電路9章正弦穩(wěn)態(tài)的分析

1、第 1 章第九章靜電場(chǎng)正弦穩(wěn)態(tài)電路的分析阻抗和導(dǎo)納阻抗的串聯(lián)并聯(lián)電路的相量圖正弦穩(wěn)態(tài)電路的分析正弦穩(wěn)態(tài)電路的功率、復(fù)功率最大功率傳輸串聯(lián)電路諧振、并聯(lián)諧振電路返 回下 頁(yè)第 1 章靜電場(chǎng)9.1 阻抗和導(dǎo)納一、阻抗等效阻抗（輸入阻抗、驅(qū)動(dòng)點(diǎn)阻抗）+U def U = U (fZ-f ) = Z f+UuiZIIIZZ為復(fù)阻抗= UZZ為阻抗模IfZ= fu -fiZjfZ為阻抗角XZZ = R + jXRI僅含線(xiàn)性電阻、電感和電容的網(wǎng)絡(luò)第 1 章靜電場(chǎng)Z( jw) = R(w) + jX (w)對(duì)于RLC串聯(lián)電路，1wCX = wL -= X+ XLCf= arctan( X )=+ X 2R2

2、ZZRR = Z cosfZ ,X = Z sinfZ當(dāng)X 0,即wL 1時(shí)，稱(chēng)Z呈感性wC當(dāng)X 0,即wL 0,即wC 1 時(shí)，稱(chēng)Y呈容性wL當(dāng)B 0,即wC 1 時(shí)，稱(chēng)Y呈感性。wL第 1 章靜電場(chǎng)三、阻抗與導(dǎo)納的轉(zhuǎn)換Y ( jw)Z ( jw) = 1fZ+ fY= 0Y ( jw)Z( jw) =1R(w)X (w)1G(w) + jB(w) =- jR(w) + jX (w)Z ( jw) 2Z ( jw) 2G(w) = R(w) R(w) = G(w) Z ( jw) 2Y ( jw) 2B(w) = - X (w) X (w) = - B(w) Z ( jw) 2Y ( jw)

3、 2第 1 章靜電場(chǎng)9.2 阻抗（導(dǎo)納）的串聯(lián)和并聯(lián)串聯(lián)等效阻抗= Z1 + Z2 + ZnZeq= Zk串聯(lián)分壓公式k = 1, 2,UU , nkZeq= Y1 + Y2 + YnYeq并聯(lián)等效導(dǎo)納= Ykk = 1, 2,II , n并聯(lián)分流公式kYeq第 1 章靜電場(chǎng)9.3 電路的相量圖RLI +R =15WC = 5mFL =12mH+URULUCUCu = 1002 cos(5000t)V求電流瞬時(shí)表和各元件的電壓相量，并畫(huà)相量圖。w = 5000ZR = 15WU =1000解：VZL = jwL = j60W1wCZ= -j= -j40WC第1 章靜電場(chǎng)= ZR + ZL +

4、ZC = 15 + j(60 - 40)=2553.130WZeq1000UI =4 - 53.130A2553.13Zeqi = 42 cos(5000t - 53.130 )AUR = RI = 60 - 53.130UL = j60I = 24036.870VVUC = -j40I = -160 -143.130V第 1 章靜電場(chǎng)UL畫(huà)相量圖：UoIURUC53.130UoIULUCUR第 1 章靜電場(chǎng)R1LI +R = 10WL = 0.5H1C = 10mFURULI2+I1R = 103W+2CRUUC2sU = 100 Vw = 314s求各支路電流和電容電壓，并畫(huà)相量圖。= 1

5、000解：令 UsV參考相量1wCZ= -j= - j318.47WZ= jwL = j157WCLZ= 1000(- j318.47) = 303.45(-72.330 )W121000 - j318.47第1章Zeq靜電場(chǎng)= Z12 + Z+ ZL = 102.11- j132.13W= 166.99 - 52.300 WR1I = Us= 0.6052.300 AZeqU= ZI = 182.07 - 20.030V1012= U10= 0.5769.970 AI1ZC= U10= 0.18 - 20.030 AI2ZR2畫(huà)相量圖第1章靜電場(chǎng)9.4 正弦穩(wěn)態(tài)電路的分析US 3Z3Z5Z1

6、+IS 5+US1UZ2Z4節(jié)點(diǎn)電壓法第 1 章靜電場(chǎng)US 3Z3+電壓源受控電流源US 2I3+013Z+4Z5U6Z1b I32第 1 章靜電場(chǎng)Z1+1+求：一端口的戴維寧等效電路。+rI2US1UocZ2Is3I20解：先求開(kāi)路電壓- r U10= (1- rY2 )(Y1US1- IS 3 )= UUUoc10ZocY + Y212(Y1 + Y2 )U10 = Y1Us1- Is3第 1 章靜電場(chǎng)再求等效阻抗將電源置零U0 = Z2 I2 - rI2 = (Z2 - r)I2I= I + Z2 I2 = (1+ Z Y )I 02212Z1Z2 - r= U0=Zeq1+ Z YI0

7、21第 1 章靜電場(chǎng)IU= 380VAA1Sf = 50HzIR1+US11jwCjwL1C = 80.95mFIC電流表A的讀數(shù)最小I = 2.59A求：電流表A1的讀數(shù)。IC = wCUs = 100p 38080.9510= 9.66A-6向量圖：電流三角形第 1 章靜電場(chǎng)9.5 正弦穩(wěn)態(tài)電路的功率瞬時(shí)功率的兩種分解形式平均功率(有功功率) ，無(wú)功功率， 視在功率，功率因素，復(fù)功率兩種分解：五個(gè)定義：四種情形：三個(gè)例子： 一個(gè)性質(zhì)：純電阻, 純電感, 純電容，RLC串聯(lián)第1章靜電場(chǎng)2U cos(wt +fu )2I cos(wt +fi )u =i =p = ui= 2UI cos(wt

8、 + fu ) cos(wt + fi )p = UI cosj +UI cos(2wt +fu +fi )恒定量第一種分解：正弦量j = fu-fi第 1 章靜電場(chǎng)p = UI cosj +UI cos(2wt + 2fu -j)p = UI cosj +UI cosj cos(2wt + 2fu ) +UI sinj sin(2wt + 2fu )p = UI cosj1+ cos2(wt +fu )+UI sinj sin2(wt +fu )可逆部分不可逆部分定義1：有功功率P = 1pdt = 1TTUIcosj + cos(2wt + f + f )dtuiTT00= UI cosj

9、第 1 章靜電場(chǎng)cosj定義2：功率因數(shù)j功率因數(shù)角定義3：無(wú)功功率def UI sin jQ實(shí)數(shù)，：乏定義4：視在功率defSUI實(shí)數(shù)，：伏安第 1 章靜電場(chǎng)四種情形瞬時(shí)功率有功功率無(wú)功功率純電阻p = UI1+ cos2(wt +fu )P = UI = I 2RQ = 0純電感p = UI sin2(wt +fu )PL = 0U 2QL = wL純電容p = -UI sin2(wt + fu )PC= 0Q= -wCU 2CRLC串聯(lián)P = I 2RQ = QL + QC第 1 章靜電場(chǎng)功率不是同頻率正弦量，無(wú)法用相量法討論；但可以用一個(gè)輔助計(jì)算的復(fù)數(shù)來(lái)統(tǒng)一上述四個(gè)定義。定義5：復(fù)功率

10、defUI * = UIf -f= P + jQSuiUI有意義嗎？S = UI * = (IZ )I * = II *Z = I 2Z = I 2R+ jI 2 XeqeqS = UI * = U (UY )* = UU *Y * = U 2Y * = U 2G- jU 2Beqeq第 1 章靜電場(chǎng)第 1 章靜電場(chǎng)已知：電感線(xiàn)圈功率為30W,電壓為50V,電流為1A,頻率50Hz求：1電感線(xiàn)圈的電阻和電感; 2電感線(xiàn)圈吸收的復(fù)功率第 1 章靜電場(chǎng)例9-11 50HZ, 380V的正弦電壓，感性負(fù)載吸收的有功功率20KW，功率因數(shù)為0.6。將功率因數(shù)提高到0.9，求在負(fù)載兩端并聯(lián)的電容值。1解

11、：2解：QCU 2wICUwC = U 2wCI= UwCC =QCCsinjold20000- Inew sinjnew= 87.72 A= Qnew- Qold= IoldQCIC= P tanjnew = 9.69QnewkVar=Iold380 0.6= P tanjold = 26.67 kVarQoldcosjcosj= IIoldoldnewnew= 58.48 AInew補(bǔ)償前后，有功不變第1 章靜電場(chǎng)9.7 最大功率傳輸II+ZeqZNsUZoc2 RU負(fù)載吸收的有功功率為： P = I 2R = oc(R + R+ ( X + X)2)2eqeqX + Xeq= 0X =

12、- XeqR = R(R + R)2deq = 0dR ReqZ = ZY = Y*或最佳匹配eqeq第 1 章靜電場(chǎng)此時(shí)獲得的最大功率為：2U=P ocmax4Req2I=(P sc)4Gmaxeq第 1 章靜電場(chǎng)= 200 A求最佳匹配時(shí)獲得最大功率。例9-12： IS2W2W1IIsj4Z2W1等效電路；先求短路電流解：求11端口的I= 1 I= 100 AY= 0.25 - j0.25U= 22450scseq2oc2= (22)R= 2W= 1 WPeqmax4 2第1章靜電場(chǎng)a IIaR1R2+1jwCUsR3bZ3解：開(kāi)路電壓= UUocsR + R+ Z1123=Z31+ jw

13、CR3UsI=(1-a )Isc R2 + Isc R1 = Us短路電流(1-a )Rsc+ R 21第 1 章靜電場(chǎng)j5I1I+1a6Wj10W+606Wb6解：開(kāi)路電壓= j5I+ 6II =Uoc11112 + j10短路電流(I1 - Isc )6 = - j5I16I =I (6 + j10) = 6 + j5I1116 + j5第 1 章靜電場(chǎng)2W1W- jW0.5U1+j0.4WZLU110- 450求負(fù)載能獲得的最大功率？第 1 章靜電場(chǎng)20W1500+- j10Wj20WZL1009000源發(fā)出的復(fù)功率。求負(fù)載獲得最大功率時(shí)各第 1 章靜電場(chǎng)*WA*II1I23+Z1Z2Z3USUS = 220VP1 = 4.4kW , I1= 44.7A(感性)P2 = 8.8kW , I2 = 50A(感性)P3 = 6.6kW ,