Hafez a radi, john o rasmussen auth principles of physics for scientists and engineers 2 33

23.4 Capacitors in Parallel and Series 793 For the two capacitors in Fig 23.12b, we have: V1 = Q1 Q = and C1 C1 V2 = Q2 Q = C2 C2 (23.27) Substituting in Eq 23.26, we get: V = Q Q + C1 C2 (23.28) The equivalent capacitor Ceq has the same charge Q and applied potential difference V ; thus: V = Q Q Q = + Ceq C1 C2 (23.29) Canceling Q, we arrive at the following relationship: 1 = + (Series combination) Ceq C1 C2 (23.30) We can extend this treatment to n capacitors connected in series as: 1 1 = + + + ··· + Ceq C1 C2 C3 Cn (Series combination) (23.31) Thus, the equivalent capacitance of a series combination of capacitors is simply the algebraic sum of the reciprocals of the individual capacitances and will always be less than any one of them Example 23.7 In Fig 23.12, let C1 = µF and C2 = µF, and and V = 18 V Find Ceq , Q, V2 Solution: The equivalent capacitance of the series combination is: 1 1 1 + = = + = Ceq C1 C2 µF µF µF Consequently: V1 = Q = Ceq ⇒ Ceq = µF V = (2 µF)(18 V) = 36 µC Q 36 µV = = V and C1 µF V2 = Q 36 µV = = 12 V C2 µF V1 , 794 23 Capacitors and Capacitance Example 23.8 For the combination of capacitors shown in Fig 23.13a, assume that C1 = µF, C2 = µF, and C3 = µF, and V = 12 V (a) Find the equivalent capacitance of the combination (b) What is the charge on C1 ? Q12 =Q123 C1 C 12 C2 ΔV C3 C3 (a) Q123 ΔV ΔV C 123 Q =Q123 (b) (c) Fig 23.13 Solution: (a) Capacitors C1 and C2 in Fig 23.13a are in parallel and their equivalent capacitance C12 is: C12 = C1 + C2 = µF + µF = µF From Fig 23.13b, we find that C12 and C3 form a series combination and their equivalent capacitance C123 is given by: 1 1 1 + = = + = C123 C12 C3 µF µF µF ⇒ C123 = µF (b) We first find the charge Q123 on C123 in Fig 23.13c as follows: Q123 = C123 V = (2 µF)(12 V) = 24 µC This same charge exists on each capacitor in the series combination of Fig 23.13b Therefore, if Q12 represents the charge on C12 , then Q12 = Q123 = 24 µC Accordingly, the potential difference across C12 is: V12 = Q12 24 µC =4V = C12 µF This same potential difference exists across C1 , i.e V1 = V12 Thus: Q1 = C1 V1 = (2 µF)(4 V) = µC 23.5 Energy Stored in a Charged Capacitor 23.5 795 Energy Stored in a Charged Capacitor When the switch S of Fig 23.14a is closed, the process of charging the capacitor starts by transferring electrons from the left plate (leaving it with an excess of positive charge) to the right plate In the process of charging this capacitor, the battery must work at the expense of its stored chemical energy +q C S -q +Q C Intermediate state S B S C Final state B ΔV (c) B ΔV (b) ΔV (a) -Q Fig 23.14 (a) A circuit consisting of a battery B, a switch S, and a capacitor C (b) An intermediate state when the magnitude of the charge on the capacitor is q (c) A final state when q = Q In principle, the charging process occurs as if positive charges were pulled off from the right plate and transferred directly to the left plate Suppose that, at a given instant during the charging process, as shown in Fig 23.14b, the charge on the capacitor is q, i.e q = C V Moreover, according to Eq 22.11, the differential applied work necessary to transfer a differential charge dq from the plate having charge −q to the plate having a charge +q is given by: dW (app) = dq V = q dq C (23.32) The total work required to charge the capacitor from a charge q = to a final charge q = Q, see Fig 23.14c, is thus: Q W (app) = q dq = C C Q q dq = Q2 2C (23.33) According to Eqs 22.6 and 22.10, this work done by the battery is stored as electrostatic potential energy U in the capacitor Thus: U= Q2 2C (Electric potential energy) (23.34) 796 23 Capacitors and Capacitance From Eq 23.1, we can write this stored electric potential energy in the following forms: U = 21 C( V )2 (Electric potential energy) (23.35) or U = 21 Q V (Electric potential energy) (23.36) It is important to note that Eqs 23.34 to 23.36 hold for any capacitor, regardless of its shape → When we neglect the fringing effect (nonuniform E ) in a parallel-plate capacitor filled with a dielectric, we know that the electric field has the same value at any point between the plates Thus, the potential energy per unit volume between the plates, known as the energy density uE , should also be uniform Then we can find uE by dividing the electric potential energy U by the volume Ad between the plates: uE = Using C = κ ◦ A/d and C( V )2 U = Ad Ad (23.37) V = Ed for parallel-plate capacitors, we get: uE = 21 κ ◦E (Electric energy density) (23.38) Although this equation is derived for a parallel-plate capacitor, it holds true for any → source of electric field When the electric field E exists at any point in a dielectric material of dielectric constant κ, the potential energy per unit volume at this point is given by Eq 23.38 When κ = 1, this relation reduces to uE = 21 ◦E Example 23.9 A capacitor C1 = µF is charged by an initial potential difference Vi = 12 V, see Fig 23.15a The charging battery is then removed, as shown in Fig 23.15b, and the capacitor is connected to the uncharged capacitor C2 = µF, as shown in Fig 23.15c (a) Find the final potential difference Vf as well as Q1f and Q2f (b) Find the stored energy before and after the switch is closed Solution: (a) The original charge is now shared by C1 and C2 , so: Q1i = Q1f + Q2f 23.5 Energy Stored in a Charged Capacitor S Δ Vi 797 S Q 1i C1 Δ Vi C2 S Q 1i C1 (a) C2 Δ Vf Q2f Q 1f (b) C1 C2 (c) Fig 23.15 Using of the relation Q = C V in each term of this equation, we get: C1 Vi = C1 Vf + C2 Vf Thus: Vf = C1 C1 + C2 Vi = (4 µF) (12 V) = V µF + µF Q1f = C1 Vf = (4 µF)(8 V) = 32 µC and: Q2f = C2 Vf = (2 µF)(8 V) = 16 µC (b) The initial potential energy is: Ui = 21 C1 ( Vi )2 = 21 (4 µF)(12 V)2 = 288 µJ The final potential energy is: Uf = 21 C1 ( Vf )2 + 21 C2 ( Vf )2 = 21 (4 µF + µF)(8 V)2 = 192 µJ Although Ui > Uf , this is not a violation of the conservation of energy principle The missing energy is transferred as thermal energy into the connecting wires and as radiated electromagnetic waves 23.6 Exercises Section 23.1 Capacitor and Capacitance (1) A capacitor has a capacitance of 15 µF How much charge must be removed to lower the potential difference between its conductors to 10 V? 798 23 Capacitors and Capacitance (2) Two identical coins carry equal but opposite charges of magnitude 1.6 µC The capacitance of this combination is 20 pF What is the potential difference between the coins? (3) A capacitor with a charge of magnitude 10−4 C has a potential difference of 50 V What charge value is needed to produce a potential difference of 15 V? Section 23.2 Calculating Capacitance (4) A computer memory chip contains a large number of capacitors, each of which has a plate area A = 20 × 10−12 m2 and a capacitance of 50 f F (50 femtofarads) Assuming a parallel-plate configuration, find the order of magnitude of the separation distance d between the plates of such a capacitor (5) A parallel-plate capacitor has a plate area A = 0.04 m2 and a vacuum separation d = × 10−3 m A potential difference of 20 V is applied between the plates of the capacitor (a) Find the capacitance of the capacitor (b) Find the magnitude of the charge and charge density on the plates of the capacitor (c) Find the magnitude of the electric field between the plates (6) An electric spark occurs if the electric field in air exceeds the value × 106 V/m Find the maximum magnitude of the charge on the plates of an air-filled parallelplate capacitor of area A = 30 cm2 such that a spark is avoided (7) A parallel-plate capacitor has circular plates, each with a radius r = cm Assume a vacuum separation d = mm exists between the plates, see Fig 23.16 How much charge is stored on each plate of the capacitor when their potential difference has the value Fig 23.16 See Exercise (7) V = 50 V d r ΔV (8) Figure 23.17 shows a set of two parallel sheets of a conductor connected together to form one plate of a capacitor, while the second set is connected 23.6 Exercises 799 together to form the other plate of the capacitor Assume that the effective area of adjacent sheets is A and that the air separation is d From the figure, confirm that the number of adjoining sheets of positive and negative charges is and the capacitor has a capacitance C = ◦ A/d Fig 23.17 See Exercise (8) Area A 2 d (9) If each set in Exercise consists of n plates, see Fig 23.18, then show that the capacitance of the capacitor will be given by: C= (2n − 1) d Fig 23.18 See Exercise (9) ◦A n Area A d n (10) A variable air capacitor used in radio tuning consists of a set of n fixed semicircular plates, each of radius r, and located a distance d from a neighboring plate of an identical yet rotatable set, see Fig 23.19 Show that when one set is rotated by an angle θ, the capacitance is: C= (2n − 1) ◦ (π − θ ) r 2d 800 23 Capacitors and Capacitance Fig 23.19 See Exercise (10) d θ r (11) A coaxial cable of length = m consists of a solid cylindrical conductor surrounded by a cylindrical conducting shell The inner conductor has a radius a = 2.5 mm and carries a charge Q, while the surrounding shell has a radius b = 8.5 mm and carries a charge −Q, see Fig 23.20 Assume that Q = +8 × 10−8 C and that air fills the gap between the conductors (a) What is the capacitance of this cable? (b) What is the magnitude of the potential difference between the two cylinders? Fig 23.20 See Exercise (11) −Q b a Q (12) An isolated spherical conductor carries a charge Q = nC, see Fig 23.21 The potential difference between the sphere and its surroundings is V = 100 V What is the capacitance formed from the sphere and its surroundings? Fig 23.21 See Exercise (12) Q 23.6 Exercises 801 (13) A capacitor consists of two concentric spheres of radii a = 30 cm and b = 36 cm, see Fig 23.22 Assume the gap between the conductors is filled with air (a) What is the capacitance of this capacitor? (b) How much charge is stored in the capacitor if the potential difference between the two spheres is Fig 23.22 See Exercise (13) V = 50 V? −Q Q b a (14) Find the capacitance of Earth by assuming that the “missing second conducting sphere” has an infinite radius The radius of Earth is R = 6.37 × 106 m (15) A spherical drop of mercury has a capacitance of 2.78 f F If two such drops combine into one, what would its capacitance be? Section 23.3 Capacitors with Dielectrics (16) Two parallel plates of area A = 0.01 m2 are separated by a distance d = × 10−3 m The region between these plates is filled with a dielectric material of κ = 3, and the plates are given equal but opposite charges of µC (a) What is the capacitance of this capacitor? (b) Find the potential difference between the plates (17) An air-filled parallel-plate capacitor of 15 µF is connected to a 50 V battery; then the battery is removed (a) Find the charge on the capacitor (b) If the air is replaced with oil having κ = 2.2, find the new values of the capacitance and the potential difference between the plates (18) A parallel-plate capacitor has an area A = cm2 (a) Find the maximum stored charge on the capacitor if air fills the space between the plates (b) Redo part (a) when paper is used instead of the air (use the dielectric strengths given in Table 23.1) 802 23 Capacitors and Capacitance (19) The charged air capacitor shown in Fig 23.23 is first placed at a pressure of atm and found to have a potential difference V = 10,376 V Then, the capacitor is placed in a vacuum chamber and the air is removed The potential difference is C -Q +Q Air +Q V◦ = 10,382 V Determine the dielectric constant of the air C° -Q Vacuum found to rise to Before ΔV After Δ V° Fig 23.23 See Exercise (19) (20) A parallel-plate capacitor having an area A = 0.2 m2 , and a plate separation d = mm filled with air as an insulator, is connected to a battery that has a potential difference V◦ = 12 V, see Fig 23.24 While the battery is still connected to the capacitor, a sheet of glass (κ = 4.5) is inserted to fill the space between the plates, see the figure (a) Determine both the initial capacitance (C◦ ) and the initial charge (Q◦ ), then find C and Q after inserting the glass (b) If σi is the magnitude of the induced surface charge density on the glass and σ◦ is the magnitude of the charge density of the plates before the insertion of the glass, then show that: σi = (κ − 1)σ◦ (c) Find the values of σi and the induced electric field Ei Section 23.4 Capacitors in Parallel and Series (21) Two capacitors, C1 = µF and C2 = µF, are connected in parallel to a battery that has a potential difference V = V (a) Find the equivalent capacitance of the combination (b) Find the charge on each capacitor (c) Find the potential difference across each capacitor 808 23 Capacitors and Capacitance (40) Confirm the relationships shown in Fig 23.35, where and V is shortened by V Fixed charge Before inserting the dielectric C° +Q C° ° −Q ⇒ ° V° E° u° ° E° Q (Fixed) κ Voltmeter Voltmeter V° V ° ⇒ Q° V° (Fixed) u° Fixed charge After inserting the dielectric C + Q° − Q° C Dielectric E Q° (Fixed) V Fixed voltage Before inserting the dielectric C° + Q° − Q° C° E E° V◦ is shortened by V◦ V° Fig 23.35 See Exercise (40) B E u Fixed voltage After inserting the dielectric C −Q +Q C Dielectric E Q κ V° B Relationships C = κ C° Q° (Fixed) V V = κ° E° κ u = u ° /κ E= Relationships C = κ C° Q = κ Q° V° (Fixed) V° (Fixed) E E = E° u u = κ u° Electric Circuits 24 In this chapter we analyze simple electric circuits that contain devices such as batteries, resistors, and capacitors in various combinations We begin by introducing steady-state electric circuits and the concept of a constant rate of flow of electric charges, known as direct current (dc) We also introduce Kirchhoff’s two rules, which are used to simplify and analyze more complicated circuits Finally, we consider circuits containing resistors and capacitors, in which currents can vary with time 24.1 Electric Current and Electric Current Density Electric Current When there is a net flow of charge across any area, we say there is an electric current (or simply current) across that area To maintain a continuous current, we must maintain a net force on the mobile charge in some way The net force may → result, for example, from an electrostatic field We assume that an electric field E is maintained within a conductor such that the charged particle q is acted on by a force → F = q E We refer to this force as the particle’s driving force → To define the current, we consider positive charges moving perpendicularly onto a surface area A as shown in Fig 24.1 Spotlight The current I across an area A is defined as the net charge flowing perpendicularly to that area per unit time H A Radi and J O Rasmussen, Principles of Physics, Undergraduate Lecture Notes in Physics, DOI: 10.1007/978-3-642-23026-4_24, © Springer-Verlag Berlin Heidelberg 2013 809 810 24 Electric Circuits Thus, if a net charge Q flows across an area A in a time t, the average current Iav across the area is: Q t Iav = + + + A + (24.1) + + A I Fig 24.1 Charged particles in motion perpendicular onto an area A The current I represents the time rate of flow of charges and has by convention the direction of the motion of positive charges When the rate of flow varies with time, we define the instantaneous current (or the current) I as: I= dQ dt (24.2) The SI unit of the current is ampere (abbreviated by A) That is: 1A = 1C 1s (24.3) Thus, A is equivalent to C of charge passing through the surface area in s Small currents are more conveniently expressed in milliamperes (1 mA = 10−3 A) or microamperes (1 µA = 10−6 A) Currents can be due to positive charges, or negative charges, or both In conductors, the current is due to the motion of only negatively charged free electrons (called conduction electrons) By convention, the direction of the current is the direction of the flow of positive charges Therefore, the direction of the current is opposite to the direction of the flow of electrons, see Fig 24.2b A moving charge, positive or negative, is usually referred to as a mobile charge carrier + + (a) I1 A I2 (b) A + - + A I = I1 + I2 (c) Fig 24.2 Direction of current due to (a) positive charges, (b) negative charges, and (c) both positive and negative charges 24.1 Electric Current and Electric Current Density 811 Electric Current Density The current across an area can be expressed in terms of the motion of the charge carriers To achieve this we consider a portion of a cylindrical rod that has a cross-sectional area A, length x, and carries a constant current I, see Fig 24.3 For convenience we consider positive charge carriers each having a charge q, and the number of carriers per unit volume in the rod is n Therefore, in this portion, the number of carriers is n A x and the total charge Q is: Q = (n A x) q (24.4) Fig 24.3 A portion of a x straight rod of uniform cross-sectional area A, + carrying a constant current I d + A The mobile charge carriers are d + assumed to be positive and move with an average speed vd I d d t Suppose that all the carriers move with an average speed vd (called the drift speed) Therefore, during a time interval t, all carriers must achieve a displacement x = vd t in the x direction Now, let us choose t such that the carriers in the cylindrical portion move through a displacement whose magnitude is equal to the length of the cylinder, see Fig 24.3 During such a time interval, all the charge carriers in this cylindrical portion must pass through the circular area A at the right end Accordingly, we write the last relation as: Q = (n A vd t) q (24.5) Therefore, the current I = Q/ t in the rod will be given by: I = n q vd A (24.6) The charge carriers in a solid conductor are all free electrons If the conductor is isolated, these electrons move with speeds of the order of 106 m/s, and because of their collisions with the scatterers (atoms or molecules in the conductor), they move randomly in all directions This results in a zero drift velocity and hence no net 812 24 Electric Circuits → charge transport, which means zero current When an electric field E is established → → across the conductor, this field exerts an electric force F = −e E on each electron, producing a current Of course, the electrons not move in a straight line along the conductor, but their resultant motion is complicated and zigzagged, see Fig 24.4 Regardless of the collisions of these electrons, they move slowly along the conductor → in a direction opposite to E with a drift velocity v→d , see Fig 24.4 Fig 24.4 A schematic Scatterer d Conductor representation of the random zigzag motion and the drift of a free electron with an average I - - speed vd in a conductor, due to J the effect of an external → E electric field E The current density J is defined as the current per unit area, i.e.: J= I A (24.7) Using the relation I = n q vd A, we get: J = n q vd (24.8) where the SI unit of the current density is A/m2 Equation 24.8 is valid only if J is uniform and the direction of I is perpendicular to the cross-sectional area A Generally, the current density is a vector quantity that has the direction of q v→d , for both signs of q; that is: → J = n q v→d (24.9) The amount of current that passes through an element of area dA, can be written → → → as J • dA , where dA is the vector area of the element The current that passes throughout the entire area A is thus: I= → → J •dA (24.10) → If the current density is uniform across the area and parallel to dA , then this equation leads to Eq 24.7 24.1 Electric Current and Electric Current Density 813 Example 24.1 Estimate the drift speed of the conduction electrons in a copper wire that is mm in diameter and carries a current of A Comment on your result The density of copper is 8.92 × 103 kg/m3 [Hint: Assume that each copper atom contributes one free conduction electron to the current.] Solution: To get the drift speed vd , we need to find the free-electron density n To get n, we need to know the volume occupied by one kmol of copper From the periodic table of elements, see Appendix C, the molar mass of copper is M(Cu) = 63.546 kg/kmol Recall that the mass of one kmol of 63.5 Cu contains Avogadro’s number of atoms (NA = 6.022 × 1026 atoms/kmol) Thus: Volume of kmol = Number of copper atoms/m3 = Mass of kmol Density ⇒ Avogadro’s number Volume of kmol V = ⇒ M ρ n= NA NA ρ = V M Therefore: n = (6.022 × 1026 atoms/kmol)(8.92 × 103 kg/m3 ) NA ρ = M 63.546 kg/kmol = 8.45 × 1028 atoms/m3 Since the density of free-electrons is equal to the density of copper atoms, then we use Eq 24.6 to find the drift speed as follows: vd = C/s I = ne A (8.45 × 1028 electrons/m3 )(1.6 × 10−19 C)(π × (10−3 m)2 ) = 2.35 × 10−5 m/s = 8.46 cm/h (Very small speed) You might ask why, even though vd is so small, that regular light bulbs light up very quickly when one turns on its circuit switch? The answer is that the electric field travels along the connecting wires of the circuit at almost the speed of light, so electrons everywhere in the wires all begin to drift at once with a small drift speed 814 24 Electric Circuits Example 24.2 One end of the copper wire in example is welded to one end of an aluminum wire with a mm diameter The composite wire carries a steady current equal to that of Example 24.1 (i.e I = A) (a) What is the current density in each wire? (b) What is the value of the drift speed vd in the aluminum? [Aluminum has one free electron per atom and density 2.7 × 103 kg/m3 ] Solution: (a) Except near the junction, the current density in a copper wire of radius rCu = mm and aluminum wire of radius rAl = mm are: I I 1A = = = 3.18 × 105 A/m2 −3 m)2 ACu π × (10 π rCu I I 1A = = = = 7.96 × 104 A/m2 AAl π × (2 × 10−3 m)2 π rAl JCu = JAl (b) From the periodic table of elements, see Appendix C, the molar mass of aluminum is M(Al) = 26.98 kg/kmol As in Example 24.1, we find: n= NA ρ (6.022 × 1026 atoms/kmol)(2.7 × 103 kg/m3 ) = M 26.98 kg/kmol = 6.03 × 1028 atoms/m3 vd = C/s I = ne A (6.03 × 1028 electrons/m3 )(1.6 × 10−19 C)(π × (2 × 10−3 m)2 ) = 8.25 × 10−6 m/s = 2.97 cm/h 24.2 Ohm’s Law and Electric Resistance As a result of maintaining a potential difference → → V across a conductor, an elec- tric field E and a current density J are established in the conductor For materials with electrical properties that are the same in all directions (isotropic materials), the electric field is found to be proportional to the current density That is: → → E =ρJ (Ohm’s law) (24.11) where the constant ρ is called the resistivity of the conductor Materials that obey this relation are said to obey Ohm’s law: Not to be confused with ρ referring to mass density or charge density 24.2 Ohm’s Law and Electric Resistance 815 Spotlight For many materials and most metals, the ratio of the magnitude of the electric field to the magnitude of the current density is a constant and does not depend on the electric field producing the current → → Since it is difficult to measure E and J directly, we need to put Ohm’s law into a more practical form This can be obtained by considering a portion of a straight conductor that has a uniform cross-sectional area A and length L, as shown in Fig 24.5 In addition, a potential difference V = Vb − Va between the ends of the conductor (denoted by a and b) will create a straight electric field and current, as also shown in Fig 24.5 Since charge carriers in conductors are electrons, they will drift from face → a to face b, against the field E Fig 24.5 A potential L b difference V = Vb − Va across a conductor of cross-sectional area A and d → q - a e A - length L sets up a field E and current I Vb E I J Va Recall that for uniform electric fields we have: V =EL (24.12) Using this relation to eliminate E from the scalar form of Eq 24.11, we get: V = ρJ L (24.13) Also, using J = I/A, the potential difference V can be written as: V = ρ L I A (24.14) The quantity in brackets is called the electrical resistance (or simply resistance) of the conductor and is denoted by the symbol R; that is: R=ρ L A ⇒ R∝ρ (24.15) 816 24 Electric Circuits We can define the resistance R as a proportionality constant to the relation V ∝ I and write the equivalent Ohm’s law as: Equivalent form V =IR of Ohm’s law The SI unit of resistance is ohm (abbreviated by = (24.16) ) That is: 1V 1A (24.17) This means that if one applies a potential difference of V across a conductor and this causes A to flow, then the resistance of the conductor is Note that according to Eq 24.15, the SI unit of resistivity is ohm-meter ( m) Also, since V = Vb − Va , we note that the direction of the current is in the direction of decreasing potential The inverse of resistivity is called the conductivity σ , thus: σ = ρ (24.18) where the SI unit of σ is ( m)−1 The resistance of a conductor can also be written in terms of the conductivity as follows: R= 1L σA (24.19) Equations 24.15 and 24.19 hold true only for isotropic conductors Additionally, the resistance in Eq 24.15 depends on the geometry of the resistor through the length L, area A, and resistivity ρ, which is a constant for a specific metallic conductor (assuming a constant temperature) A material obeying Ohm’s law is called an ohmic material or a linear material If a material does not obeys Ohm’s law, the material is called a non-ohmic or a nonlinear material Variation of Resistance with Temperature The variation of resistivity with temperature is mostly linear over a broad range Since R ∝ ρ, then for most engineering purposes a good empirical linear approximation for ρ and R can be written as: ρ = ρ◦ [1 + α(T − T◦ )] or R = R◦ [1 + α(T − T◦ )] (24.20) 24.2 Ohm’s Law and Electric Resistance 817 where ρ is the resistivity at temperature T (in degrees Celsius), ρ◦ is the resistivity at a reference temperature T◦ (usually selected to be 20 ◦ C), and α is the temperature coefficient of resistivity The same applies for the resistance The coefficient α is selected such that Eq 24.20 matches best with experimental measurements for the selected range of temperatures From Eq 24.20, we find that: ⎧ ⎪ ⎪ ⎨ ρ = ρ − ρ◦ R ρ = with α= (24.21) R = R − R◦ ⎪ ρ◦ T R◦ T ⎪ ⎩ T =T −T ◦ Table 24.1 lists the resistivity ρ and the temperature coefficient of resistivity α for some materials at 20 ◦ C Table 24.1 The resistivity and temperature coefficient of resistivity for various materials at 20◦ Celsius Material Resistivity ρ ( m) Temperature coefficient of resistivity α [(C◦ )−1 ] Silver 1.59 × 10−8 3.8 × 10−3 Copper 1.7 × 10−8 3.9 × 10−3 Gold 2.44 × 10−8 3.4 × 10−3 Aluminum 2.82 × 10−8 3.9 × 10−3 Tungsten 5.6 × 10−8 4.5 × 10−3 Iron 10 × 10−8 5.0 × 10−3 Platinum 11 × 10−8 3.98 × 10−3 Lead 22 × 10−8 3.9 × 10−3 Nichromea 1.50 × 10−6 0.4 × 10−3 Carbon 3.5 × 10−5 −0.5 × 10−3 Germanium 0.46 −48 × 10−3 Silicon 640 −75 × 10−3 Glass 1010 –1014 Hard rubber ∼1013 Sulfur 1015 Fused quartz 75 × 1016 a A nickel–chromium alloy commonly used in heating elements Most electric circuits use elements called resistors to control the current flowing through the circuit Values of the resistance are normally indicated by color-coding as shown in Tables 24.2 and 24.3 818 Table 24.2 Color-coding for resistors Table 24.3 Tolerance-coding 24 Electric Circuits Color Number Multiplier Black Brown 101 Red 102 Orange 103 Yellow 104 Green 105 Blue 106 Violet 107 Gray 108 White 109 Gold – 10−1 Silver – 10−2 Color Number Multiplier Tolerance – 10−1 5% Silver – 10−2 10% Colorless – – 20% Gold How to Read the Color-coding • First find the tolerance band; it will typically be gold (5%) or silver (10%), and sometimes colorless (20%), see the example shown in Fig 24.6 In this example, the color is Gold, so 5% tolerance Fig 24.6 1st digit 2nd digit Multiplier Tolerance Quality 24.2 Ohm’s Law and Electric Resistance 819 • Starting from the other end, identify the first band, write down the number associated with that color; in this example Blue is ‘6’ • Now read the next color, in this example it is Red, so write down a ‘2’ next to the six (You should have ‘62’ so far) • Now read the third color, which indicates the multiplier exponent band, and write that down as the power of ten for the multiplier of the resistance value In this example the multiplier is Yellow which represents ‘four’, so we get ‘62 × 104 ’ • If the resistor has one extra band past the tolerance band, it is a quality band Read the number as the % Failure rate per 1,000 h In this example it is Red, so that we can expect a 2% failure rate per 1,000 h All Ohmic resistors have a linear-potential-difference relationship over a broad band of applied potential differences The slope of the I versus V curve in the linear region yields a value for 1/R, see Fig 24.7 Fig 24.7 I S l ope 1/ R V Example 24.3 At 20 ◦ C, a copper wire has a diameter of mm, a length of 10 m, a resistivity of 1.7 × 10−8 m, a temperature coefficient of resistivity of 3.9 × 10−3 (C◦ )−1 , and carries a current of A (a) What is the current density in the wire? (b) What is the magnitude of the electric field applied to the wire? (c) What is the potential difference between the two ends of the wire? (d) What is the resistance of the wire? (e) When the wire is used in a thermometer for measuring the melting point of indium, the resistance calculated in part (d) increases to 0.0207 Find the melting point temperature of indium Solution: (a) The current density in a copper wire of radius mm is: J= I I 1A = = = 7.96 × 104 A/m2 A π r2 π × (2 × 10−3 m)2 820 24 Electric Circuits (b) From Eq 24.11, the electric field is given by: E = ρ J = (1.7 × 10−8 m)(7.96 × 104 A/m2 ) = 1.353 × 10−3 V/m (c) Using Eq 24.12, the potential difference will be given by: V = E L = (1.353 × 10−3 V/m)(10 m) = 1.353 × 10−2 V (d) From Eq 24.15, the resistance of the wire is: R=ρ L = 1.7 × 10−8 A m 10 m = 0.0135 π × (2 × 10−3 m)2 (e) Solving Eq 24.21 for T and then using: (1) the calculated resistance R◦ = 0.0135 at the reference temperature T◦ = 20 ◦ C, (2) the value of α, and (3) the final resistance R = 0.0207 , we obtain: T= R R − R◦ 0.0207 − 0.0135 = 136.8 C◦ = = αR◦ αR◦ [3.9 × 10−3 (C◦ )−1 ](0.0135 ) Since T◦ = 20 ◦ C, we find that the melting point of indium is: T = T◦ + T = 20 ◦ C + 136.8 C◦ = 156.8 ◦ C Example 24.4 A cylindrical shell of length L = 20 cm is made of aluminum and has an inner radius a = mm and an outer radius b = mm, see Fig 24.8 Assume that the shell has a uniform current density J = × 105 A/m2 in the direction of the wire’s length (a) What is the current through the shell? (b) What is the resistance of the shell and the potential difference V ? Solution: (a) Since the current density is uniform across any plane perpendicular to the length of the shell, we can use the relation I = J A to find the current First, we calculate the cross-sectional area of the shell as follows: A = π b2 − π a2 = π [b2 − a2 ] = π [(4 × 10−3 m)2 − (2 × 10−3 m)2 ] = 3.77 × 10−5 m2 Then, we use this result to find I as follows: I = J A = (2 × 105 A/m2 )(3.77 × 10−5 m2 ) = 7.54 A 24.2 Ohm’s Law and Electric Resistance 821 Uniform Current density J J b S a V L I Fig 24.8 (b) From Table 24.1, the resistivity of aluminum is 2.82 × 10−8 m We use Eq 24.15 to find the resistance of the shell, and then use Eq 24.16 to find the potential difference V as follows: R=ρ L = 2.82 × 10−8 A m 0.2 m = 1.5 × 10−4 3.77 × 10−5 m2 V = I R = (7.54 A)(1.5 × 10−4 ) = 1.13 × 10−3 V Example 24.5 A conducting rod of radius a = mm is concentric with a conducting cylindrical shell that has a radius b = mm and length L = 2.94 cm, see Fig 24.9a The space between the rod and the shell is tightly packed with silicon of resistivity ρ = 640 m A battery of potential difference V = 12 V is connected in such a way that the current through the silicon flows in the radial direction (a) Find the resistance of the silicon between the rod and the shell (b) Find the radial current in the circuit (c) Find the radial current density and electric field at the inner and outer surfaces of the silicon 822 24 Electric Circuits b a Radial direction of the current a S b I Silicon V r L dr Ja Conducting shell Silicon Conducting wire Jb (a) (b) Fig 24.9 Solution: (a) Cylindrical symmetry of the silicon suggests a radial flow of the cur→ rent density J Equation 24.15 cannot be used directly because the cross section through which the charge travels varies from π a L (at the inner cylindrical face) to π b L (at the outer cylindrical face) Therefore, we consider a cylindrical silicon shell element of an inner radius r, height L, face area A = π r L, and thickness d r, see Fig 24.9b This shell element has a resistance d R In this case, Eq 24.15 will take the following form: dR=ρ dr 2π rL To find the total resistance across the entire silicon, we must integrate the previous expression from r = a to r = b Thus: b b dR= R= a ρ a ρ dr = 2π rL 2π L b a ρ b dr = ln r 2π L a Now, substituting with the given values, we get: R= b ρ ln 2π L a = 640 m mm ln −2 π (2.94 × 10 m) mm = 2.4 × 103 (b) Knowing the resistance R and the potential difference V , we use Ohm’s law given by Eq 24.16 to find the total current in the silicon (which is the current in the circuit) as follows: [...]... equivalent capacitance of two capacitors in series, one capacitor with a dielectric of thickness a and the second an air-filled capacitor of thickness d − a ( 32) A parallel-plate capacitor of plate area A and separation d is filled in two different ways with two dielectrics κ1 and 2 as shown in parts (a) and (b) of Fig 23 . 32 Show that the capacitances of the two capacitors of parts (a) and (b) are:... volume occupied by one kmol of copper From the periodic table of elements, see Appendix C, the molar mass of copper is M(Cu) = 63.546 kg/kmol Recall that the mass of one kmol of 63.5 Cu contains Avogadro’s number of atoms (NA = 6. 022 × 1 026 atoms/kmol) Thus: Volume of 1 kmol = Number of copper atoms/m3 = Mass of 1 kmol Density ⇒ Avogadro’s number Volume of 1 kmol V = ⇒ M ρ n= NA NA ρ = V M Therefore:... Show that the capacitor is equivalent to two capacitors in series, each having a plate separation (d − a) /2, as shown in part (b) of the figure, and show that the capacitance after inserting the slab is given by: C= A d a 806 23 Capacitors and Capacitance Fig 23 .31 See Exercise (30) A A (d -a) / 2 d Copper a (d -a) / 2 (a) (b) (31) Show that the capacitance of the capacitor in Fig 23 .10b can be obtained... conductivity as follows: R= 1L A (24 .19) Equations 24 .15 and 24 .19 hold true only for isotropic conductors Additionally, the resistance in Eq 24 .15 depends on the geometry of the resistor through the length L, area A, and resistivity ρ, which is a constant for a specific metallic conductor (assuming a constant temperature) A material obeying Ohm’s law is called an ohmic material or a linear material If a material... 23 .26 In all configurations, the potential difference is 22 V How many coulombs of charge pass from the battery to each combination? (25 ) When the three capacitors C1 = 2 µF, C2 = 1 µF, and C3 = 4 µF are connected to a source of a potential difference V, as shown Fig 23 .27 , the charge Q2 on C2 is found to be 10 µC (a) Find the values of the charges on the two capacitors C1 and C3 (b) Determine the value... material does not obeys Ohm’s law, the material is called a non-ohmic or a nonlinear material Variation of Resistance with Temperature The variation of resistivity with temperature is mostly linear over a broad range Since R ∝ ρ, then for most engineering purposes a good empirical linear approximation for ρ and R can be written as: ρ = ρ◦ [1 + α(T − T◦ )] or R = R◦ [1 + α(T − T◦ )] (24 .20 ) 24 .2 Ohm’s Law... (26 ) C1 C3 C2 C4 ΔV (27 ) For each of the combinations shown in Fig 23 .29 , find a formula that represents the equivalent capacitance between the terminals A and B (28 ) Assume that in Exercise 27 , C = 12 µF and VBA = 12 V For each combination, find the magnitude of the total charge that the source between A and B will distribute on the capacitors 23 .6 Exercises 805 C A A C C C C C C C C C A C C A (a) ... What is the charge Q on the equivalent capacitor? (c) What is the potential difference VBA between A and B? (d) Find the final charge on each capacitor Q1 C1 Q2 C2 Q1 f C1 Ceq A A B Q B Q2 f ΔV ΔV C2 (a) (b) Fig 23 .30 See Exercise (29 ) (30) A parallel-plate capacitor has an area A and separation d A slab of copper of thickness a is inserted midway between the plates, see part (a) of Fig 23 .31 Show... current across an area can be expressed in terms of the motion of the charge carriers To achieve this we consider a portion of a cylindrical rod that has a cross-sectional area A, length x, and carries a constant current I, see Fig 24 .3 For convenience we consider positive charge carriers each having a charge q, and the number of carriers per unit volume in the rod is n Therefore, in this portion, the... number of carriers is n A x and the total charge Q is: Q = (n A x) q (24 .4) Fig 24 .3 A portion of a x straight rod of uniform cross-sectional area A, + carrying a constant current I d + A The mobile charge carriers are d + assumed to be positive and move with an average speed vd I d d t Suppose that all the carriers move with an average speed vd (called the drift speed) Therefore, during a time interval