Solution Manual For Electrical Engineering: Concepts and Applications, 1st Edition

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Chapter 2 SolutionsSection 2.1Introduction2.1Current source2.2Voltage source2.3Resistor2.4Capacitor2.5InductorSection 2.2Charge and Current2.6b)The current direction is designated as the direction of the movement of positivecharges.2.7The relationship of charge and current is( )( )()00tqdttitqtt+=∫so( )()()0010sin2tqdtttqtt+=∫π( )()()010cos1020tqttqtt+⎥⎦⎤⎢⎣⎡ −=ππ2.8The coulomb of one electron is denoted byeand( )( )()00tqdttitqtt+=∫So( )()00121/)(tqdtteetqtntt+==∫Ift0= 0 and()00=tq,( )26tetn=2.9

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( )∫=idttq( )tdtqt∫=05( )ttq5=2.10( )[]( )( )250555550=−==ttqCoulombs2.11Using the definition of current-charge relationship, the equation can be rewrittenas:etndtdqiΔΔ==Thus, the current flow within t1and t2time interval is,Ai3)106.1(210)275.5(1919−=×−×−=−The negative sign shows the current flow in the opposite direction with respect to theelectric charge.2.12Assuming the area of the metal surface isS, The mass of the nickel with depthd= 0.15mm ism=ρ×d×SMeanwhile, using the electro-chemical equivalent, the mass of the nickel can beexpressed asm=k×I×twhereI=σ×S.Equating the two expressions of the mass, the coating time is found:t =ρ×d /σ= 1.24×105s≈34.4 hourSection 2.3Voltage2.13By the definition of voltage, when a positive charge moves from high voltage tolow voltage, its potential energy decreases.So a is “+”, b is “-”. In other words,uab=1V.2.14The currenti(t) is defined as:( )⎭⎬⎫⎩⎨⎧≤<=elsewheretti0103Therefore, the charge

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3310∫==dtgCThe energy in Joules is given by:JCVJ1535=×=×=2.151 electron =19106.1−×−Coulombs.Therefore, there are181025.6×electrons in a coulomb.Coulombs of16105×electrons =318161081025.6105−×=××Therefore, the voltage is1875108153=×==−CJVV2.16( )dtttq⎟⎠⎞⎜⎝⎛=∫π23sin210C4244.023cos3410=⎥⎦⎤⎢⎣⎡⎟⎠⎞⎜⎝⎛−=tππJ122.24244.05=×=×=CVJ2.17CVJq10220==AsCdtdqi5.2410===Section 2.4 Respective Direction of Voltage and Current2.18True2.19False2.20TrueSection 2.5 Kirchoff’s Current Law2.21According to KCL

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3702−−=iTherefore,Ai42=2.22Notice that4230i+−=Therefore,Ai14−=It can be seen thatAi25−=i6can now be found.61250i−++=Ai86=2.23i= 10mA5kΩ2kΩ13kΩ+v-i1i2Figure S2.23: Circuit for Problem 2.23.According to KCL,2110iimA+=and21182kiki=Therefore,mAi91=,mAi12=VmAkRIV13113=×=×=2.24Using the KCL method, we have

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321iii+=Also, at the other end of the node, we have,324iii+=Then, combine these two equations we obtain,41ii=2.25The algebraic sum of currents flowing into the node in Figure 2.11 isi1+i2-i3-i4wherei1andi2flow in, andi3andi4flow outAs current is defined as the rate of variation of the charge,i1+i2-i3-i4= dq/dtBy law of the conservation of charge, charges cannot be stored in the node and chargescan neither be destroyed nor created. Therefore the variation rate of the charge is zero andfurther the KCL law holds,i1+i2-i3-i4= 02.26The resistance across the voltage source isΩ=+×+=+=50120401204020120//4020RSo the current flowing through the voltage source isAVRVI1.0505=Ω==Therefore,mAi251.016040=×=2.27By KCL,xIVV23530−+−=Note that,xxIVVI33=→=Therefore,xxxIII25330−+−=xxII5330−−=

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A5.123−=→−=xxII2.28The voltage across node C and D (node C is positive) isU2- U1, the voltageacross node C and B isUAB- U1, and the voltage across node D and B isUAB- U2The KCL equation for node C isU1/R = (U2- U1) /R + (UAB- U1) /2Rwhich is5U1- 2U2= UABThe KCL equation for node D isU2/2R + (U2- U1) /R = (UAB- U2) /Rwhich is5U2- 2U1= 2UABSolvingU2andU1,U1= 9UAB/21, U2= 12UAB/21Thus, by the KCL equation at node AIAB= U1/R + U2/2RRAB=UAB/IAB= 7R /5R2RR2RRA+-BCDIABUAB+-+-U1U2Figure S2.28: Circuit for Problem 2.28.Section 2.6 Kirchoff’s Voltage Law2.29By KVL,DCBAvvvv+−−−=06620++−−=AvThenVvA10=2.30Using KVL,146120+−+−=RvThen4=RvV

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2.31Using KVL,530++−=DvThen8=DvVUsing KVL,8280++−−=CvThen2=CvV2.32Using KVL,Cv−+−=490Then5−=CvVUsing KVL,65350−−−+−=EvThen13−=EvV2.33Using KVL,340−+−=DvThen1=DvVUsing KVL,Cv+++=1310Then5−=CvVUsing KVL,035=−+EvThen8=EvV2.34Using KVL,170+−=EvThen8=EvVUsing KVL,850++−=BvThen13=BvVUsing KVL,750−−=DvThen

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2−=DvVUsing KVL2130−+=AvThen11−=AvV2.35Using KVL,VVVHH8210=→+=VVVVIIH552=→+=+VVVCC264=→=+VVVVDCD145=→+=+2.36V= 6V+-R1R24Ω6Ω3Ωii1i2i3i4Figure S2.36: Circuit for Problem 2.36.a)R1=6Ω,R2=3ΩBy KCL,i1+i = i2i4+i = i3Applying KVL on the loop made by 6V source,R1andR2,6i1+ 3i2= 6Applying KVL on the loop made by 6V source, 3Ωresistor and 6Ωresistor,3i3+ 6i4= 6Applying KVL on the loop made by 3Ωresistor, 4Ωresistor andR1,3i3+ 4i =6i1Now we have 5 equations for 5 variables andican be solved using Cramer’s ruleor variable elimination. Then we geti =1/4A,i1= 7/12A,i2= 5/6A,i3= 5/6A,i4= 7/12Ab)i =0 =>i1= i2,i3= i4.

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Moreover, the voltage across the 4Ωresistor is zero. This also leads to the voltageacrossR1equals to the voltage across the 3Ωresistor and the voltage acrossR2equals to the voltage across the 6Ωresistor,R1i1=3i3R2i2=6i4Therefore,R1/R2= 3/6 = 1/22.37The current from node A to node C isIAB- I1, the current from node D to node BisI1- I2, and the current from node C to node B isIAB- I1+ I2.Then the KVL equation for loop ADC isI12R + I2R = (IAB- I1)Rwhich is3I1+ I2= IABThe KVL equation for loop DBC is(I1- I2)R = I2R + (IAB- I1+ I2)2Rwhich is3I1- 4I2= 2IABSolvingI1andI2, I1= 2IAB/5, I2= -IAB/5Thus,UAB= I12R + (I1- I2)R=7IABR /5RAB=UAB/IAB= 7R /5R2RR2RRA+-BCDIABUABI1I2Figure S2.37: Circuit for Problem 2.37.Section 2.7 Ohm’s Law2.38By KCL, the current through the 4Ωresistor is 3A. Using Ohm’s law,1234=×=×=RIVV2.39By KVL , the voltage across the resistor is 5V.Using Ohm’s law:

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Ω===5.225IVR2.40Using Ohm’s law:ARVI3515===2.41The resistivity of copper is,m−Ω×=−81072.1ρThe resistance of the copper wire is,Ω=×××==−−70102286.15.01072.1108ALRρUsing the definition of Ohm’s Law, we can obtain the maximum allowable voltage,which is:VRIV70701maxmax=⋅==2.42Using Ohm’s law, the resistance of the tissue isΩ=×=−kR5.8041043.12106Because,AρL=RTherefore,2310468.37515.048mLmΩ=kΩ−××⋅94.15=Lm2.43The resistance across the voltage source is32321321//RRRRRRRRR+⋅+=+=Using Ohm’s lawRVI=, and3232132322322)()(RRRRRRRVRRRIRRRis+++⋅⋅+=⋅+=323212)(RRRRRRVs++⋅=

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2.44By KCL:A520=ITherefore, by Ohm’s Law:V100520=⋅=AVV30103=⋅=BVV30152=⋅=CVBy KVL:V130=SV2.45Using Ohm’s law:2.1102002403=×=ImA2.46For short circuit,v(t) = 0Using ohm’s lawRshort=v(t)/i(t) = 0For open circuiti(t) = 0Using ohm’s lawRopen=v(t)/i(t) = infinity2.47Vout= Vs(Rs /(Rs + Rc)) - Vs(Rb /(Ra+Rb))By product rule of derivationdVout/dRs=VsRc/(Rs+Rc)2WhenRs= 0, the above expression is minimum. However, you cannot control oversensor resistance, but you can chooseRcto achieve the maximum slope.We differentiate dVout/dRswith respect toRc,d(dVout/dRs) /dRc =Vs(Rs-Rc) /(Rs+Rc)3The maximum slope is achieved whenRc = Rs.Rccan be chosen to have the same value as the nominal resistance of the sensor.2.48Using data pairs in the table, we get three equations inA, B, C(1/273)=A+Bln(16330) +C(ln(16330))3(1/298)=A+Bln(5000) +C(ln(5000))3(1/323)=A+Bln(1801) +C(ln(1801))3Solving those equations, we getA= 0.001284B= 2.364×10-4C= 9.304×10-8

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Section 2.8 Power and Energy2.49a)The voltage and the current are in associated direction,P=u×i= -3×1 = -3W=>active elementb)The voltage and the current are not in associated direction,P= -u×i= 2×2 = 4W=>passive elementc)The voltage and the current are not in associated direction,P= -u×i= -2×3 = -6W=>active element2.50a)The voltage and the current are in associated direction,P=u×i= 20×2.5e-2t= 50e-2tW=>active elementb)The voltage and the current are not in associated direction,P= -u×i= -20×2sint= -40sintW=>t> 0, active element;t< 0, passiveelement2.51The current can be found using the power formula:5.0105===VPIAAnd the resistance can then be found using Ohm’s law:Ω===205.010IVR2.522IRIVP×=×=5005202=×=W2.53Applying KCL, the current through each resistor is 4mA.Given2IRP×=,()mWP801041052331=×××=−()mWP22410410142332=×××=−()mWP51210410322333=×××=−2.54According to KVL:2135.1kiki=By KCL:213iimA+=mAi12=

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VkmAV111=Ω×=2.55From the problem 2.54, we havemA21=iandmA12=i.RecallRIP2=. Therefore,()mW6105.11023235.1=×⋅×=−kP()mW21021013232=×⋅×=−kP()mW11011013231=×⋅×=−kP2.56+-VIFigure S2.56: Resistor with current direction and voltage polarity.Consider Figure S2.56, the power absorbed by the resistor isP=VIBy Ohm’s law,V=RIThusP=I2R=V2/RThereforePis always a nonnegative number whenRis positive.2.57The voltage and the current are in associated direction, soP=u×iand the waveform of the power is shown belowP(W)12010-10t(s)Figure S2.57: Power curve.The consumed energy is∫=20PW=0

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which is resulted from the negative symmetry ofParoundt= 1.2.58a)Let the current through the -5V voltage source be in associated direction with thevoltage, which is from the positive side to the negative side.By KCL at the node below the -5V voltage source, the current through it equals2A+4A=6A.Therefore the power of the -5V voltage source isP= -5V×6A =30Wb)To zero the power above, the current through the -5V voltage has to be zero. Sothe 4A current source needs to be changed to -2A by KCL.Section 2.9 Independent and Dependent Sources2.59In Figure S2.59,iaxis is the current through the source anduaxis is the voltageacross the source.usdenotes the voltage provided by the ideal voltage source andisdenotes the current provided by the ideal current source. It is shown that for idealvoltage source the voltage does not change with the current through it; for idealcurrent source the current does not change with the voltage.iuusisIdeal voltage sourceIdeal current sourceFigure S2.59: Voltage VS current curves.2.60Because the two current sources are in series, the currents through them are thesame,i= 1A2.61Because the 3Ωresistor is in series with the current source, the current throughthe resistor is 1A and, by Ohm’s law, its voltage isu= 3Ω×1A = 3VBy KVL, the voltage across the current source isu+ 2V = 5VThe power supplied by the current source is (the current and the voltage are not inassociated direction)

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P= -5V×1A = -5W2.62Using KVL:xxVV3100++−=Therefore,5.2=xV2.63Using KVL and Ohm’s law,()xxII20105.3603+××−−=xI31048.36×−=mAIx724.1−=2.64The voltage across the 2Ωresistor is -1V, where the voltage polarity is associatedwith the current direction. So by Ohm’s lawI= -1V/2Ω= -0.5ATherefore, the current of the dependent source is2I= -1A2.65According to KCL, the current through R3 isxiβAccording to Ohm’s law,30Rivxβ=ixis given by:21RRvisx+=Therefore,2130RRvRvs+=β2.66According to Ohm’s law, the current flow (in associated direction with thevoltage) through the 5Ωresistor isi1=4.9V/5Ω= 0.98A.Because the CCCS is in series with the 5Ωresistor,i1=0.98i=>i= 1Awhich is the current through the 6Ω.By KCL, the current through the 0.1Ωresistor isi-0.98i =0.02A.So the voltage across the 0.1Ωresistor

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