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电路分析基础习题解答

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电路分析基础习题解答1-1在图题1-1所示电路中。元件A吸收功率30W,元件B吸收功率15W,元件C产生功率30W,分别求出三个元件中的电流I1、I2、I3。解A,A,A1-5在图题1-5所示电路中,求电流I和电压UAB。解A,V1-6在图题1-6所示电路中,求电压U。解,即有V1-8在图题1-8所示电路中,求各元件的功率。解电阻功率:W,W电流源功率:,W电压源功率:W,W2-7电路如图题2-7所示。求电路中的未知量。解VAAA2-9电路如图题2-9所示。求电路中的电流。解从图中可知,2(与3(并联,由分流公式,得A所以,有解得A2...

电路分析基础习题解答
1-1在图题1-1所示电路中。元件A吸收功率30W,元件B吸收功率15W,元件C产生功率30W,分别求出三个元件中的电流I1、I2、I3。解A,A,A1-5在图题1-5所示电路中,求电流I和电压UAB。解A,V1-6在图题1-6所示电路中,求电压U。解,即有V1-8在图题1-8所示电路中,求各元件的功率。解电阻功率:W,W电流源功率:,W电压源功率:W,W2-7电路如图题2-7所示。求电路中的未知量。解VAAA2-9电路如图题2-9所示。求电路中的电流。解从图中可知,2(与3(并联,由分流公式,得A所以,有解得A2-8电路如图题2-8所示。已知,求电路中的电阻。解KCL:解得mA,mA.R为k(解(a)由于有短路线,,(b)等效电阻为2-12电路如图题2-12所示。求电路AB间的等效电阻。解(a)(b)3-4用电源变换的 方法 快递客服问题件处理详细方法山木方法pdf计算方法pdf华与华方法下载八字理论方法下载 求如图题3-4所示电路中的电流I。解电路通过电源等效变换如图题解(a)、(b)、(c)、(d)所示。所以,电流为A3-6求如图题3-6所示电路中的电压。并作出可以求的最简单的等效电路。解V,最简单的等效电路如图题解3-6所示3-8求图题3-8所示电路中的电流。解KVL:或由KCL:联立解得A3-14求图题3-14所示电路中电流表的读数。(设电流表的内阻为零)解电路是一个平衡电桥,电流表的读数为0。4-2用网孔电流法求如图题4-2所示电路中的电流。解先将电流源模型变换成电压源模型,设网孔电流如图所示。列网孔方程解得:A,A,A所以A4-3用网孔电流法求如图题4-3所示电路中的功率损耗。解显然,有一个超网孔,应用KVL即电流源与网孔电流的关系解得:A,A电路中各元件的功率为W,W,W,W显然,功率平衡。电路中的损耗功率为740W。4-10用节点电压法求如图题4-10所示电路中的电压。解只需列两个节点方程解得V,V所以V4-13电路如图题4-13所示,求电路中开关S打开和闭合时的电压。解由弥尔曼定理求解开关S打开时:V开关S闭合时5-4用叠加定理求如图题5-4所示电路中的电压U。解应用叠加定理可求得10V电压源单独作用时:5A电流源单独作用时:电压为5-8图题5-8所示无源网络N外接US=2V,IS=2A时,响应I=10A。当US=2V,IS=0A时,响应I=5A。现若US=4V,IS=2A时,则响应I为多少?解根据叠加定理:I=K1US+K2IS当US=2A.IS=0A时I=5A∴K1=5/2当US=2V.IS=2A时I=10A∴K2=5/2当US=4V.IS=2A时响应为I=5/2×4+5/2×2=15A5-10求如图题5-10所示电路的戴维南等效电路。解用叠加定理求戴维南电压V戴维南等效电阻为5-16用诺顿定理求图题5-16示电路中的电流I。解短路电流ISC=120/40=3A等效电阻R0=80//80//40//60//30=10(5-18电路如图题5-18所示。求RL为何值时,RL消耗的功率最大?最大功率为多少?解用戴维南定理有,开路电压:V戴维南等效电阻为所以,RL=R0=4.8(时,RL可获得最大功率,其最大功率为5-20如图题5-20所示电路中,电阻RL可调,当RL=2(时,有最大功率Pmax=4.5W,求R=?解:先将RL移去,求戴维南等效电阻:R0=(2+R)//4(由最大传输定理:用叠加定理求开路电压:由最大传输定理:,故有US=16V6-1参见图题6-1:(a)画出ms时随时间变化的曲线;(b)求电感吸收功率达到最大时的时刻;(c)求电感提供最大功率时的时刻;(d)求ms时电感贮存的能量。解(a)的波形如图题解6-1所示。(b),吸收功率,吸收功率达到最大时的时刻为。(c)提供功率,提供最大功率时的时刻为。(d)ms时电感贮存的能量:J6-5如图题6-5所示电路原已稳定,t=0时将开关S打开,求及。解=2/5×6=2.4A=2.4×3=7.2V画出初态等效电路如图题解6-5所示,用叠加定理得:A;6-7在图题6-7的电路中,电压和电流的表达式是V,A,求:(a)R;(b);(c)L;(d)电感的初始贮能。解(a)由欧姆定律(b)时间常数为s(c)求L:,即有H。(d)电感的初始贮能J6-8图题6-8所示电路中,开关S断开前电路已达稳态,时S断开,求电流。解初始值A终值时间常数:伏安关系法s所以,电流为6-9如图题6-9所示电路中,换路前电路已处于稳态,t=0时开关闭合,求uC(t)、iC(t),并画出它们的波形。解:初始值:uC(0+)=uC(0-)=10V,稳态值:uC(()=-5V时间常数:(=RC=10(1=10s;故有波形如图题解6-9所示。6-11图题6-11所示电路原已稳定,t=0时断开开关S后,求电压u(t)。解:电感中的电流时,电感元件相当于开路,故有V稳态时,电感元件相当于短路,故有V时间常数:s所以,V6-13如图题6-13所示电路中,开关S打开前电路已稳定,求S打开后的i1(t)、iL(t)。解:初始值:uC(0+)=uC(0-)=20V,iL(0+)=iL(0-)=1A;i1(0+)=(40-20)/20=1A稳态值:iL(()=0;i1(()=0时间常数:(1=RC=20(0.1=2s;(2=L/R=0.2/20=0.01s故有6-15如图题6-15所示电路原已达稳态。开关S在t=0时闭合。求电容电压的零输入响应,零状态响应及全响应。解初始值:;稳态值:时间常数:零输入响应:零状态响应:全响应:8-1求下列各对正弦波的相位差,并判断各对正弦波的超前与滞后。(a)和;(b)和;(c)和。解(a)相位差为,前者超前后者162((b)相位差为,前者滞后后者90((c)相位差为,前者超前后者90(8-3已知电流相量A,频率Hz,求s时电流的瞬时值。解A,时域表达式为As时,A8-6某元件上的电压为V,电流为A。判断该元件是什么类型的元件,并计算它的值。解电压和电流的相量为V,A元件的阻抗为显然,该元件是电容元件。由,则电容值为8-8二端无源网络N如图题8-8所示,已知:V,。求N的阻抗,并画出最简等效电路。解电压和电流的相量为V,AN的阻抗为即R=35.355(,H。画出最简等效电路如图题解8-8所示。8-9在如图题8-9所示的正弦稳态电路中,已知V,电流表A的读数为2A。电压表V1、V2的读数均为200V。求R、XL和XC。解根据题意,可得根据阻抗三角形,有解得:R=50(,XL=50(8-14计算图题8-14所示两电路的阻抗和导纳。解(a)S(b)S8-17在如图题8-17所示电路中,已知:V,求电压。解由分压公式:8-19图题8-19所示电路中,已知电源V,,求电流。解感抗和容抗为电路的阻抗为总电流和分电流为A,A。9-2图题9-2所示电路中,已知负载两端电压,电流,求:(a)负载阻抗Z,并指明性质。(b)负载的功率因数、有功功率和无功功率。解(a)负载阻抗Z为负载是感性的。(b)负载的功率因数有功功率和无功功率为WVar9-5两组负载并联,一组S1=1000KVA,功率因数为0.6,另一组S2=500KVA,功率因数为1,求总视在功率和总有功功率。解根据题意,第一组负载有,kWkVar第二组负载有kW所以,总的有功功率kW总视在功率kVA9-6如图题9-6所示为日光灯与白炽灯并联的电路,图中R1为灯管电阻,XL为镇流器感抗,R2为白炽灯电阻。已知U=220V,镇流器电阻不计,灯管功率为40W,功率因数为0.5;白炽灯功率为60W。求I1、I2、I及总的功率因数。解由灯管中的电流为A由,白炽灯中的电流为A电路的总功率W日光灯的无功功率为Var总视在功率为故总电流A总功率因数9-13在如图题9-13所示电路中,假定阻抗Z上允许得到的功率为任意时,求阻抗Z能得到的最大功率。解从Z的左边用戴维南等效戴维南电压:V当时,Z可获得最大功率。最大功率为W10-2如图题10-2所示对称三相电路,已知,,,求各线电流。解用抽出一相计算法,可得A相线电流为:;其它两相为。10-4对称三相电源的相电压为125V,对称Y形负载阻抗为(19.9+j14.2)(,线路阻抗为(0.1+j0.8)(,以电源的A相电压为参考,求(a)三个相电流。(b)电源处的三个线电压。(c)负载处的三个相电压。(d)负载处的三个线电压。解用抽出一相的计算方法,设V(a)A相电流为AAA(b)电源处的三个线电压为VVV(c)负载处的三个相电压VV(d)负载处的三个线电压VVV10-5对称(形负载阻抗为(60+j45)(,线路阻抗为(0.8+j0.6)(,负载两端的相电压为480V,以负载为参考,求(a)负载的三个相电流。(b)三个线电流。(c)线路发送端的三个线电压。解(a)负载的三个相电流,设相电压VAAA(b)三个线电流AAA(c)线路发送端的三个线电压,先将(负载转换成Y负载,则可以求得电源处的相电压线电压为VVV10-11某三相对称负载,、,接于电源电压为380V的电源上。求:(a)负载接成星形时,线电流、相电流和有功功率。(b)负载接成三角形时,线电流、相电流和有功功率。解(a)相电压,由于是Y接法,线电流或相电流为有功功率为(b)(接法时,相电流、线电流为,有功功率为11-2用网孔 分析 定性数据统计分析pdf销售业绩分析模板建筑结构震害分析销售进度分析表京东商城竞争战略分析 法求如图题11-2所示电路中的电流和。解列网孔方程为(1)(2)由式(2)得,并代入式(1)得故有A11-4求如图题11-4所示电路中的电流和。解列网孔方程解得AA11-8如图题11-8所示电路,已知iS=2cos2tA。求原边电压u1。解副边阻抗为所以,Z2=1(,副边折合到原边的阻抗为原边电压为V故有u1=25.6cos2tV11-11电路如图题11-11所示。使负载电阻10(能获得最大功率,确定理想变压器的匝比n,并求负载的最大功率。解用戴维南定理求副边等效电路。等效阻抗为当负载电阻时,可获得最大功率。即开路电压为V最大功率为W另解:也可以将副边阻抗折合到原边回路来计算。12-2如图12-2所示为串联谐振电路。已知:,为电路谐振角频率,,上消耗功率为,电路通频带。求(a)电感值。(b)谐振频率。(c)谐振时电容电压的峰值。解(a)谐振时,R上的功率为,即由于,可得(b)谐振频率为(c)品质因数为电容电压的最大值为12-3RLC串联电路中,R=4(和L=25mH。(a)若品质因数Q=50,计算电容C的值。(b)求半功率点频率,和带宽BW。(c)计算,,时的电路的平均功率,电源电压峰值V。解(a)根据品质因数Q的定义,所以(b)谐振频率为rad/s带宽为rad/s半功率点频率,为rad/srad/s(c)谐振时,kW当频率为半功率点频率时13-1电路如图题13-1所示,求双口网络的Z参数矩阵和Y参数矩阵。解用网孔方程即可求得Z参数矩阵为(Y参数矩阵为S13-3求如图题13-3所示双口网络的Z参数矩阵和Y参数矩阵。解对双口网络中的受控源进行电源变换,如图题解13-3所示。与Z参数的等效电路比较,可得:,,解得,解得,解得所以,Z参数矩阵为(Y参数矩阵为S�EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3���图题1-1�EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3���图题1-7图题1-6�EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3���图题1-8�EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3���图题2-7�EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3���图题2-9�EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3���图题2-12(a)(b)�EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3���图题2-8�EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3���图题3-4�EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3���图题3-6图题解3-6�EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3���图题8-8�EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3���图题3-8�EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3���图题解8-8图题3-14�EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3���图题解4-3�EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3���图题解4-2�EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3���图题4-10�EMBEDEquation.3����EMBEDEquation.3���图题解4-13�EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEquation.3����EMBEDEqu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