Corrientes en el Cuerpo
Description 
Variables
Calculations
Calculations
Equations
None
(ID 3214)
None
(ID 3215)
None
(ID 3843)
(ID 3846)
None
(ID 3856)
Examples
The electric eield ($E$), together with the electron Charge ($e$), generates a force that, through the mass of the electron ($m_e$), results in the acceleration of charge in the conductor ($a$). This relationship can be expressed as:
| $ a =\displaystyle\frac{ e E }{ m_e }$ |
(ID 3843)
The conductance ($G$) is defined as the inverse of the resistance ($R$). This relationship is expressed as:
| $ G =\displaystyle\frac{1}{ R }$ |
(ID 3847)
The resistivity ($\rho_e$) is defined as the inverse of the conductivity ($\kappa_e$). This relationship is expressed as:
| $ \rho_e =\displaystyle\frac{1}{ \kappa_e } $ |
(ID 3848)
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)
The current ($I$) can be calculated from the electric eield ($E$), in combination with the electron Charge ($e$), the charge concentration ($c$), the mass of the electron ($m_e$), the time between collisions ($\tau$), and the section of Conductors ($S$), using the following relationship:
| $ I =\displaystyle\frac{ e ^2 E }{2 m_e } \tau c S $ |
(ID 3837)
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)
The electric eield ($E$) is generated by the potential difference ($\Delta\varphi$) between two electrodes, separated by a distance of a conductor length ($L$). This value can be calculated using the following expression:
| $ E =\displaystyle\frac{ \Delta\varphi }{ L }$ |
(ID 3838)
The parallel connection of the hydraulic conductance 1 ($G_{h1}$), and the hydraulic conductance 2 ($G_{h2}$) results in an equivalent combination of the parallel total hydraulic conductance ($G_{pt}$):
| $ G_{pt} = G_{h1} + G_{h2} $ |
(ID 3856)
The parallel connection of the hydraulic conductance 1 ($G_{h1}$), the hydraulic conductance 2 ($G_{h2}$), and the hydraulic conductance 3 ($G_{h3}$) results in an equivalent combination of the parallel total hydraulic conductance ($G_{pt}$):
| $ G_{pt} = G_{h1} + G_{h2} + G_{h3} $ |
(ID 3857)
The series combination of the hydraulic conductance 1 ($G_{h1}$) and the hydraulic conductance 2 ($G_{h2}$) results in a total sum of the total Series Hydraulic Conductance ($G_{st}$):
| $\displaystyle\frac{1}{ G_{st} }=\displaystyle\frac{1}{ G_{h1} }+\displaystyle\frac{1}{ G_{h2} }$ |
(ID 3860)
The series combination of the hydraulic conductance 1 ($G_{h1}$), the hydraulic conductance 2 ($G_{h2}$) and the hydraulic conductance 3 ($G_{h3}$) results in a total sum of the total Series Hydraulic Conductance ($G_{st}$):
| $\displaystyle\frac{1}{ G_{st} }=\displaystyle\frac{1}{ G_{h1} }+\displaystyle\frac{1}{ G_{h2} }+\displaystyle\frac{1}{ G_{h3} }$ |
(ID 3861)
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)
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)
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)
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)
In a time between collisions ($\tau$), the electron is accelerated by the electric eield ($E$), in combination with the electron Charge ($e$) and the mass of the electron ($m_e$), until it reaches the maximum Speed ($v_{max}$). This process is described by the following relationship:
| $ v_{max} =\displaystyle\frac{ e E }{ m_e } \tau $ |
(ID 3836)
If the current ($I$) is expressed using the potential difference ($\Delta\varphi$) instead of the electric eield ($E$), the microscopic form of Ohm's law is obtained. This equation involves the electron Charge ($e$), the charge concentration ($c$), the mass of the electron ($m_e$), the time between collisions ($\tau$), the section of Conductors ($S$), and the conductor length ($L$), through the following relationship:
| $ \Delta\varphi =\displaystyle\frac{2 m_e }{ e ^2 \tau c }\displaystyle\frac{ L }{ S } I $ |
(ID 3839)
Traditional Ohm's law establishes a relationship between the potential difference ($\Delta\varphi$) and the current ($I$) through the resistance ($R$), using the following expression:
| $ \Delta\varphi = R I $ |
(ID 3214)
Si los extremos del conductor est n a los potenciales
| $ \Delta\varphi = \varphi_2 - \varphi_1 $ |
(ID 3845)
Using the resistivity ($\rho_e$) along with the geometric parameters the conductor length ($L$) and the section of Conductors ($S$), the resistance ($R$) can be defined through the following relationship:
| $ R = \rho_e \displaystyle\frac{ L }{ S }$ |
(ID 3841)
Al conectarse resistencias
$\Delta\varphi=\displaystyle\sum_i \Delta\varphi_i$
Como la corriente
$\Delta\varphi_i=R_i I$
Si se reemplaza esta expresi n en la suma de las diferencias de potencial se obtiene
$\Delta\varphi=\displaystyle\sum_i R_iI$
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)
From the microscopic form of Ohm's law, a factor specific to the material of the conductor can be identified. This allows the resistivity ($\rho_e$) to be defined in terms of the electron Charge ($e$), the charge concentration ($c$), the mass of the electron ($m_e$), and the time between collisions ($\tau$), using the following relationship:
| $ \rho_e =\displaystyle\frac{2 m_e }{ e ^2 \tau c }$ |
(ID 3840)
The parallel combination of the hydraulic Resistance 1 ($R_{h1}$) and the hydraulic Resistance 2 ($R_{h2}$) results in a total equivalent of the total hydraulic resistance in series ($R_{st}$):
| $\displaystyle\frac{1}{ R_{pt} }=\displaystyle\frac{1}{ R_{h1} }+\displaystyle\frac{1}{ R_{h2} }$ |
(ID 3858)
The parallel combination of the hydraulic Resistance 1 ($R_{h1}$), the hydraulic Resistance 2 ($R_{h2}$), and the hydraulic Resistance 3 ($R_{h3}$) results in a total equivalent of the total hydraulic resistance in series ($R_{st}$):
| $\displaystyle\frac{1}{ R_{pt} }=\displaystyle\frac{1}{ R_{h1} }+\displaystyle\frac{1}{ R_{h2} }+\displaystyle\frac{1}{ R_{h3} }$ |
(ID 3859)
The series combination of the hydraulic Resistance 1 ($R_{h1}$) and the hydraulic Resistance 2 ($R_{h2}$) results in a total sum of the total hydraulic resistance in series ($R_{st}$):
| $ R_{st} = R_{h1} + R_{h2} $ |
(ID 3854)
The series combination of the hydraulic Resistance 1 ($R_{h1}$), the hydraulic Resistance 2 ($R_{h2}$) and the hydraulic Resistance 3 ($R_{h3}$) results in a total sum of the total hydraulic resistance in series ($R_{st}$):
| $ R_{st} = R_{h1} + R_{h2} + R_{h3} $ |
(ID 3855)
Como la conductividad es proporcional a la concentraci n de los iones
| $ \kappa_i = \Lambda_i c_i $ |
se puede definir una conductividad total como la suma de las conductividades de los distintos iones. Con la definici n de la conductividad molar
| $ \Lambda_i =\displaystyle\frac{ Q_i ^2 \tau_i }{2 m_i } $ |
se tiene que
| $ \kappa_e =\displaystyle\sum_i \Lambda_i c_i $ |
(ID 3849)
(ID 772)
ID:(333, 0)
