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Δ C 1 V eq = = C QC 1 eq 1 +   ,   C Δ 1 2 V 1 = QC 1   ,   Δ V 2 = QC 2

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Símbolos(de(circuitos:(

(( Condensador((inicialmente(descargados)(

(( ((

(( Batería((la(terminal(positiva(está(a(mayor(

potencial)((

Dos(o(mas(condensadores(se(pueden(combinar(en(

circuitos(eléctricos.(

(Carga(total(Q(almacenada(en(los(condensadores:(

Q(=(Q1(+(Q2(

(La(carga(de(cada(condensador(es:(

Q1(=(C1(ΔV((((((y((((((Q2(=(C2(ΔV(

(

Combinación(en(paralelo:( Combinación(en(paralelo:(

El(condensador(equivalente(

debe(almacenar(Q(unidades(

de(carga((con(Q(=(Q1(+(Q2):(

(Qeq(=(Ceq(ΔV(

(Ceq(=(C1(+(C2(

La&carga&en&los&condensadores&conectados&en&

serie&es&la&misma:&&Q1&=&Q2&

(

El(voltaje(de(la(batería(está(dividido(entre(los(dos(

condensadores:(((((((ΔV&=&ΔV1&+&ΔV2&

(Condensador(equivalente(a(los(dos(

condensadores(en(serie:(carga(Q(por(ΔV(

Combinación(en(serie:(

ΔV = Q

Ceq , ΔV1=Q

C1 , ΔV2=Q C2 1

Ceq= 1 C1+ 1

C2

Ejemplo:&

Encuentre(la(capacitancia(equivalente(entre(a(y(b(

para(la(combinación(de(condensadores(que(se(

muestre(en(la(figura.(Todas(las(capacitancias(están(

en(microfaradios.(

(2)

Definición:&

Cantidad(de(cargas(eléctricas(que(pasan(por(

una(sección(transversal(de(un(condensador(por(

segundo.(

I = ΔQ

Δt   [I ] = A = C s  

The speed of the charge carriers vdis an average speed called the drift speed. To

understand the meaning of drift speed, consider a conductor in which the charge car- riers are free electrons. If the conductor is isolated—that is, the potential difference across it is zero—then these electrons undergo random motion that is analogous to the motion of gas molecules. As we discussed earlier, when a potential difference is applied across the conductor (for example, by means of a battery), an electric field is set up in the conductor; this field exerts an electric force on the electrons, producing a current.

However, the electrons do not move in straight lines along the conductor. Instead, they collide repeatedly with the metal atoms, and their resultant motion is complicated and zigzag (Fig. 27.3). Despite the collisions, the electrons move slowly along the conduc- tor (in a direction opposite that of E) at the drift velocity vd.

We can think of the atom–electron collisions in a conductor as an effective internal friction (or drag force) similar to that experienced by the molecules of a liquid flowing through a pipe stuffed with steel wool. The energy transferred from the electrons to the metal atoms during collisions causes an increase in the vibrational energy of the atoms and a corresponding increase in the temperature of the conductor.

834 CHAPTE R 27 • Current and Resistance

Figure 27.3 A schematic representation of the zigzag motion of an electron in a conductor. The changes in direction are the result of collisions between the electron and atoms in the conductor. Note that the net motion of the electron is opposite the direction of the electric field.

Because of the acceleration of the charge carriers due to the electric force, the paths are actually parabolic. However, the drift speed is much smaller than the average speed, so the parabolic shape is not visible on this scale.

Quick Quiz 27.1

Consider positive and negative charges moving horizon- tally through the four regions shown in Figure 27.4. Rank the current in these four regions, from lowest to highest.

Quick Quiz 27.2

Electric charge is conserved. As a consequence, when current arrives at a junction of wires, the charges can take either of two paths out of the junction and the numerical sum of the currents in the two paths equals the current that entered the junction. Thus, current is (a) a vector (b) a scalar (c) neither a vector nor a scalar.

vd

E

Figure 27.4 (Quick Quiz 27.1) Charges move through four regions.

(a) –

– ++

+

++ + +

++ –

– –

(b) (c) (d)

Example 27.1 Drift Speed in a Copper Wire

The 12-gauge copper wire in a typical residential building has a cross-sectional area of 3.31 ! 10"6m2. If it carries a current of 10.0 A, what is the drift speed of the electrons?

Assume that each copper atom contributes one free electron to the current. The density of copper is 8.95 g/cm3.

Solution From the periodic table of the elements in Appendix C, we find that the molar mass of copper is 63.5 g/mol. Recall that 1 mol of any substance contains Avogadro’s number of atoms (6.02 ! 1023). Knowing the density of copper, we can calculate the volume occupied by 63.5 g (# 1 mol) of copper:

Because each copper atom contributes one free electron to the current, we have

V # m

$ # 63.5 g

8.95 g/cm3 #7.09 cm3

From Equation 27.4, we find that the drift speed is

where q is the absolute value of the charge on each electron.

Thus,

2.22 ! 10"4 m/s

#

# 10.0 C/s

(8.49 ! 1028 m"3)(1.60 ! 10"19 C)(3.31 ! 10"6 m2) vd# I

nqA

vd# I nqA # 8.49 ! 1028 electrons/m3 n # 6.02 ! 1023 electrons

7.09 cm3

!

1.00 ! 101 m36 cm3

"

Ordenar(la(corriente(de(la(mas(grande(a(la(mas(

pequeña.((

(

Ley&de&Ohm:&

( ( ( (

(

Resistencia&/&Conductancia:&

(

ΔV = R ⋅ I   con R = resistencia [R] = V

A = Ω (Ohmio) 

R = ρ L

A G = 1 R = 1

ρ A L = σ A

L

[ρ] = Ω.m [σ ] = Ω

−1

.m

−1

Calcular(la(corriente(eléctrica(a(través(del(agua(

destilada(contenida(en(un(tubo(aislante(de(10(

cm(de(longitud(y(5(cm(de(diámetro(cuando(se(

establece(una(diferencia(de(potencial((voltaje)(

de(100(V(entre(los(extremos(del(tubo.(

(agua(distilada:(ρ~2*10

5

(Ω.m)(

(

Calcular(la(capacitancia(y(resistencia(de(un(axón(

(con(constante(dieléctrica(κ).(

Contenido Corriente el´ectrica

Resistencia y ley de Ohm

Circuitos el´ectricos y leyes de Kirchho↵

Corrientes en Membranas El´ectricas, Circuito RC Ecuaci´on de Nerst

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 =✏0A

d =✏02⇡aL

d ; R =⇢d

A = ⇢d 2⇡aL

Asuma los valores dados en los ejercicios del taller. Ax´on sin mielina:  = 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, ⇢ = 1,6⇥ 107⌦.m.

Bioelectricidad y Biomagnetismo

C = ε

0

A

d C = κε

0

A

d y( A = 2 π aL

(3)

Calcular(la(capacitancia(y(resistencia(de(un(axón(

(con(constante(dieléctrica(κ).(

Contenido Corriente el´ectrica

Resistencia y ley de Ohm

Circuitos el´ectricos y leyes de Kirchho↵

Corrientes en Membranas El´ectricas, Circuito RC Ecuaci´on de Nerst

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 =✏0A

d =✏02⇡aL

d ; R =⇢d

A = ⇢d 2⇡aL

Asuma los valores dados en los ejercicios del taller. Ax´on sin mielina:  = 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, ⇢ = 1,6⇥ 107⌦.m.

Bioelectricidad y Biomagnetismo

R = ρ d A = ρ d

2 π aL

2(resistencias(R1(y(R2(&

&

&

&

Resistencias&en&Serie:(

( ( ( ( ( ( (

R

eq

= R

1

+ R

2

  V = V

1

+V

2

 = R

1

⋅ I + R

2

⋅ I V = (R

1

+ R

2

)⋅ I

V = R

eq

⋅ I

2(resistencias(R1(y(R2(&

&

( Resistencias&en&Paralelo:&

( (

( 1

R

eq

= 1 R

1

+ 1

R

2

  I = I

1

+ I

2

 = V

R

1

+ V

R

2

= V ⋅ ( 1 R

1

+ 1

R

2

) I = V

R

eq

Ejemplo:&

Encuentre(la(resistencia(equivalente(entre(a(y(c(

para(la(combinación(de(resistores(que(se(muestre(

en(la(figura.(

Referencias

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