电路习题答案VIP免费

例3)(:),5cos(2120)(tittu求已知解00120U20j54jjLXΩ10j02.051jjCX相量模型下页上页+_15u4H0.02FiUj20-j101I2I3II+_15返回A9.36106812681012011511200jjjjjA)9.365cos(210)(0tti下页上页CLCLRjjXUXURUIIIIUj20-j101I2I3II+_15返回已知：f=50Hz,U=220V,P=10kW,cos1=0.6，要使功率因数提高到0.9,求并联电容C，并联前后电路的总电流各为多大？o1113.536.0cosφφ例解o2284.259.0cosφφF557)84.25tg13.53tg(2203141010)tgtg(23212φφUPCA8.756.02201010cos31UPIIL未并电容时：并联电容后：A5.509.02201010cos32UPI下页上页LRCUILICI+_返回求电路各支路的复功率V)1.37(236010ooZU例解1)15j5//()25j10(ZVA1424j1882010)1.37(236oo发SVA1920j768)25j101(236*2*121YUS吸VA3345j1113*222YUS吸发吸吸SSS21下页上页+_U100∠oA10j255-j151I2I返回A)3.105(77.815j525j1015j5010oo1I解2A5.3494.14o12IIISVA1923j769)25j10(77.821211ZIS吸VA3348j1116)15j5(94.1422222ZIS吸VA1423j1885)25j10)(3.105(77.810o*11SIZIS发下页上页返回求ZL=?时能获得最大功率，并求最大功率。例解45j15)30//30j(30jiZ045260)30//30j(j4SU45j15*iLZZ当W120154)260(2maxP有上页SUZi+-IZLI490∠oAZL-j3030-j30返回例1已知电源内阻RS=1k，负载电阻RL=10。为使RL获得最大功率，求理想变压器的变比n。当n2RL=RS时匹配，即10n2=1000n2=100，n=10.下页上页RLuSRS**n:1+_n2RL+–uSRS解应用变阻抗性质返回例某收音机输入回路L=0.3mH，R=10，为收到中央电台560kHz信号，求：(1)调谐电容C值；(2)如输入电压为1.5V,求谐振电流和此时的电容电压。Aμ15.0105.1)2(0RUIpF269)2(1)1(2LfC解Vμ5.1Vμ5.1580CCXIUURLQUUo0Cr+_LCRu下页上页返回例一接收器的电路参数为：U=10V=5103rad/s,调C使电路中的电流最大，Imax=200mA，测得电容电压为600V，求R、L、C及Q。解50102001030IUR6010600UUQQUUCCmH60105605030RQLμF67.61C20L+_LCRuV下页上页返回例1对称三相电源线电压为380V，Z=6.4+j4.8，Zl=6.4+j4.8。负载Z的相、线电压和电流。解画出一相计算图ZZla+–AUAIANN’+–AUBUCU+–+–AIBICIZZACBZlZlZlZNN’acb下页上页返回求A3.171.1743.188.1203220j8.89.403220ooooANAlZZUIV2.368.1369.3683.171.17oooANaZIUV2.69.236V2.68.1363303oooanabUUV03220V0380oAoABUU则设ZZla+–AUAIANN’下页上页返回求：(1)线电流和电源发出的总功率；(2)用两表法测三相负载的功率，画接线图求两表读数。解例1已知Ul=380V，Z1=30+j40，电动机P=1700W，cos=0.8(感性)。A1.5341.440j300220Z1NA1AUIV0220NAU(1)DABCZ11AI2AI电动机AI下页上页返回电动机负载：W1700cos3A2φIUPlA9.3623.39.36,8.0cosA23.38.03803cos3A2A2IφφPφUPIl总电流：A2.4656.79.3623.31.5341.4A2A1AIIIkW44.32.46cos56.73803cos3A总总φIUPlkW74.13041.4332121A1RIPZ下页上页返回表W1的读数P1：(2)两表接法如图P1=UACIAcos1=3807.56cos(46.2–30)=2758.73W表W2的读数P2：P2=UBCIBcos2=3807.56cos(30+46.2)=685.26W=P-P1DABCZ11AI2AI电动机AIW1****W2下页上页返回例10)0(Li(2)画运算电路sL1ss11s11sCV1)0(cu解(1)计算初值下页上页电路原处于稳态，t=0时开关闭合，试用运...