61 CHAPTER 3 Exercises E3.1 V )10sin(5.0)102/()10sin(10/)()( 5656 ttCtqtv =×== −− A )10cos(1.0)10cos()105.0)(102()( 5556 tt dt dv Cti =××== − E3.2 Because the capacitor voltage is zero at t = 0, the charge on the capacitor is zero at t = 0. ms 4ms 2 for 101041010 ms 20 for 1010 0)()( 3-6 32E 3 32E 0 3 3 0 3 0 ≤≤−×=−+= ≤≤== += − − − − − −− ∫∫ ∫ ∫ ttdxdx ttdx dxxitq t t t ms 4ms 2 for 1040 ms 20 for 10 /)()( 4 4 ≤≤−= ≤≤= = tt tt Ctqtv ms 4ms 2 for 101040 ms 20 for 10 )()()( 3 ≤≤+×−= ≤≤= = − tt tt tvtitp ms 4 ms 2 for )1040(105.0 ms 20 for 5 2/)()( 247 2 2 ≤≤−×= ≤≤= = − tt tt tCvtw in which the units of charge, electrical potential, power, and energy are coulombs, volts, watts and joules, respectively. Plots of these quantities are shown in Figure 3.8 in the book. E3.3 Refer to Figure 3.10 in the book. Applying KVL, we have 321 vvvv ++= Then using Equation 3.8 to substitute for the voltages we have 62 )0()( 1 )0()( 1 )0()( 1 )( 3 03 2 02 1 01 vdtti C vdtti C vdtti C tv ttt +++++= ∫∫∫ This can be written as )0()0()0()( 111 )( 321 0321 vvvdtti CCC tv t +++ ++= ∫ (1) Now if we define )0()0()0()0( and 1111 321 321eq vvvv CCCC ++= ++= we can write Equation (1) as )0()( 1 )( 0 vdtti C tv t eq += ∫ Thus the three capacitances in series have an equivalent capacitance given by Equation 3.25 in the book. E3.4 (a) For series capacitances: F 3/2 1/12/1 1 /1/1 1 21 eq µ= + = + = CC C (b) For parallel capacitances: F 321 21eq µ=+=+= CCC E3.5 From Table 3.1 we find that the relative dielectric constant of polyester is 3.4. We solve Equation 3.26 for the area of each sheet: 2 12 66 0 m 4985.0 1085.84.3 101510 = ×× ×× === − −− εεε r CdCd A Then the length of the strip is m 93.24)102/(4985.0/ 2 =×== − WAL E3.6 ()[]()V 10sin1010cos1.0)1010( )( )( 443 tt dt d dt tdi Ltv −=×== − ( )[ ] ()J 10cos105010cos1.0105)()( 426 2 432 2 1 tttLitw −− ×=××== 63 E3.7 s 20 forV 1025105.76667 )( 10150 1 )0()( 1 )( 29 0 6 0 6 0 µ≤≤×=×= × =+= ∫ ∫∫ − ttxdx dxxvidxxv L ti t tt s 4s2 forV 1.0105.76667 -6E2 0 6 µµ ≤≤=×= ∫ txdx () s 5s4 forV 105.015105.76667 5 t -6E4 -6E2 0 6 µµ ≤≤−= −+×= ∫∫ ttdxxdx A plot of i(t) versus t is shown in Figure 3.19b in the book. E3.8 Refer to Figure 3.20a in the book. Using KVL we can write: )()()()( 321 tvtvtvtv ++= Using Equation 3.28 to substitute, this becomes dt tdi L dt tdi L dt tdi Ltv )()()( )( 321 ++= (1) Then if we define 321eq LLLL ++= , Equation (1) becomes: dt tdi Ltv )( )( eq = which shows that the series combination of the three inductances has the same terminal equation as the equivalent inductance. E3.9 Refer to Figure 3.20b in the book. Using KCL we can write: )()()()( 321 titititi ++= Using Equation 3.32 to substitute, this becomes )0()( 1 )0()( 1 )0()( 1 )( 3 03 2 02 1 01 idttv L idttv L idttv L ti ttt +++++= ∫∫∫ This can be written as )0()0()0()( 111 )( 321 0321 iiidttv LLL tv t +++ ++= ∫ (1) Now if we define )0()0()0()0( and 1111 321 321eq iiii LLLL ++= ++= we can write Equation (1) as 64 )0()( 1 )( 0 idttv L ti t eq += ∫ Thus, the three inductances in parallel have the equivalent inductance shown in Figure 3.20b in the book. E3.10 Refer to Figure 3.21 in the book. (a) The 2-H and 3-H inductances are in series and are equivalent to a 5- H inductance, which in turn is in parallel with the other 5-H inductance. This combination has an equivalent inductance of 1/(1/5 + 1/5) = 2.5 H. Finally the 1-H inductance is in series with the combination of the other inductances so the equivalent inductance is 1 + 2.5 = 3.5 H. (b) The 2-H and 3-H inductances are in series and have an equivalent inductance of 5 H. This equivalent inductance is in parallel with both the 5-H and 4-H inductances. The equivalent inductance of the parallel combination is 1/(1/5 + 1/4 + 1/5) = 1.538 H. This combination is in series with the 1-H and 6-H inductances so the overall equivalent inductance is 1.538 + 1 + 6 = 8.538 H. Problems P3.1 Capacitors consist of two conductors separated by an insulating material. Frequently the conductors are sheets of metal that are separated by a thin layer of the insulating material. P3.2 A dielectric material is an electrical insulator through which virtually no current flows assuming normal operating voltages. Some examples of dielectrics mentioned in the text are air, Mylar, polyester, polypropylene, and mica. Some others are porcelain, glass, and certain types of oil. P3.3 Because we have dtCdvi /= for a capacitance, the current is zero if the voltage is constant. Thus we say that capacitances act as open circuits for constant dc voltages. 65 P3.4 The net charge on each plate is 02.0200)10100( 6 =××== − CVQ C. One plate has a net positive charge and the other has a net negative charge so the net charge for both plates is zero. P3.5* s 2000 05.0 100 V/s 05.0 102000 10100 6 6 === = × × == = − − dtdv v t C i dt dv dt dv Ci ∆ ∆ P3.6 mC 510105 36 =××== − CvQ ( ) J 5.210105 2 1 2 1 2 362 =×××== − CvW MW 5.2 10 5.2 6 === − t W P ∆ ∆ P3.7* () dt dv Cti = ()t dt d 1000sin10010 5− = ()t1000cos= () ()()titvtp = ()( )tt 1000sin1000cos100= ()t2000sin50= ()[] 2 2 1 )( tvCtw = ()t1000sin05.0 2 = 66 P3.8 () dt dv Cti = () t e dt d 1006 10010 −− = A 01.0 100t e − −= () ()()titvtp = W 200t e − −= () ()[] 2 2 1 tvCtw = J 105 2003 t e −− ××= P3.9 67 P3.10 () () () ∫ += t vdtti C tv 0 0 1 () () ∫ ×= t dttitv 0 6 102 () ()()titvtp = () ()tCvtw 2 2 1 = ( )tv 26 1025.0 ××= − P3.11 () () () ∫ += t vdtti C tv 0 0 1 () () ∫ +×= t dttitv 0 6 1010333.0 () ()()titvtp = 68 () ()tCvtw 2 2 1 = ()tv 26 105.1 ××= − P3.12 () () () ∫∫ =+××=+= − tt tdtvdtti C tv 0 33 0 1000)105(10200 1 V 2)1020( 3 =× − v tvip 5.0== mW 10)1020( 3 =× − p () () 22 25.0 2 1 ttCvtw == Jw µ 100)1020( 3 =× − P3.13 We can write 200105 2 1 262 =×== − vCvw Solving, we find V. 6325 =v Then because the stored energy is decreasing, the power is negative. Thus, we have mA 79.06- 6325 500 = − == v p i The minus sign shows that the current actually flows opposite to the passive convention. Thus, current flows out of the positive terminal of the capacitor. P3.14 () () [][])200sin(1010010200)()( 66 t dt d tvtC dt d dt tdq ti −− ×+×=== () mA )200cos(20 tti = P3.15 By definition, the voltage across a short circuit must be zero. For an initially uncharged capacitance, we have () () ∫ = t dtti C tv 0 1 For the voltage to be zero for all values of current and time, the capacitance must be infinite. Thus, an infinite initially uncharged capacitance is equivalent to a short circuit. 69 For an open circuit, the current must be zero. For a capacitance, we have () () dt tdv Cti = Thus, a capacitance of zero is equivalent to an open circuit. P3.16 A capacitance initially charged to 10 V has () () ∫ += t dtti C tv 0 10 1 However, if the capacitance is infinite, this becomes () V 10=tv which describes a 10-V voltage source. Thus, a very large capacitance initially charged to 10 V is an approximate 10-V voltage source. P3.17* time power×=W s 3600hp/W 746hp 5 ××= J 104.13 6 ×= kV 8.51 01.0 104.1322 6 = ×× == C W V It turns out that a 0.01-F capacitor rated for this voltage would be much too large and massive for powering an automobile. Besides, to have reasonable performance, an automobile would need much more than 5 hp for an hour. P3.18 () () dt tdv Cti = ( ) ( )[] tt dt d 556 10sin210cos31020 +×= − ( ) ( )tt 55 10cos410sin6 +−= () ()()titvtp = ( ) ( )[ ] ( ) ( )[ ]ttt t 5555 10cos410sin610sin210cos3 +−+= Evaluating at 0=t , we have ( ) W 120 =p . Because ( )0p is positive, we know that the capacitor is absorbing energy at 0=t . Evaluating at 5 2 102 − ×= πt , we have ( ) W 12 2 −=tp . Because ( ) 2 tp is negative, we know that the capacitor is supplying energy at 2 tt = . 70 P3.19 () () () ∫ += t vdtti C tv 0 0 1 () ∫ −×= − t dttv 0 34 2010510 () V 2050 −= ttv () ()()tvtitp = ()W 2050105 3 −×= − t Evaluating at 0=t , we have ( ) W 1.00 −=p . Because the power has a negative value, the capacitor is delivering energy. At s 1=t , we have () W 15.01 =p . Because the power is positive, we know that the capacitor is absorbing energy. P3.20 Capacitances in parallel are combined by adding their values. Thus capacitances in parallel are combined as are resistances in series. Capacitances in series are combined by taking the reciprocal of the sum of the reciprocals of the individual capacitances. Thus capacitances in series are combined as are resistances in parallel. P3.21* (a) F 2 2121 1 1 µ= + += eq C (b) The two 4- µF capacitances are in series and have an equivalent capacitance of F 2 4141 1 µ= + . This combination is a parallel with the 2- µF capacitance, giving an equivalent of 4 µF. Then the 6 µF is in series, giving a capacitance of F 4.2 4161 1 µ= + . Finally, the 3 µF is in parallel, giving an equivalent capacitance of F 4.54.23 µ=+= eq C . P3.22 (a) () F 667.4 11121 1 1121 1 3 µ= ++ + + += eq C (b) ()() F 2 241121 1 µ= +++ = eq C 71 P3.23 FC eq µ= ++ = 5.1 )21/(13/1 1 P3.24 We obtain the maximum capacitance of 4 µF by connecting all four 1- µF capacitors in parallel. We obtain the minimum capacitance of 1/4 µF by connecting all four 1- µF capacitors in series. P3.25 F 32 11 1 21 µ= + = CC C eq The charges stored on each capacitor and on the equivalent capacitance are equal. V 4 V 8 C 8V 12 2 2 1 1 == == =×= C Q v C Q v CQ eq µ As a check, we verify that V 12 21 =+vv . P3.26* As shown below, the two capacitors are placed in series with the heart to produce the output pulse. While the capacitors are connected, the average voltage supplied to the heart is 4.95 V. Thus, the average current is mA 9.950095.4 == pulse I . The charge removed from each capacitor during the pulse is C 9.9ms 1mA 9.9 µ=×=∆Q . This results in a 0.l V change in voltage, so we have F 99 1.0 109.9 2 6 µ= × === − v QC C eq ∆ ∆ . Thus, each capacitor has a 72 capacitance of F 198 µ=C . Then as shown below, the capacitors are placed in parallel with the 2.5 V battery to recharge them. The battery must supply 9.9 µC to each battery. Thus, the average current supplied by the battery is A 8.19 1s C 9.92 µ µ = × = battery I . The ampere-hour rating of the battery is hours Ampere 867.0243655108.19 6 =×××× − . P3.27 d WL d A C rr 00 εεεε == (a) Thus if W and L are both doubled, the capacitance is increased by a factor of four resulting in 400=C pF. (b) If d is doubled the capacitance is cut in half resulting in 50=C pF. (c) The relative dielectric constant of air is approximately unity. Thus replacing air with oil increases ε r by a factor of 25 increasing the capacitance to 2500 pF. P3.28* F 398.0 1001.0 103010101085.815 3 2212 0 µ εε = × ×××××× == − −−− d A C r P3.29 Using d WL d A C rr 00 εεεε == and KdV = max to substitute into 2 maxmax 2 1 CVW = , we have WLdKdK d WL W r r 2 0 220 max 2 1 2 1 εε εε == . However, the volume of the dielectric is Vol = WLd, so we have Vol)( 2 1 2 0max KW r εε= Thus, we conclude that the maximum energy stored is independent of W, L, and d if the volume is constant and if both W and L are much larger than d. To achieve large energy storage per unit volume, we should look for a dielectric having a large value for . 2 K r ε The dielectric should have high relative dielectric constant and high breakdown strength. 73 P3.30* The charge Q remains constant because the terminals of the capacitor are open-circuited. ()() J 50021 C 11000101000 2 111 12 11 µ µ == =××== − VCW VCQ After the distance between the plates is doubled, the capacitance becomes pF 500 2 =C . The voltage increases to V 2000 10500 10 12 6 2 2 = × == − − C Q V and the stored energy is ()( ) J 100021 2 222 µ== VCW . The additional energy is supplied by the force needed to pull the plates apart. P3.31 Before the switch closes, the energies are ()() ()() J 500021 J 500021 2 222 2 111 µ µ == == VCW VCW Thus, the total stored energy is 10,000 µJ. The charge on the top plate of 1 C is C 100 111 µ+== VCQ . The charge on the top plate of 2 C is C 100 222 µ−== VCQ . Thus, the total charge on the top plates is zero. When the switch closes, the charges cancel, the voltage becomes zero, and the stored energy becomes zero. Where did the energy go? Usually, the resistance of the wires absorbs it. If the superconductors are used so that the resistance is zero, the energy can be accounted for by considering the inductance of the circuit. (It is not possible to have a real circuit that is precisely modeled by Figure P3.31; there is always resistance and inductance associated with the wires that connect the capacitances.) P3.32 Refering to Figure P3.32 in the book, we see that the transducer consists of two capacitors in parallel: one above the surface of the liquid and one below. Furthermore, the capacitance of each portion is proprotional to its length and the relative dielectric constant of the material between the plates. Thus for the portion above the liquid, the capacitance in pF is 100 100 200 x C above − = 74 in which x is the height of the liquid in cm. For the portion of the plates below the surface of the liquid: 100 )25(200 x C below = Then the total capacitance is: pF48200 xC CCC belowabove += += P3.33 The capacitance of the microphone is )]200sin(01.01[1085.8 10)]200sin(01.01[ 101085.8 11 3 212 0 t td A C −×≅ + ×× = ε = − − −− The current flowing through the microphone is A)200cos(104.35 ][)( )( 9 t dt Cvd dt tdq ti − ×−≅== P3.34 () () ()[]()tt dt d dt tdv Cti c c 100sin10100cos1010 47 −− −=== () () ( ) () () () () () ()tttv tvtvtv ttRitv rc cr 100sin10100cos10 100sin10 3 3 − − −= += −== Thus, () ()tvtv c = to within 1% accuracy, and the resistance can be neglected. Repeating for () ( )ttv c 7 10cos1.0= , we find () ( ) () () () () () ()()tttvtvtv tv tti rc r c 77 7 7 10sin10cos1.0 10sin 10sin1.0 −=+= −= −= Thus, in this case, the voltage across the parasitic resistance is larger in magnitude than the voltage across the capacitance. P3.35* For square plates, we have ,WL = the plate area is , 2 LA = and the volume of the dielectric is Vol dL 2 = . The minimum thickness of the dielectric is mm 3125.0 1032 1000 5 max min = × == K V d 75 The required volume is () 36 2 512 3 2 0 max m 1007.22 10321085.8 1022 Vol − − − ×= ×× × === K W Ad r εε and the area is 2 m 07062.0Vol == dA The length of each side of the square plate is m 2657.0== AL P3.36 Inductors consist of coils of wire wound on coil forms such as toriods. P3.37 Because the energy stored in an inductor is 2 2 1 Liw = , the energy stored in the inductor decreases when the current magnitude decreases. Therefore, energy is flowing out of the inductor. P3.38 Because we have dt tdi Ltv L L )( )( = , the voltage is zero when the current is constant. Thus we say that inductors act as short circuits for steady dc currents. P3.39 A fluid flow analogy for an inductor consists of an incompressible fluid flowing through a frictionless pipe of constant diameter. The pressure differential between the ends of the pipe is analogous to the voltage across the inductor and the flow rate of the liquid is proportional to the current. (If the pipe had friction, the electrical analog would have series resistance. If the ends of the pipe had different diameters, a pressure differential would exist for constant flow rate, whereas an inductance has zero voltage for constant flow rate.) P3.40* H 2=L 76 () () dt tdi Ltv L L = () ()()titvtp LL = () ()[] 2 2 1 tiLtw L = P3.41 H 1.0=L () ()A 1000sin5.0 tti L = () () ()V 1000cos50 t dt tdi Ltv L L = = () ()() ()() ()W 2000sin5.12 1000sin1000cos25 t tt titvtp LL = = = 77 () ()[] ()J 1000sin0125.0 2 1 2 2 t tiLtw L = = P3.42 H 2=L () t L eti 20 5 − = () () V 200 20t L L e dt tdi Ltv − −= = () ()() ()() W 1000 5200 40 2020 t tt LL e ee titvtp − −− −= −= = () ()[] J 25 2 1 40 2 t L e tiLtw − = = P3.43 H 2=L 78 () () () () ∫ ∫ = += t L t LLL dttv idttv L ti 0 0 2 1 0 1 () ()()titvtp LL = () ()[] ()[] 2 2 2 1 ti tiLtw L L = = P3.44 H 10 µ=L () ( )ttv L 6 10sin5= () () () () ()A 10cos5.0 5.010sin510 0 1 6 0 65 0 t dtt idttv L ti t t LLL −= −= += ∫ ∫ () ()() ()() ()t tt titvtp LL 6 66 102sin25.1 10cos10sin5.2 ×−= −= = 79 () ()[] ()J 10cos25.1 2 1 62 2 µt tiLtw L = = P3.45* () () () A 1.0102 1.0101020 0 1 5 0 3 0 −×= −×= += ∫ ∫ t dt idttv L ti t t LLL Solving for the time that the current reaches mA 100+ , we have () s 1 1.01021.0 0 5 µ= −×== x xL t tti P3.46* V 10 2.0 4 5.0 === dt di Lv L P3.47 80 P3.48 () () () ∫ += 0 0 0 0 1 t LLL idttv L ti A 333.305)2( 2 0 3 1 =+= ∫ dti L () ()() W 67.16222 == LL ivp J 67.16)2()2( 2 2 1 == L Liw P3.49 A 14.14)5(J 200)5()5( 2 2 1 =⇒== LL iLiw Since a reference is not specified, we can choose A. 14.14)5( += L i Also, because the stored energy is increasing, the power for the inductor carries a plus sign. Thus )5()5(100)5( LL ivp =+= and we have V. 071.7)5( += L v Finally, because the current and voltage have the same algebraic signs, the current flows into the positive polarity. P3.50 Because we the current through an open circuit is zero by definition and we have () ()dttv L ti t t ∫ = 0 1 (assuming zero initial current), infinite inductance corresponds to an open circuit. Because the voltage across a short circuit is zero by definition and dt tdi Lv L L )( = , we see that 0=L corresponds to a short circuit. P3.51 For an inductor with an initial current of 10 A, we have () () 10 1 0 += ∫ dttv L ti t t . However for an infinite inductance. this becomes () 10+=ti which is the specification for a 10-A current source. Thus, a very large inductance with an initial current of 10 A is an approximation to a 10-A current source. P3.52 Inductances are combined in the same way as resistances. Inductances in series are added. Inductances in parallel are combined by taking the reciprocal of the sum of the reciprocals of the several inductances. P3.53* (a) () H 333.2 31121 1 1 = ++ += eq L (b) 2 H in parallel with 2 H is equivalent to 1H. Also, 2 H in parallel with 6 H is equivalent to 1.5 H. Finally, we have () H 538.1 5.11141 1 = ++ = eq L 81 P3.54 (a) The 2 H inductors and 0.5 H inductor have no effect because they are in parallel with a short circuit. Thus, H 1= eq L . (b) The two 2-H inductances in parallel are equivalent to 1 H. Also, the 1 H in parallel with the 3 H inductance is equivalent to 0.75 H. Thus, ()( ) H 158.2 75.021111 1 1 = +++ += eq L . P3.55 If all four inductors are connected in series, we obtain the maximum inductance: H 8 max =L By connecting all four inductors in parallel, we obtain the minimum inductance: H 5.0 21212121 1 min = +++ =L P3.56 Ordinarilly negative inductance is not practical. Thus, adding inductance in series always increases the equivalent inductance and placing inductance in parallel results in smaller inductance. Thus, we need to consider a parallel inductance such that 2 5/1/1 1 = +L Solving we find that 333.3=L H. P3.57 In this case we need to place 3 H in series with the original 5-H inductance. P3.58* () () () ∫∫ + == t t eq dttv LL LL dttv L ti 0021 21 1 () () ∫ = t dttv L ti 01 1 1 Thus, we can write () () ()titi LL L ti 3 2 21 2 1 = + = . Similarly, we have () () ()titi LL L ti 3 1 21 1 2 = + = . This is similar to the current-division principle for resistances. Keep in mind that these formulas assume that the initial currents are zero. 82 P3.59 (a) In this case to obtain 1% accuracy, the resistance can be neglected. (b) For () ()tti 10cos1.0= , we have () () () () () () ()tttv ttv ttv L R 10sin01.010cos1.0 10sin01.0 10cos1.0 −= −= = Thus in this case, the parasitic resistance cannot be neglected. P3.60 See Figure 3.22 in the book. P3.61 When a time-varying current flows in a coil, a time-varying magnetic field is produced. If some of this field links a second coil, voltage is induced in it. Thus time-varying current in one coil results in a contribution to the voltage across a second coil. P3.62 Refer to Figures 3.23 and P3.62. For the dots as shown in Figure P3.62, we have () ( ) ( ) () () () () ()t dt tdi L dt tdi Mtv t dt tdi M dt tdi Ltv 10cos20 10cos15 2 2 1 2 21 11 =+= =+= P3.63* With the dot moved to the bottom of 2 L , we have () ( ) ( ) () () () () 0 10cos5 2 2 1 2 21 11 =+−= =−= dt tdi L dt tdi Mtv t dt tdi M dt tdi Ltv () () ( ) () () () () () () () ()tt tvtvtv dt tdi Ltv tRitv LR L R 55 5 5 10sin10010cos1.0 10sin100 10cos1.0 −= += −== == 83 P3.64* (a) As in Figure 3.23a, we can write () ( ) ( ) () () () dt tdi L dt tdi Mtv dt tdi M dt tdi Ltv 2 2 1 2 21 11 += += However, for the circuit at hand, we have ( ) ( )()tititi 21 == . Thus, () () ( ) () () () dt tdi MLtv dt tdi MLtv += += 22 11 Also, we have () ( ) ( )tvtvtv 21 += . Substituting, we obtain () () ( ) dt tdi LMLtv 21 2 ++= . Thus, we can write () ( ) dt tdi Ltv eq = , in which 21 2 LMLL eq ++= . (b) Similarly, for the dot at the bottom end of 2 L , we have 21 2 LMLL eq +−= P3.65 In general, we have dt di L dt di Mtv dt di M dt di Ltv 2 2 1 2 21 11 )( )( +±= ±= Substituting the given information, we have 84 )1000sin(10)1000sin(10 )1000sin(102)( 44 4 1 tMt ttv m= ×−= We deduce that H. 1=M Furthermore, because the lower of the two algebraic signs applies, we know that the currents are referenced into unlike terminals. Owner Microsoft Word - Chapter_03.doc