ |
ID:(3216, 0)
Current per conductor (classic model)
Equation
Si asumimos que la velocidad media es la mitad de la velocidad máxima es con ,
$ \bar{v} =\displaystyle\frac{1}{2} v_{max} $ |
El flujo total de cargas será con
$ I = e S c \bar{v} $ |
Con la expresión para la velocidad máxima con electric eield $V/m$, electron Charge $C$, mass of the electron $kg$, maximum Speed $m/s$ and time between collisions $s$
$ v_{max} =\displaystyle\frac{ e E }{ m } \tau $ |
se obtiene la expresión para la corriente con electric eield $V/m$, electron Charge $C$, mass of the electron $kg$, maximum Speed $m/s$ and time between collisions $s$
$ I =\displaystyle\frac{ e ^2 E }{2 m_e } \tau c S $ |
ID:(3837, 0)
Distance between ends of the conductor
Equation
La distancia entre los extremos del conductor, a lo largo de este, dan la distancia sobre la cual esta actuando la diferencia de potencial. Si los extremos se encuentran en
$ dx = x_2 - x_1 $ |
ID:(3846, 0)
Field in the conductor
Equation
Si la diferencia de potencial
$ \varphi =\varphi_0 - \displaystyle\int_C \vec{E}\cdot d\vec{s}$ |
donde
$ E =\displaystyle\frac{ \Delta\varphi }{ L }$ |
ID:(3838, 0)
Ion conductivity (2)
Equation
Given that
$ \kappa_e =\displaystyle\sum_i \Lambda_i c_i $ |
In the case of two types of ions, it is:
$ \kappa_e = \Lambda_1 c_1 + \Lambda_2 c_2 $ |
ID:(3850, 0)
Ion conductivity (3)
Equation
Given that
$ \kappa_e =\displaystyle\sum_i \Lambda_i c_i $ |
In the case of three types of ions, it is:
$ \kappa_e = \Lambda_1 c_1 + \Lambda_2 c_2 + \Lambda_3 c_3 $ |
ID:(3851, 0)
Ion conductivity (4)
Equation
Given that
$ \kappa_e =\displaystyle\sum_i \Lambda_i c_i $ |
In the case of two types of ions, it is:
$ \kappa_e = \Lambda_1 c_1 + \Lambda_2 c_2 + \Lambda_3 c_3 + \Lambda_4 c_4 $ |
ID:(3852, 0)
Ion conductivity (5)
Equation
Given that
$ \kappa_e =\displaystyle\sum_i \Lambda_i c_i $ |
In the case of two types of ions, it is:
$ \kappa_e = \Lambda_1 c_1 + \Lambda_2 c_2 + \Lambda_3 c_3 + \Lambda_4 c_4 + \Lambda_5 c_5 $ |
ID:(3853, 0)
Microscopic Ohm's law
Equation
Como la corriente en el conductor es con charge concentration $1/m^3$, current $A$, electric eield $V/m$, electron Charge $C$, mass of the electron $kg$, section of Conductors $m^2$ and time between collisions $s$
$ I =\displaystyle\frac{ e ^2 E }{2 m_e } \tau c S $ |
El campo eléctrico en el conductor es con conductor length $m$, electric field, infinite wire $V/m$ and potential difference $V$
$ E =\displaystyle\frac{ \Delta\varphi }{ L }$ |
\\n\\ncon
$I=\displaystyle\frac{e^2\tau c}{2m_e}\displaystyle\frac{S}{L}\Delta\varphi$
Esta expresión se puede despejar en función del potencial tomando la forma microscópica de la ley de Ohm con conductor length $m$, electric field, infinite wire $V/m$ and potential difference $V$:
$ \varphi =\displaystyle\frac{2 m_e }{ e ^2 \tau c }\displaystyle\frac{ L }{ S } I $ |
ID:(3839, 0)
Ohm's Law
Equation
La ecuación de Ohm microscópica es con charge concentration $1/m^3$, conductor length $m$, current $A$, electron Charge $C$, mass of the electron $kg$, potential difference $V$, section of Conductors $m^2$ and time between collisions $s$
$ \varphi =\displaystyle\frac{2 m_e }{ e ^2 \tau c }\displaystyle\frac{ L }{ S } I $ |
La definición de la resistividad con charge concentration $1/m^3$, electron Charge $C$, mass of the electron $kg$, resistivity $Ohm m$ and time between collisions $s$ es
$ \rho_e =\displaystyle\frac{2 m_e }{ e ^2 \tau c }$ |
y la resistencia con conductor length $m$, resistance $Ohm$, resistivity $Ohm m$ and section of Conductors $m^2$
$ R = \rho_e \displaystyle\frac{ L }{ S }$ |
se obtiene finalmente la ley de Ohm con conductor length $m$, resistance $Ohm$, resistivity $Ohm m$ and section of Conductors $m^2$
$ \Delta\varphi_R = R I_R $ |
ID:(3214, 0)
Potential energy difference
Equation
Si los extremos del conductor están a los potenciales
$ \Delta\varphi = \varphi_2 - \varphi_1 $ |
ID:(3845, 0)
Resistance
Equation
La ecuación de Ohm microscópica es con charge concentration $1/m^3$, conductor length $m$, current $A$, electron Charge $C$, mass of the electron $kg$, potential difference $V$, section of Conductors $m^2$ and time between collisions $s$
$ \varphi =\displaystyle\frac{2 m_e }{ e ^2 \tau c }\displaystyle\frac{ L }{ S } I $ |
El factor
$ \rho_e =\displaystyle\frac{2 m_e }{ e ^2 \tau c }$ |
y es parte, con la parte geométrica
$ R = \rho_e \displaystyle\frac{ L }{ S }$ |
ID:(3841, 0)
Resistance in the conductor
Equation
Los átomos del conductor representan obstáculos contra los que estos impactaran. Por este motivo, la aceleración generada por un campo
$ a =\displaystyle\frac{ e E }{ m_e }$ |
donde
Si el tiempo entre dos choques es
$ v_{max} =\displaystyle\frac{ e E }{ m } \tau $ |
ID:(3836, 0)
Series resistance
Equation
Al conectarse resistencias
$\Delta\varphi=\displaystyle\sum_i \Delta\varphi_i$
\\n\\nComo la corriente
$\Delta\varphi_i=R_i I$
\\n\\nSi se reemplaza esta expresión en la suma de las diferencias de potencial se obtiene\\n\\n
$\Delta\varphi=\displaystyle\sum_i R_iI$
por lo que la resistencia en serie se calcula como la suma de las resistencias individuales con :
$ R_s =\displaystyle\sum_ i R_i $ |
ID:(3215, 0)
Simple model for a conductor
Equation
Si suponemos que los electrones se pueden mover en un conductor, la aplicación de un campo eléctrico
$ F = q E $ |
Dada esta fuerza y la masa de los electrones
$ \vec{F} = m_i \vec{a} $ |
estimar la aceleración que estos experimentan con
$ a =\displaystyle\frac{ e E }{ m_e }$ |
ID:(3843, 0)
Specific resistivity
Equation
La ecuación de Ohm microscópica es con charge concentration $1/m^3$, conductor length $m$, current $A$, electron Charge $C$, mass of the electron $kg$, potential difference $V$, section of Conductors $m^2$ and time between collisions $s$
$ \varphi =\displaystyle\frac{2 m_e }{ e ^2 \tau c }\displaystyle\frac{ L }{ S } I $ |
El factor
$ \rho_e =\displaystyle\frac{2 m_e }{ e ^2 \tau c }$ |
ID:(3840, 0)
Sum of resistors in parallel (2)
Equation
Since the sum of resistors in parallel is
$\displaystyle\frac{1}{ R_p }=\displaystyle\sum_i\displaystyle\frac{1}{ R_i }$ |
You have to in the case of two resistors:
$\displaystyle\frac{1}{ R_p }=\displaystyle\frac{1}{ R_1 }+\displaystyle\frac{1}{ R_2 }$ |
ID:(3858, 0)
Sum of resistors in parallel (3)
Equation
Since the sum of resistors in parallel is
$\displaystyle\frac{1}{ R_p }=\displaystyle\sum_i\displaystyle\frac{1}{ R_i }$ |
You have to in the case of three resistors:
$\displaystyle\frac{1}{ R_{p3} }=\displaystyle\frac{1}{ R_{1p3} }+\displaystyle\frac{1}{ R_{2p3} }+\displaystyle\frac{1}{ R_{3p3} }$ |
ID:(3859, 0)
Sum of resistors in parallel (4)
Equation
Since the sum of resistors in parallel is
$\displaystyle\frac{1}{ R_p }=\displaystyle\sum_i\displaystyle\frac{1}{ R_i }$ |
it is necessary that in the case of four resistances:
$\displaystyle\frac{1}{ R_p }=\displaystyle\frac{1}{ R_1 }+\displaystyle\frac{1}{ R_2 }+\displaystyle\frac{1}{ R_3 }+\displaystyle\frac{1}{ R_4 }$ |
ID:(3860, 0)
Sum of resistors in parallel (5)
Equation
Since the sum of resistors in parallel is
$\displaystyle\frac{1}{ R_p }=\displaystyle\sum_i\displaystyle\frac{1}{ R_i }$ |
You have to in the case of five resistors:
$\displaystyle\frac{1}{ R_p }=\displaystyle\frac{1}{ R_1 }+\displaystyle\frac{1}{ R_2 }+\displaystyle\frac{1}{ R_3 }+\displaystyle\frac{1}{ R_4 }+\displaystyle\frac{1}{ R_5 }$ |
ID:(3861, 0)
Sum of resistors in series (2)
Equation
Since the sum of series resistors is
$ R_s =\displaystyle\sum_ i R_i $ |
You have to in the case of two resistors:
$ R_s = R_1 + R_2 $ |
ID:(3854, 0)
Sum of resistors in series (3)
Equation
Since the sum of series resistors is
$ R_s =\displaystyle\sum_ i R_i $ |
You have to in the case of three resistors:
$ R_s = R_1 + R_2 + R_3 $ |
ID:(3855, 0)
Sum of resistors in series (4)
Equation
Since the sum of series resistors is
$ R_s =\displaystyle\sum_ i R_i $ |
it is necessary that in the case of four resistances:
$ R_s = R_1 + R_2 + R_3 + R_4 $ |
ID:(3856, 0)
Sum of resistors in series (5)
Equation
Since the sum of series resistors is
$ R_s =\displaystyle\sum_ i R_i $ |
You have to in the case of five resistors:
$ R_s = R_1 + R_2 + R_3 + R_4 + R_5 $ |
ID:(3857, 0)
Total conductivity
Equation
Como la conductividad es proporcional a la concentración de los iones con
$ \kappa_i =\displaystyle\frac{ Q_i ^2 \tau_i }{2 m_i } c_i $ |
se puede definir una conductividad total como la suma de las conductividades de los distintos iones. Con la definición de la conductividad molar con
$ \Lambda_i =\displaystyle\frac{ Q_i ^2 \tau_i }{2 m_i } $ |
se tiene que con
$ \kappa_e =\displaystyle\sum_i \Lambda_i c_i $ |
ID:(3849, 0)