第十一講元件復(fù)阻抗課件

3.1正弦量及其三要素（最大值）（頻率、周期）(Um)(f、T)1、有效值2、角頻率3、初相位UI(Im)u

、i一一對應(yīng)正弦量三要素4、相位差同頻率正弦量的相位之差=(t+u)-(t+i)=u－i等于兩個正弦量的初相位之差常數(shù)i=ImSin(t+i)?u=UmSin(t+u)=u-i＞0＜0=0同相位電壓超前電流電壓滯后電流

=±π/2

=±π正交反相3.2相量法----正弦量用復(fù)數(shù)表示和運算的方法相量只是用復(fù)數(shù)表示正弦量的一種表示方法,它是用正弦量的有效值與初相位來表示。u=2ISin(t+u)√=IejiI?Ii=i=2ISin(t+i)√一一對應(yīng)U

?Uu=相量圖)uU·3.3單一元件的正弦交流電路i+u–u=Ri

i=2ISin(t+i)√

=2RISin(t+i)√

=2USin(t+u)√U=RIu

=i

I?Ii=U

?Uu==RIi=RI?RRI?+–U

?相量模型相量關(guān)系U

?一、電阻元件的交流電路+1+j0相量圖I?U

?i+u–L

i=2ISin(t+i)√u=LdidtCos(t+i

)√I2=L×=LISin(t+i+90°

)√2

=2USin(t+u)√U=LIu

=i

+90°二、電感元件的交流電路jLIi=I?Uu==LIi+90°=L90°Ii

×=jLI?I?+–U

?相量模型相量關(guān)系U

?+1+j0相量圖

i=2ISin(t+i)√=LISin(t+i+90°)√2uU

?I?U

?電壓超前電流90°感抗，對照U=RI可見：XL單位為Ω變電所，變壓器與輸電線之間加幾匝線圈？----防高頻閃電U=LI=XLIXL=

Lf高，XL大，UL一定時，iL小----阻交流f=0，XL=0(相當(dāng)于短路)-------通直流=jLI?U

?與f及L成正比

u=2USin(t+u)√i=CdudtCos(t+u

)√U2=C×=

CUSin(t+u+90°

)√2I=CUi

=u

+90°=2ISin(t+i)√U=IC1Ci+-u三、電容元件的交流電路Ii=I?U

?Uu==CUu+90°=C90°Uu×=jC+–U

?相量模型相量關(guān)系U

?jC

u=2USin(t+u)√I?I?–jC1+–U

?=

CUSin(t+u+90°)√2iI?–jC1U

?I?=+1+j0相量圖I?U

?容抗，對照U=RI可見：XC單位為Ωf高，XC小，UC一定時，Ic大----通交流f=0，XC→∞(相當(dāng)于開路)-----隔直流C1XC=U=IC1U

?=jCI?–jC1U

?I?=I=CU=XCI與f及C成反比i+u–Ru=RiRI?+–U

?=RI?U

?時域復(fù)數(shù)域U=RIu

=

i

+1+j0I?U

?時域復(fù)數(shù)域i+u–Lu=LdidtjLI?+–U

?=jLI?U

?=

j

XLI?U=LIu

=

i

+90°+1+j0I?U

?時域復(fù)數(shù)域i+u–Ci=CdudtI?–jC1+–U

?–jC1U

?I?==

－jXCI?i

=

u

+90°U=IC1+1+j0I?U

?下列各式是否正確？iu=XLIU=XL??IU=jLiu=jCIU=–XC??=jCIU??××××××RI?+–U

?=RI?U

?jLI?+–U

?=jLI?U

?=jXLI?I?–jC1+–U

?–jC1U

?I?==－jXCI?一、RLC串聯(lián)電路3.4復(fù)阻抗和復(fù)導(dǎo)納LRCU=

?UR

?+–UC

?+–UL

?+–jLR–jC1+–uR+–uC+–uLu=uR+uL+uCUR

?UC

?UL

?++=RI?I?++jLI?–jC1I?()=I?–jC1(R+jL)–jC1(R+jL)令：Z=復(fù)阻抗U

?+–+–ui–jC1(R+jL)Z==R+jXL–jXC=R+j(XL–XC)=R+jX電阻電抗=ZZ=√R2+X2=arctgRX模阻抗角＞0＜0=

0感性電阻性容性=jXL=–jXC=RI?U

?=jLI?U

?=jXLI?–jC1U

?I?==–jXCI?=RI?U

?=ZZI?U

?+–=ZI?U

?復(fù)數(shù)形式的歐姆定律復(fù)阻抗無源網(wǎng)絡(luò)i+–u無源網(wǎng)絡(luò)I?U

?+–

i=2ISin(t+i)√u

=2USin(t+u)√I?Ii=U

?Uu=I?U

?=ZI?U

?=ZIiUuIU–iu=R==R+jX=Z=√R2+X2arctgRX=ZCos=ZSinXZI?U

?+–Z=u–iIU=ZI?U

?+–=ZI?U

?u=0i=？=ZZ已知若令：問：i=0u=？若令：問：例：260°Z=i=0i=u=60°u=0–60°260°Z1=i=–u=

U=

?UR

?UC

?UL

?++UR

?+–UC

?+–UL

?+–jLR–jC1I?U

?+–I?U

?UR

?UC

?UL

?U2=UR2+(UL–UC)2電壓三角形V2V1V0RC10V10VV0V1V2R40V100VL例：求V0的讀數(shù)I?UC

?UL

?UR

?U

?I?UR

?U

?二、RLC并聯(lián)電路iRiCiLi+–uiLiCiRi=+++–U?I?IC?IL?IR?）））RLC）））RjXL–jXC?IC?IL?IR?I=++=++–jXC1U?jXL1U?R1U?=()U?++–jXC1jXL1R1–jXC1jXL1R1=G=jL1=–jL1?I==jBC=–jBL=–jC11=jCU

?[G+(–jBL)+jBC]=()++–jXC1jXL1R1?IU

?U

?=[G+j(BC–BL)]BLBC感納容納G+j(BC–BL)Y==G+jB令：復(fù)導(dǎo)納=Y模導(dǎo)納角Y=√G2+B2=arctgGB＞0＜0=0感性電阻性容性電導(dǎo)電納U?I

?=YYI?U

?+–=YI?U

?復(fù)數(shù)形式的歐姆定律復(fù)導(dǎo)納U

?–jXC1I?=jXL1I?=U

?R1I?=U

?=–jBL=jBC–jXC1jXL1R1無源網(wǎng)絡(luò)i+–u無源網(wǎng)絡(luò)I?U

?+–

i=2ISin(t+i)√u

=2USin(t+u)√I?Ii=U

?Uu=I?U

?=Y+–U?I?IC?IL?IR?）））RjXL–jXC?IC?IL?IR?I=++U

??IC?IL?IR?II2=IR2+(IL–IC)2電流三角形例：求A0的讀數(shù)I?IR

?U

?I?LRA0A2A110A6ARCA1A0A24A3AIC

?U

?IR

?IL

?三、復(fù)阻抗串聯(lián)U1?+–+–U?+–Un?+–U2?I?Z1ZnZ2ZeqU?+–I?Un?U2?U1?U=?+++……=Z1+Z2+……+ZnI?I?I?=(Z1+Z2+……+Zn)I?=(Z1+Z2+……+Zn)ZeqUK?U?=ZeqZKUR

?+–UC

?+–UL

?+–jLR–jC1I?U

?+––jC1Z=R+jL=ZR+ZL+ZC四、復(fù)阻抗并聯(lián)Z1Z2

ZnY1Y2YnIn?I?I2?I1?I1?I?In?I2?=++……+=++……+YnU?Y2U?Y1U?+

–U?U?=(++……+)Y1Y2Yn=(++……+)Y1Y2YnYeqI?+

–U?YeqZeqIK?I?=YeqYK+–U?I?IC?IL?IR?）））RjXL–jXCY=G–jL1+jC=YR+Y