Bioelectricidad y Biomagnetismo

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Contenido Corriente el´ectrica

Bioelectricidad y Biomagnetismo

Universidad Antonio Nari˜ no

23 de abril de 2014

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Contenido

1 Corriente el´ ectrica

Resistencia y ley de Ohm

Circuitos el´ ectricos y leyes de Kirchhoff

Corrientes en Membranas El´ ectricas, Circuito RC

Ecuaci´ on de Nerst

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Andr´ e-Marie Amp` ere

Ampere

Figura: Andr´ e-Marie Amp` ere (Enero 20 de 1775, Junio 10 de 1836)

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Resistencia y ley de Ohm

Circuitos eléctricos y leyes de Kirchhoff Corrientes en Membranas Eléctricas, Circuito RC Ecuación de Nerst

Corriente el´ ectrica y resistencia

Corriente el´ ectrica

I = ∆Q

∆t [I ] = C

s = A (Amperios)

El sentido de la corriente en un conductor se asume en la direcci´ on contraria al flujo de electrones. Ej.: A trav´ es de un alambre conductor fluye una cantidad de electrones que portan una carga de 10

⁻⁸

C en un tiempo t = 1 µ s. La corriente es:

I = Q

t = 10

⁻⁸

C

10

⁻⁶

s = 10

⁻²

A

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Corriente el´ ectrica y resistencia

Corriente el´ ectrica

I = ∆Q

∆t [I ] = C

s = A (Amperios)

El sentido de la corriente en un conductor se asume en la direcci´ on contraria al flujo de electrones. Ej.: A trav´ es de un alambre conductor fluye una cantidad de electrones que portan una carga de 10

⁻⁸

C en un tiempo t = 1 µ s. La corriente es:

I = Q

t = 10

⁻⁸

C

10

⁻⁶

s = 10

⁻²

A

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Corriente el´ ectrica y resistencia

Corriente el´ ectrica

I = ∆Q

∆t [I ] = C

s = A (Amperios)

El sentido de la corriente en un conductor se asume en la direcci´ on contraria al flujo de electrones.

Ej.: A trav´ es de un alambre conductor fluye una cantidad de electrones que portan una carga de 10

⁻⁸

C en un tiempo t = 1 µ s. La corriente es:

I = Q

t = 10

⁻⁸

C

10

⁻⁶

s = 10

⁻²

A

(7)

Corriente el´ ectrica y resistencia

Corriente el´ ectrica

I = ∆Q

∆t [I ] = C

s = A (Amperios)

El sentido de la corriente en un conductor se asume en la direcci´ on contraria al flujo de electrones. Ej.: A trav´ es de un alambre conductor fluye una cantidad de electrones que portan una carga de 10

⁻⁸

C en un tiempo t = 1 µ s. La corriente es:

I = Q

t = 10

⁻⁸

C

10

⁻⁶

s = 10

⁻²

A

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Corriente el´ ectrica y resistencia

Resistencia y ley de Ohm.

V = IR ; R : resistencia ; [R] = V

A = Ω (Ohmios)

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Corriente el´ ectrica y resistencia

Cargas en un campo el´ ectrico

Electr´ on en un campo el´ ectrico constante

a = F m = eE

m ⇒ v

_deriva

= aτ = eE m τ = eV

mL τ τ → tiempo promedio entre colisiones, L → distancia promedio recorrida.

I = ∆Q

∆t = (n e)(A l ) l /v

deriva

= (n e)A v

deriva

= (n e

²

)Aτ V m L σ = n e

²

τ

m Conductividad. ⇒ I = σ A

L V = GV (G : Conductancia) V = IR (R : Resistencia) R = 1

G = 1 σ

L A = ρ L

A (ρ : Resistividad)

Nota. La conductividad y la resistividad dependen de caracter´ısticas intr´ınsecas del

material, la conductancia y la resistencia combinan factores intr´ınsecos con factores

geom´ etricos.

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Corriente el´ ectrica y resistencia

Cargas en un campo el´ ectrico

Electr´ on en un campo el´ ectrico constante

a = F m = eE

m

⇒ v

_deriva

= aτ = eE m τ = eV

mL τ τ → tiempo promedio entre colisiones, L → distancia promedio recorrida.

I = ∆Q

∆t = (n e)(A l ) l /v

deriva

= (n e)A v

deriva

= (n e

²

)Aτ V m L σ = n e

²

τ

m Conductividad. ⇒ I = σ A

L V = GV (G : Conductancia) V = IR (R : Resistencia) R = 1

G = 1 σ

L A = ρ L

A (ρ : Resistividad)

Nota. La conductividad y la resistividad dependen de caracter´ısticas intr´ınsecas del

material, la conductancia y la resistencia combinan factores intr´ınsecos con factores

geom´ etricos.

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Corriente el´ ectrica y resistencia

Cargas en un campo el´ ectrico

Electr´ on en un campo el´ ectrico constante

a = F m = eE

m ⇒ v

_deriva

= aτ = eE m τ

= eV mL τ τ → tiempo promedio entre colisiones, L → distancia promedio recorrida.

I = ∆Q

∆t = (n e)(A l ) l /v

deriva

= (n e)A v

deriva

= (n e

²

)Aτ V m L σ = n e

²

τ

m Conductividad. ⇒ I = σ A

L V = GV (G : Conductancia) V = IR (R : Resistencia) R = 1

G = 1 σ

L A = ρ L

A (ρ : Resistividad)

Nota. La conductividad y la resistividad dependen de caracter´ısticas intr´ınsecas del

material, la conductancia y la resistencia combinan factores intr´ınsecos con factores

geom´ etricos.

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Corriente el´ ectrica y resistencia

Cargas en un campo el´ ectrico

Electr´ on en un campo el´ ectrico constante

a = F m = eE

m ⇒ v

_deriva

= aτ = eE m τ = eV

mL τ

τ → tiempo promedio entre colisiones, L → distancia promedio recorrida.

I = ∆Q

∆t = (n e)(A l ) l /v

deriva

= (n e)A v

deriva

= (n e

²

)Aτ V m L σ = n e

²

τ

m Conductividad. ⇒ I = σ A

L V = GV (G : Conductancia) V = IR (R : Resistencia) R = 1

G = 1 σ

L A = ρ L

A (ρ : Resistividad)

Nota. La conductividad y la resistividad dependen de caracter´ısticas intr´ınsecas del

material, la conductancia y la resistencia combinan factores intr´ınsecos con factores

geom´ etricos.

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Corriente el´ ectrica y resistencia

Cargas en un campo el´ ectrico

Electr´ on en un campo el´ ectrico constante

a = F m = eE

m ⇒ v

_deriva

= aτ = eE m τ = eV

mL τ τ → tiempo promedio entre colisiones, L → distancia promedio recorrida.

I = ∆Q

∆t = (n e)(A l ) l /v

deriva

= (n e)A v

deriva

= (n e

²

)Aτ V m L σ = n e

²

τ

m Conductividad. ⇒ I = σ A

L V = GV (G : Conductancia) V = IR (R : Resistencia) R = 1

G = 1 σ

L A = ρ L

A (ρ : Resistividad)

Nota. La conductividad y la resistividad dependen de caracter´ısticas intr´ınsecas del

material, la conductancia y la resistencia combinan factores intr´ınsecos con factores

geom´ etricos.

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Corriente el´ ectrica y resistencia

Cargas en un campo el´ ectrico

Electr´ on en un campo el´ ectrico constante

a = F m = eE

m ⇒ v

_deriva

= aτ = eE m τ = eV

mL τ τ → tiempo promedio entre colisiones, L → distancia promedio recorrida.

I = ∆Q

∆t = (n e)(A l ) l /v

deriva

= (n e)A v

deriva

= (n e

²

)Aτ V m L

σ = n e

²

τ

m Conductividad. ⇒ I = σ A

L V = GV (G : Conductancia) V = IR (R : Resistencia) R = 1

G = 1 σ

L A = ρ L

A (ρ : Resistividad)

Nota. La conductividad y la resistividad dependen de caracter´ısticas intr´ınsecas del

material, la conductancia y la resistencia combinan factores intr´ınsecos con factores

geom´ etricos.

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Corriente el´ ectrica y resistencia

Cargas en un campo el´ ectrico

Electr´ on en un campo el´ ectrico constante

a = F m = eE

m ⇒ v

_deriva

= aτ = eE m τ = eV

mL τ τ → tiempo promedio entre colisiones, L → distancia promedio recorrida.

I = ∆Q

∆t = (n e)(A l ) l /v

deriva

= (n e)A v

deriva

= (n e

²

)Aτ V m L σ = n e

²

τ

m Conductividad.

⇒ I = σ A

L V = GV (G : Conductancia) V = IR (R : Resistencia) R = 1

G = 1 σ

L A = ρ L

A (ρ : Resistividad)

Nota. La conductividad y la resistividad dependen de caracter´ısticas intr´ınsecas del

material, la conductancia y la resistencia combinan factores intr´ınsecos con factores

geom´ etricos.

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Corriente el´ ectrica y resistencia

Cargas en un campo el´ ectrico

Electr´ on en un campo el´ ectrico constante

a = F m = eE

m ⇒ v

_deriva

= aτ = eE m τ = eV

mL τ τ → tiempo promedio entre colisiones, L → distancia promedio recorrida.

I = ∆Q

∆t = (n e)(A l ) l /v

deriva

= (n e)A v

deriva

= (n e

²

)Aτ V m L σ = n e

²

τ

m Conductividad. ⇒ I = σ A

L V = GV (G : Conductancia)

V = IR (R : Resistencia) R = 1 G = 1

σ L A = ρ L

A (ρ : Resistividad)

Nota. La conductividad y la resistividad dependen de caracter´ısticas intr´ınsecas del

material, la conductancia y la resistencia combinan factores intr´ınsecos con factores

geom´ etricos.

(17)

Corriente el´ ectrica y resistencia

Cargas en un campo el´ ectrico

Electr´ on en un campo el´ ectrico constante

a = F m = eE

m ⇒ v

_deriva

= aτ = eE m τ = eV

mL τ τ → tiempo promedio entre colisiones, L → distancia promedio recorrida.

I = ∆Q

∆t = (n e)(A l ) l /v

deriva

= (n e)A v

deriva

= (n e

²

)Aτ V m L σ = n e

²

τ

m Conductividad. ⇒ I = σ A

L V = GV (G : Conductancia) V = IR (R : Resistencia) R = 1

G = 1 σ

L A = ρ L

A (ρ : Resistividad)

Nota. La conductividad y la resistividad dependen de caracter´ısticas intr´ınsecas del

material, la conductancia y la resistencia combinan factores intr´ınsecos con factores

geom´ etricos.

(18)

Corriente el´ ectrica y resistencia

Cargas en un campo el´ ectrico

Electr´ on en un campo el´ ectrico constante

a = F m = eE

m ⇒ v

_deriva

= aτ = eE m τ = eV

mL τ τ → tiempo promedio entre colisiones, L → distancia promedio recorrida.

I = ∆Q

∆t = (n e)(A l ) l /v

deriva

= (n e)A v

deriva

= (n e

²

)Aτ V m L σ = n e

²

τ

m Conductividad. ⇒ I = σ A

L V = GV (G : Conductancia) V = IR (R : Resistencia) R = 1

G = 1 σ

L A = ρ L

A (ρ : Resistividad)

Nota. La conductividad y la resistividad dependen de caracter´ısticas intr´ınsecas del

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Corriente el´ ectrica y resistencia

Resistividad de algunos materiales

The SI unit for resistance is the ohm (!), where 1 V/A "

1 ! (read as 1 ohm). Units for resistivity are then given as

!-m and for conductivity as (!-m)^#1. The unit for conduc- tance, the reciprocal of resistance, is the !^#1which is also known as the siemens (S). Table 16.1 lists some values for resistivity of various materials. A wire made from a metal will have a very low resistance value. For example, a 1 m length of 1 mm diameter copper wire has a resistance of only 0.02 !.

Simple devices known as resistors (shown in Figure 16.5) are manufactured to have various resistance values. The symbol is used to represent a resistor in a schematic or circuit diagram such as the one shown in Figure 16.6. Connecting wires have negligible resistance, so that their length and shape are usually not important in a circuit diagram or in the actual circuit itself.

Table 16.1Resistivities of Various Materials (20°C) Material Resistivity, $ (! .. m) Conductors

Aluminum 2.8 % 10#8

Copper 1.7 % 10#8

Iron 10. % 10^#8

Mercury 96. % 10^#8

Silver 1.6 % 10^#8

Tungsten 5.6 % 10#8

Ionic materials

Water (distilled) ~2 % 105

Fresh water ~5 % 10²

Sea water ~0.3

Cytoplasm ~0.5

Fatty tissue ~15

Semiconductors

Germanium ~0.5

Silicon ~2. % 10³

Insulators

Air (dry) 4 % 10¹³

Glass 10¹⁰#10¹⁴

Rubber 1013#1016

Example 16.1How much electric current flows through water contained in an insulating tube 10 cm long and 5 cm in diameter when a 100 V potential differ- ence is applied across the ends of the tube using electrodes inserted at either end? Ignore any complications from the metal electrode–water contact and do the calculation using the three entries in Table 16.1 for different purities of water.

FIGURE 16.5An assortment of resistors.

R₁

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Corriente el´ ectrica y resistencia

Ejemplo. Calcular la corriente el´ ectrica a trav´ es del agua destilada contenida en un tubo aislante de 10 cm de longitud y 5 cm de di´ ametro cuando se establece una diferencia de potencial (voltaje) de 100 V entre los extremos del tubo.

Sol. El tubo tiene forma cil´ındrica con ´ area trasversal A = πr

²

= π(0,05/2m)

²

. La resistencia es entonces

(recuerden, factores intr´ınsecos y factores geom´ etricos!):

R = ρL

A = ρ 0,1m

19,6 × 10

⁻⁴

m

²

= 51 ρ m

⁻¹

R = 2 × 10

⁵

Ω.m × 51m

⁻¹

= 102 × 10

⁵

Ω = 1,02 × 10

⁷

Ω ' 10

⁷

Ω

I = V

R = 100V

10

⁷

Ω = 10

⁻⁵

A = 10 × 10

⁻⁶

A = 10µA

(21)

Corriente el´ ectrica y resistencia

Ejemplo. Calcular la corriente el´ ectrica a trav´ es del agua destilada contenida en un tubo aislante de 10 cm de longitud y 5 cm de di´ ametro cuando se establece una diferencia de potencial (voltaje) de 100 V entre los extremos del tubo.

Sol. El tubo tiene forma cil´ındrica con ´ area trasversal A = πr

²

= π(0,05/2m)

²

. La resistencia es entonces (recuerden, factores intr´ınsecos y factores geom´ etricos!):

R = ρL

A = ρ 0,1m

19,6 × 10

⁻⁴

m

²

= 51 ρ m

⁻¹

R = 2 × 10

⁵

Ω.m × 51m

⁻¹

= 102 × 10

⁵

Ω = 1,02 × 10

⁷

Ω ' 10

⁷

Ω

I = V

R = 100V

10

⁷

Ω = 10

⁻⁵

A = 10 × 10

⁻⁶

A = 10µA

(22)

Corriente el´ ectrica y resistencia

Ejemplo. Calcular la corriente el´ ectrica a trav´ es del agua destilada contenida en un tubo aislante de 10 cm de longitud y 5 cm de di´ ametro cuando se establece una diferencia de potencial (voltaje) de 100 V entre los extremos del tubo.

Sol. El tubo tiene forma cil´ındrica con ´ area trasversal A = πr

²

= π(0,05/2m)

²

. La resistencia es entonces (recuerden, factores intr´ınsecos y factores geom´ etricos!):

R = ρL

A = ρ 0,1m

19,6 × 10

⁻⁴

m

²

= 51 ρ m

⁻¹

R = 2 × 10

⁵

Ω.m × 51m

⁻¹

= 102 × 10

⁵

Ω = 1,02 × 10

⁷

Ω ' 10

⁷

Ω

I = V

R = 100V

10

⁷

Ω = 10

⁻⁵

A = 10 × 10

⁻⁶

A = 10µA

(23)

Corriente el´ ectrica y resistencia

Ejemplo. Calcular la corriente el´ ectrica a trav´ es del agua destilada contenida en un tubo aislante de 10 cm de longitud y 5 cm de di´ ametro cuando se establece una diferencia de potencial (voltaje) de 100 V entre los extremos del tubo.

Sol. El tubo tiene forma cil´ındrica con ´ area trasversal A = πr

²

= π(0,05/2m)

²

. La resistencia es entonces (recuerden, factores intr´ınsecos y factores geom´ etricos!):

R = ρL

A = ρ 0,1m

19,6 × 10

⁻⁴

m

²

= 51 ρ m

⁻¹

R = 2 × 10

⁵

Ω.m × 51m

⁻¹

= 102 × 10

⁵

Ω = 1,02 × 10

⁷

Ω ' 10

⁷

Ω

I = V

R = 100V

10

⁷

Ω = 10

⁻⁵

A = 10 × 10

⁻⁶

A = 10µA

(24)

Corriente el´ ectrica y resistencia

Ejemplo. Calcular la corriente el´ ectrica a trav´ es del agua destilada contenida en un tubo aislante de 10 cm de longitud y 5 cm de di´ ametro cuando se establece una diferencia de potencial (voltaje) de 100 V entre los extremos del tubo.

Sol. El tubo tiene forma cil´ındrica con ´ area trasversal A = πr

²

= π(0,05/2m)

²

. La resistencia es entonces (recuerden, factores intr´ınsecos y factores geom´ etricos!):

R = ρL

A = ρ 0,1m

19,6 × 10

⁻⁴

m

²

= 51 ρ m

⁻¹

R = 2 × 10

⁵

Ω.m × 51m

⁻¹

= 102 × 10

⁵

Ω = 1,02 × 10

⁷

Ω ' 10

⁷

Ω

I = V

R = 100V

10

⁷

Ω = 10

⁻⁵

A

= 10 × 10

⁻⁶

A = 10µA

(25)

Corriente el´ ectrica y resistencia

Ejemplo. Calcular la corriente el´ ectrica a trav´ es del agua destilada contenida en un tubo aislante de 10 cm de longitud y 5 cm de di´ ametro cuando se establece una diferencia de potencial (voltaje) de 100 V entre los extremos del tubo.

Sol. El tubo tiene forma cil´ındrica con ´ area trasversal A = πr

²

= π(0,05/2m)

²

. La resistencia es entonces (recuerden, factores intr´ınsecos y factores geom´ etricos!):

R = ρL

A = ρ 0,1m

19,6 × 10

⁻⁴

m

²

= 51 ρ m

⁻¹

R = 2 × 10

⁵

Ω.m × 51m

⁻¹

= 102 × 10

⁵

Ω = 1,02 × 10

⁷

Ω ' 10

⁷

Ω

I = V

R = 100V

10

⁷

Ω = 10

⁻⁵

A = 10 × 10

⁻⁶

A = 10µA

(26)

Corriente el´ ectrica y resistencia

Ejemplo. Capacitancia y resistencia de un ax´ on. Calcular la capacitancia y la resistencia en un ax´ on considerando que tiene la geometr´ıa de un cable como el de la siguiente figura:

a d

L

Sol. La capacitancia y la resistencia del ax´ on son respectivamente:

C = κ

0

A

d = κ

0

2πaL

d ; R = ρd

A = ρd 2πaL

Asuma los valores dados en los ejercicios del taller. Ax´ on sin mielina: κ = 7, L = 1 m ,

a = 2,5µ m, d = 10 nm, V = 70 mV. Secci´ on del ax´ on con mielina: D = 1,4 mm

d = 2µ m, ρ = 1,6 × 10

⁷

Ω.m.

(27)

Corriente el´ ectrica y resistencia

Ejemplo. Capacitancia y resistencia de un ax´ on. Calcular la capacitancia y la resistencia en un ax´ on considerando que tiene la geometr´ıa de un cable como el de la siguiente figura:

a d

L

Sol. La capacitancia y la resistencia del ax´ on son respectivamente:

C = κ

0

A

d = κ

0

2πaL

d ; R = ρd

A = ρd 2πaL

Asuma los valores dados en los ejercicios del taller. Ax´ on sin mielina: κ = 7, L = 1 m ,

a = 2,5µ m, d = 10 nm, V = 70 mV. Secci´ on del ax´ on con mielina: D = 1,4 mm

d = 2µ m, ρ = 1,6 × 10

⁷

Ω.m.

(28)

Corriente el´ ectrica y resistencia

Ejemplo. Capacitancia y resistencia de un ax´ on. Calcular la capacitancia y la resistencia en un ax´ on considerando que tiene la geometr´ıa de un cable como el de la siguiente figura:

a d

L

Sol. La capacitancia y la resistencia del ax´ on son respectivamente:

C = κ

0

A

d = κ

0

2πaL

d ; R = ρd

A = ρd

2πaL

(29)

Resistencias en serie y en paralelo

Resistencias en serie Resistencias

a trav´ es de las cuales circula la misma corriente V = IR

1

+ IR

2

+ · · · = IR

eq

⇒ R

_eq

= R

1

+ R

2

+ · · · Resistencias en paralelo

Resistencias

sometidas a la misma diferencia de potencial

I

1

+I

2

+· · · = V R

1

+ V R

2

+· · · = V R

eq

⇒ 1

R

eq

= 1 R

1

+ 1 R

2

+· · ·

(30)

Resistencias en serie y en paralelo

Resistencias en serie Resistencias

a trav´ es de las cuales circula la misma corriente V = IR

1

+ IR

2

+ · · · = IR

eq

⇒ R

_eq

= R

1

+ R

2

+ · · · Resistencias en paralelo

Resistencias

sometidas a la misma diferencia de potencial

I

1

+I

2

+· · · = V R

1

+ V R

2

+· · · = V R

eq

⇒ 1

R

eq

= 1 R

1

+ 1 R

2

+· · ·

(31)

Leyes (reglas) de Kirchhoff

Ley de nudos La corriente

neta en un nudo del circuito es nula.

n

X

i =1

I

i

= 0

Se sigue la siguiente convenci´ on: Las corrientes

que entran al nudo son negativas. Las corrientes

que salen del nudo son positivas. Conservaci´ on de la carga!

Ejemplo La suma de las corrientes en el nudo de la figura es:

i

1

+ i

4

− i

2

− i

3

= 0 ⇒ i

1

+ i

4

= i

2

+ i

3

(32)

Leyes (reglas) de Kirchhoff

Ley de nudos La corriente

neta en un nudo del circuito es nula.

n

X

i =1

I

i

= 0

Se sigue la siguiente convenci´ on:

Las corrientes

que entran al nudo son negativas.

Las corrientes

que salen del nudo son positivas.

Conservaci´ on de la carga!

Ejemplo La suma de las corrientes en el nudo de la figura es:

i

1

+ i

4

− i

2

− i

3

= 0 ⇒ i

1

+ i

4

= i

2

+ i

3

(33)

Leyes (reglas) de Kirchhoff

Ley de nudos La corriente

neta en un nudo del circuito es nula.

n

X

i =1

I

i

= 0

Se sigue la siguiente convenci´ on:

Las corrientes

que entran al nudo son negativas.

Las corrientes

que salen del nudo son positivas.

Conservaci´ on de la carga!

Ejemplo La suma de las corrientes en el nudo de la figura es:

i

1

+ i

4

− i

2

− i

3

= 0 ⇒ i

1

+ i

4

= i

2

+ i

3

(34)

Leyes (reglas) de Kirchhoff

Ley de nudos La corriente

neta en un nudo del circuito es nula.

n

X

i =1

I

i

= 0

Se sigue la siguiente convenci´ on:

Las corrientes

que entran al nudo son negativas.

Las corrientes

que salen del nudo son positivas.

Conservaci´ on de la carga!

Ejemplo La suma de las corrientes en el nudo de la figura es:

i

1

+ i

4

− i

2

− i

3

= 0

⇒ i

1

+ i

4

= i

2

+ i

3

(35)

Leyes (reglas) de Kirchhoff

Ley de nudos La corriente

neta en un nudo del circuito es nula.

n

X

i =1

I

i

= 0

Se sigue la siguiente convenci´ on:

Las corrientes

que entran al nudo son negativas.

Las corrientes

que salen del nudo son positivas.

Conservaci´ on de la carga!

Ejemplo La suma de las corrientes en el nudo de la figura es:

i

1

+ i

4

− i

2

− i

3

= 0 ⇒ i

1

+ i

4

= i

2

+ i

3

(36)

Leyes (reglas) de Kirchhoff

Ley de mallas (lazos)

Figura: Suma de los voltajes en el lazo

V

1

+ V

2

+ V

3

− V

₄

= 0 La diferencia de potencial

(caida de voltaje) en un lazo del circuito es nula.

n

X

i =1

V

i

= 0

Se sigue la siguiente convenci´ on: La diferencia de potencial

a trav´ es de una resistencia es negativa

(positiva) si la atravezamos en el mismo

sentido (sentido contrario) de la corriente.

La diferencia de potencial a trav´ es de una

bater´ıa (fuerza electromotriz: fem) es positiva

(negativa) cuando vamos en el mismo sentido

(sentido contrario) al que actua la bater´ıa.

Conservaci´ on de la energ´ıa!

(37)

Leyes (reglas) de Kirchhoff

Ley de mallas (lazos)

Figura: Suma de los voltajes en el lazo

V

1

+ V

2

+ V

3

− V

₄

= 0 La diferencia de potencial

(caida de voltaje) en un lazo del circuito es nula.

n

X

i =1

V

i

= 0

Se sigue la siguiente convenci´ on:

La diferencia de potencial

a trav´ es de una resistencia es negativa (positiva) si la atravezamos en el mismo sentido (sentido contrario) de la corriente.

La diferencia de potencial a trav´ es de una

bater´ıa (fuerza electromotriz: fem) es positiva

(negativa) cuando vamos en el mismo sentido

(sentido contrario) al que actua la bater´ıa.

(38)

Leyes (reglas) de Kirchhoff

Ejemplo

Nudo A: −I

1

+ I

2

+ I

3

= 0; Malla 1: −I

2

R

2

+ I

3

R

3

+ I

4

R

4

− V

ε2

= 0

Nudo B: −I

₃

+ I

4

+ I

5

= 0; Malla 2: I

5

R

5

− I

₆

R

6

− I

₄

R

4

= 0

Nudo C: −I

₂

− I

₄

+ I

6

= 0; Malla 3: I

1

R

1

+ I

2

R

2

+ I

6

R

6

− V

_ε1

+ V

ε2

= 0

(39)

Leyes (reglas) de Kirchhoff

Ejemplo

Nudo A: −I

₁

+ I

2

+ I

3

= 0; Malla 1: −I

₂

R

2

+ I

3

R

3

+ I

4

R

4

− V

_ε2

= 0

Nudo B: −I

₃

+ I

4

+ I

5

= 0; Malla 2: I

5

R

5

− I

₆

R

6

− I

₄

R

4

= 0

Nudo C: −I

₂

− I

₄

+ I

6

= 0; Malla 3: I

1

R

1

+ I

2

R

2

+ I

6

R

6

− V

_ε1

+ V

ε2

= 0

(40)

Potencia El´ ectrica

Potencia

La potencia se define como la tasa de gasto de la energ´ıa por unidad de tiempo

P ≡ ∆U

∆t En el caso de la energ´ıa el´ ectrica U = q V

P = V ∆q

∆t = VI Unidades [P] = J/s ≡ W (Watt).

En una resistencia:

P = VI = I

²

R = V

²

R

(41)

Potencia El´ ectrica

Potencia

La potencia se define como la tasa de gasto de la energ´ıa por unidad de tiempo

P ≡ ∆U

∆t

En el caso de la energ´ıa el´ ectrica U = q V

P = V ∆q

∆t = VI Unidades [P] = J/s ≡ W (Watt).

En una resistencia:

P = VI = I

²

R = V

²

R

(42)

Potencia El´ ectrica

Potencia

La potencia se define como la tasa de gasto de la energ´ıa por unidad de tiempo

P ≡ ∆U

∆t En el caso de la energ´ıa el´ ectrica U = q V

P = V ∆q

∆t = VI Unidades [P] = J/s ≡ W (Watt).

En una resistencia:

P = VI = I

²

R = V

²

R

(43)

Potencia El´ ectrica

Potencia

La potencia se define como la tasa de gasto de la energ´ıa por unidad de tiempo

P ≡ ∆U

∆t En el caso de la energ´ıa el´ ectrica U = q V

P = V ∆q

∆t

= VI

Unidades [P] = J/s ≡ W (Watt). En una resistencia:

P = VI = I

²

R = V

²

R

(44)

Potencia El´ ectrica

Potencia

La potencia se define como la tasa de gasto de la energ´ıa por unidad de tiempo

P ≡ ∆U

∆t En el caso de la energ´ıa el´ ectrica U = q V

P = V ∆q

∆t = VI

Unidades [P] = J/s ≡ W (Watt). En una resistencia:

P = VI = I

²

R = V

²

R

(45)

Potencia El´ ectrica

Potencia

La potencia se define como la tasa de gasto de la energ´ıa por unidad de tiempo

P ≡ ∆U

∆t En el caso de la energ´ıa el´ ectrica U = q V

P = V ∆q

∆t = VI Unidades [P] = J/s ≡ W (Watt).

En una resistencia:

P = VI = I

²

R = V

²

R

(46)

Potencia El´ ectrica

Potencia

La potencia se define como la tasa de gasto de la energ´ıa por unidad de tiempo

P ≡ ∆U

∆t En el caso de la energ´ıa el´ ectrica U = q V

P = V ∆q

∆t = VI Unidades [P] = J/s ≡ W (Watt).

En una resistencia:

P = VI

= I

²

R = V

²

R

(47)

Potencia El´ ectrica

Potencia

La potencia se define como la tasa de gasto de la energ´ıa por unidad de tiempo

P ≡ ∆U

∆t En el caso de la energ´ıa el´ ectrica U = q V

P = V ∆q

∆t = VI Unidades [P] = J/s ≡ W (Watt).

En una resistencia:

P = VI = I

²

R =

V

²

R

(48)

Potencia El´ ectrica

Potencia

La potencia se define como la tasa de gasto de la energ´ıa por unidad de tiempo

P ≡ ∆U

∆t En el caso de la energ´ıa el´ ectrica U = q V

P = V ∆q

∆t = VI Unidades [P] = J/s ≡ W (Watt).

En una resistencia:

P = VI = I

²

R = V

²

R

(49)

Corrientes en Membranas El´ ectricas, Circuito RC

Membranas Celulares

Las membranas celulares pueden ser modeladas idealmente como condensadores con capacitancia espec´ıfica ¯ c = 1µF /cm

²

.

Bicapa fosfolip´ıdica con una resistividad muy grande ρ ∼ 10

¹⁵

Ω.cm. La elevada resistencia previene que la carga se escape a trav´ es del l´ıpido y mantiene la carga almacenada como si las capas fueran un condensador ideal.

Las proteinas en una membrana actuan como un canal que permite el flujo de corriente el´ ectrica en la membrana.

Este proceso se modela idealmente con un circuito RC.

and again replacing the two capacitors with a single capacitor C and noting that the voltages across each capacitor are the same because they are in paral- lel (V ₁ ! V ₂ ! V ), we find

(16.18) Remember that, just as for resistors, these results for combining two capacitors in series or parallel can easily be generalized to larger arrays of capacitors using the same tools as in the above discussion. Circuits with only resistors or only capacitors present are ideals. In the next section we turn to a presentation of more realistic circuits with both resistors and capacitors present. Such circuits are more realistic because there is always a small amount of resistance (in the conducting wires themselves) or stray capacitance (between different conducting surfaces) present in any circuit regardless of whether an actual resistor or capacitor device is present in the circuit. We approach this topic using a model for cell membranes.

3. MEMBRANE ELECTRICAL CURRENTS

In the last chapter membranes were considered as ideal capacitors with a specific capacitance (capacitance per unit area) of about 1 "F/cm ² . This turns out to be a very good approximation for a pure phospholipid bilayer which has an extremely high resistivity of about 10 ¹⁵ #-cm, comparable to a very good insulator. The very high equivalent resis- tance prevents charge from crossing the lipid region and maintains the stored charge as if the bilayer were an ideal capacitor. However, as dis- cussed in the last chapter, biological membranes are full of proteins that act as channels allowing ionic currents to flow across a membrane.

The simplest model, or equivalent circuit, for a biological mem- brane in the resting state is shown in Figure 16.13 and is known as an RC series circuit. For now, we ignore how the equivalent capacitor was charged (to a voltage V ₀ ! Q ₀ /C) and we imagine that at time zero the switch S is closed (corresponding to the membrane channels opening), discharging the capacitor. The capacitor does not discharge instanta- neously, but follows a time course that depends on the values of R and C. The resistance R represents the effective resistance to current flow across the membrane and is discussed further below.

To analyze this circuit, we use Kirchhoff’s loop method, discussed in the last section. Let’s write a loop equation for the circuit in Figure 16.13 after the switch is closed and a path is provided for current flow. When the switch is closed current will flow from the $Q ₀ side of the capacitor clockwise around the circuit. Starting at the switch S and mentally going clockwise around the loop, we find

(16.19) Because both Q and I vary with time, it turns out that we need calculus to solve this equation (see box) to find that the charge on the capacitor and the current through the resistor are given by

(16.20a)

(16.20b) I ! I ₀ e

^%

t RC

, Q ! Q ₀ e

^%

t RC

,

%IR $ Q C ! 0.

C ! C ₁ $ C ₂ . (capacitors in parallel) Q ! CV ! C ₁ V $ C ₂ V or

4 1 2 E

L E C T R I C

C

U R R E N T A N D

C

E L L

M

E M B R A N E S

R C

Q

_o

–Q

_o

S

Starting from Equation (16.19), and substi- tuting from the definition of

(the minus sign is needed to make the cur- rent positive because it is equal to the time rate of decrease of the capacitor charge), the equation becomes

. Rewriting, we have

.

Integrating both sides of this equation from t ! 0 to time t and from Q(t ! 0) ! Q ₀ to a value of Q(t), written simply as Q, we find

, so that

. Taking the antilog of both sides, remembering that these logarithms are to the base e, we find

,

or Equation (16.20a). To then find the cur- rent as a function of time, we again use its definition, so that

, or Equation (16.20b). This same procedure can be used to analyze any electrical circuit consisting of batteries, capacitors, and resistors via the loop equation.

I ! % dQ

dt ! % Q ₀ d(e ^%

t RC

) dt ! Q ₀

RC e ^%

t RC

Q Q ₀ ! e ^%

t RC

log Q % log Q ₀ ! log Q Q ₀ !% t

RC L _Q

₀

Q dQ Q !% 1

RC L

t 0

dt dQ

Q ! % dt RC R dQ

dt $ Q C ! 0 I ! % dQ

dt

FIGURE 16.13 An RC series circuit

with the capacitor initially charged

before closing the switch S

connected to the resistor.

(50)

Corrientes en Membranas El´ ectricas, Circuito RC

Membranas Celulares

Las membranas celulares pueden ser modeladas idealmente como condensadores con capacitancia espec´ıfica ¯ c = 1µF /cm

²

.

Bicapa fosfolip´ıdica con una resistividad muy grande ρ ∼ 10

¹⁵

Ω.cm. La elevada resistencia previene que la carga se escape a trav´ es del l´ıpido y mantiene la carga almacenada como si las capas fueran un condensador ideal.

Las proteinas en una membrana actuan como un canal que permite el flujo de corriente el´ ectrica en la membrana.

Este proceso se modela idealmente con un circuito RC.

and again replacing the two capacitors with a single capacitor C and noting that the voltages across each capacitor are the same because they are in paral- lel (V ₁ ! V ₂ ! V ), we find

(16.18) Remember that, just as for resistors, these results for combining two capacitors in series or parallel can easily be generalized to larger arrays of capacitors using the same tools as in the above discussion. Circuits with only resistors or only capacitors present are ideals. In the next section we turn to a presentation of more realistic circuits with both resistors and capacitors present. Such circuits are more realistic because there is always a small amount of resistance (in the conducting wires themselves) or stray capacitance (between different conducting surfaces) present in any circuit regardless of whether an actual resistor or capacitor device is present in the circuit. We approach this topic using a model for cell membranes.

3. MEMBRANE ELECTRICAL CURRENTS

In the last chapter membranes were considered as ideal capacitors with a specific capacitance (capacitance per unit area) of about 1 "F/cm ² . This turns out to be a very good approximation for a pure phospholipid bilayer which has an extremely high resistivity of about 10 ¹⁵ #-cm, comparable to a very good insulator. The very high equivalent resis- tance prevents charge from crossing the lipid region and maintains the stored charge as if the bilayer were an ideal capacitor. However, as dis- cussed in the last chapter, biological membranes are full of proteins that act as channels allowing ionic currents to flow across a membrane.

The simplest model, or equivalent circuit, for a biological mem- brane in the resting state is shown in Figure 16.13 and is known as an RC series circuit. For now, we ignore how the equivalent capacitor was charged (to a voltage V ₀ ! Q ₀ /C) and we imagine that at time zero the switch S is closed (corresponding to the membrane channels opening), discharging the capacitor. The capacitor does not discharge instanta- neously, but follows a time course that depends on the values of R and C. The resistance R represents the effective resistance to current flow across the membrane and is discussed further below.

To analyze this circuit, we use Kirchhoff’s loop method, discussed in the last section. Let’s write a loop equation for the circuit in Figure 16.13 after the switch is closed and a path is provided for current flow. When the switch is closed current will flow from the $Q ₀ side of the capacitor clockwise around the circuit. Starting at the switch S and mentally going clockwise around the loop, we find

(16.19) Because both Q and I vary with time, it turns out that we need calculus to solve this equation (see box) to find that the charge on the capacitor and the current through the resistor are given by

(16.20a)

(16.20b) I ! I ₀ e

^%

t RC

, Q ! Q ₀ e

^%

t RC

,

%IR $ Q C ! 0.

C ! C ₁ $ C ₂ . (capacitors in parallel) Q ! CV ! C ₁ V $ C ₂ V or

4 1 2 E

L E C T R I C

C

U R R E N T A N D

C

E L L

M

E M B R A N E S

R C

Q

_o

–Q

_o

S

Starting from Equation (16.19), and substi- tuting from the definition of

(the minus sign is needed to make the cur- rent positive because it is equal to the time rate of decrease of the capacitor charge), the equation becomes

. Rewriting, we have

.

Integrating both sides of this equation from t ! 0 to time t and from Q(t ! 0) ! Q ₀ to a value of Q(t), written simply as Q, we find

, so that

. Taking the antilog of both sides, remembering that these logarithms are to the base e, we find

,

or Equation (16.20a). To then find the cur- rent as a function of time, we again use its definition, so that

, or Equation (16.20b). This same procedure can be used to analyze any electrical circuit consisting of batteries, capacitors, and resistors via the loop equation.

I ! % dQ

dt ! % Q ₀ d(e ^%

t RC

) dt ! Q ₀

RC e ^%

t RC

Q Q ₀ ! e ^%

t RC

log Q % log Q ₀ ! log Q Q ₀ !% t

RC L _Q

₀

Q dQ Q !% 1

RC L

t 0

dt dQ

Q ! % dt RC R dQ

dt $ Q C ! 0 I ! % dQ

dt

FIGURE 16.13 An RC series circuit

with the capacitor initially charged

before closing the switch S

connected to the resistor.

(51)

Corrientes en Membranas El´ ectricas, Circuito RC

Membranas Celulares

Las membranas celulares pueden ser modeladas idealmente como condensadores con capacitancia espec´ıfica ¯ c = 1µF /cm

²

.

Bicapa fosfolip´ıdica con una resistividad muy grande ρ ∼ 10

¹⁵

Ω.cm.

La elevada resistencia previene que la carga se escape a trav´ es del l´ıpido y mantiene la carga almacenada como si las capas fueran un condensador ideal.

Las proteinas en una membrana actuan como un canal que permite el flujo de corriente el´ ectrica en la membrana.

Este proceso se modela idealmente con un circuito RC.

and again replacing the two capacitors with a single capacitor C and noting that the voltages across each capacitor are the same because they are in paral- lel (V ₁ ! V ₂ ! V ), we find

(16.18) Remember that, just as for resistors, these results for combining two capacitors in series or parallel can easily be generalized to larger arrays of capacitors using the same tools as in the above discussion. Circuits with only resistors or only capacitors present are ideals. In the next section we turn to a presentation of more realistic circuits with both resistors and capacitors present. Such circuits are more realistic because there is always a small amount of resistance (in the conducting wires themselves) or stray capacitance (between different conducting surfaces) present in any circuit regardless of whether an actual resistor or capacitor device is present in the circuit. We approach this topic using a model for cell membranes.

3. MEMBRANE ELECTRICAL CURRENTS

In the last chapter membranes were considered as ideal capacitors with a specific capacitance (capacitance per unit area) of about 1 "F/cm ² . This turns out to be a very good approximation for a pure phospholipid bilayer which has an extremely high resistivity of about 10 ¹⁵ #-cm, comparable to a very good insulator. The very high equivalent resis- tance prevents charge from crossing the lipid region and maintains the stored charge as if the bilayer were an ideal capacitor. However, as dis- cussed in the last chapter, biological membranes are full of proteins that act as channels allowing ionic currents to flow across a membrane.

The simplest model, or equivalent circuit, for a biological mem- brane in the resting state is shown in Figure 16.13 and is known as an RC series circuit. For now, we ignore how the equivalent capacitor was charged (to a voltage V ₀ ! Q ₀ /C) and we imagine that at time zero the switch S is closed (corresponding to the membrane channels opening), discharging the capacitor. The capacitor does not discharge instanta- neously, but follows a time course that depends on the values of R and C. The resistance R represents the effective resistance to current flow across the membrane and is discussed further below.

To analyze this circuit, we use Kirchhoff’s loop method, discussed in the last section. Let’s write a loop equation for the circuit in Figure 16.13 after the switch is closed and a path is provided for current flow. When the switch is closed current will flow from the $Q ₀ side of the capacitor clockwise around the circuit. Starting at the switch S and mentally going clockwise around the loop, we find

(16.19) Because both Q and I vary with time, it turns out that we need calculus to solve this equation (see box) to find that the charge on the capacitor and the current through the resistor are given by

(16.20a)

(16.20b) I ! I ₀ e

^%

t RC

, Q ! Q ₀ e

^%

t RC

,

%IR $ Q C ! 0.

C ! C ₁ $ C ₂ . (capacitors in parallel) Q ! CV ! C ₁ V $ C ₂ V or

4 1 2 E

L E C T R I C

C

U R R E N T A N D

C

E L L

M

E M B R A N E S

R C

Q

_o

–Q

_o

S

Starting from Equation (16.19), and substi- tuting from the definition of

(the minus sign is needed to make the cur- rent positive because it is equal to the time rate of decrease of the capacitor charge), the equation becomes

. Rewriting, we have

.

Integrating both sides of this equation from t ! 0 to time t and from Q(t ! 0) ! Q ₀ to a value of Q(t), written simply as Q, we find

, so that

. Taking the antilog of both sides, remembering that these logarithms are to the base e, we find

,

or Equation (16.20a). To then find the cur- rent as a function of time, we again use its definition, so that

, or Equation (16.20b). This same procedure can be used to analyze any electrical circuit consisting of batteries, capacitors, and resistors via the loop equation.

I ! % dQ

dt ! % Q ₀ d(e ^%

t RC

) dt ! Q ₀

RC e ^%

t RC

Q Q ₀ ! e ^%

t RC

log Q % log Q ₀ ! log Q Q ₀ !% t

RC L _Q

₀

Q dQ Q !% 1

RC L

t 0

dt dQ

Q ! % dt RC R dQ

dt $ Q C ! 0 I ! % dQ

dt

FIGURE 16.13 An RC series circuit

with the capacitor initially charged

before closing the switch S

connected to the resistor.

(52)

Corrientes en Membranas El´ ectricas, Circuito RC

Membranas Celulares

Las membranas celulares pueden ser modeladas idealmente como condensadores con capacitancia espec´ıfica ¯ c = 1µF /cm

²

.

Bicapa fosfolip´ıdica con una resistividad muy grande ρ ∼ 10

¹⁵

Ω.cm. La elevada resistencia previene que la carga se escape a trav´ es del l´ıpido y mantiene la carga almacenada como si las capas fueran un condensador ideal.

Las proteinas en una membrana actuan como un canal que permite el flujo de corriente el´ ectrica en la membrana.

Este proceso se modela idealmente con un circuito RC.

and again replacing the two capacitors with a single capacitor C and noting that the voltages across each capacitor are the same because they are in paral- lel (V ₁ ! V ₂ ! V ), we find

(16.18) Remember that, just as for resistors, these results for combining two capacitors in series or parallel can easily be generalized to larger arrays of capacitors using the same tools as in the above discussion. Circuits with only resistors or only capacitors present are ideals. In the next section we turn to a presentation of more realistic circuits with both resistors and capacitors present. Such circuits are more realistic because there is always a small amount of resistance (in the conducting wires themselves) or stray capacitance (between different conducting surfaces) present in any circuit regardless of whether an actual resistor or capacitor device is present in the circuit. We approach this topic using a model for cell membranes.

3. MEMBRANE ELECTRICAL CURRENTS

In the last chapter membranes were considered as ideal capacitors with a specific capacitance (capacitance per unit area) of about 1 "F/cm ² . This turns out to be a very good approximation for a pure phospholipid bilayer which has an extremely high resistivity of about 10 ¹⁵ #-cm, comparable to a very good insulator. The very high equivalent resis- tance prevents charge from crossing the lipid region and maintains the stored charge as if the bilayer were an ideal capacitor. However, as dis- cussed in the last chapter, biological membranes are full of proteins that act as channels allowing ionic currents to flow across a membrane.

The simplest model, or equivalent circuit, for a biological mem- brane in the resting state is shown in Figure 16.13 and is known as an RC series circuit. For now, we ignore how the equivalent capacitor was charged (to a voltage V ₀ ! Q ₀ /C) and we imagine that at time zero the switch S is closed (corresponding to the membrane channels opening), discharging the capacitor. The capacitor does not discharge instanta- neously, but follows a time course that depends on the values of R and C. The resistance R represents the effective resistance to current flow across the membrane and is discussed further below.

To analyze this circuit, we use Kirchhoff’s loop method, discussed in the last section. Let’s write a loop equation for the circuit in Figure 16.13 after the switch is closed and a path is provided for current flow. When the switch is closed current will flow from the $Q ₀ side of the capacitor clockwise around the circuit. Starting at the switch S and mentally going clockwise around the loop, we find

(16.19) Because both Q and I vary with time, it turns out that we need calculus to solve this equation (see box) to find that the charge on the capacitor and the current through the resistor are given by

(16.20a)

(16.20b) I ! I ₀ e

^%

t RC

, Q ! Q ₀ e

^%

t RC

,

%IR $ Q C ! 0.

C ! C ₁ $ C ₂ . (capacitors in parallel) Q ! CV ! C ₁ V $ C ₂ V or

4 1 2 E

L E C T R I C

C

U R R E N T A N D

C

E L L

M

E M B R A N E S

R C

Q

_o

–Q

_o

S

Starting from Equation (16.19), and substi- tuting from the definition of

(the minus sign is needed to make the cur- rent positive because it is equal to the time rate of decrease of the capacitor charge), the equation becomes

. Rewriting, we have

.

Integrating both sides of this equation from t ! 0 to time t and from Q(t ! 0) ! Q ₀ to a value of Q(t), written simply as Q, we find

, so that

. Taking the antilog of both sides, remembering that these logarithms are to the base e, we find

,

or Equation (16.20a). To then find the cur- rent as a function of time, we again use its definition, so that

, or Equation (16.20b). This same procedure can be used to analyze any electrical circuit consisting of batteries, capacitors, and resistors via the loop equation.

I ! % dQ

dt ! % Q ₀ d(e ^%

t RC

) dt ! Q ₀

RC e ^%

t RC

Q Q ₀ ! e ^%

t RC

log Q % log Q ₀ ! log Q Q ₀ !% t

RC L _Q

₀