电路_习题答案_05

电路 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 _习 题 快递公司问题件快递公司问题件货款处理关于圆的周长面积重点题型关于解方程组的题及答案关于南海问题 答案 八年级地理上册填图题岩土工程勘察试题省略号的作用及举例应急救援安全知识车间5s试题及答案 _第五章 5-0 在图示电压u和电流i的参考方向下，(1)写出图(a)和(b)中元件u和i的VCR方程；(2)已知uC=10sin(10t+30o)V，iL=5cos(10t-30o)A，求图(c)所示的电流iC和图(d)所示的电压uL。 解： (a) (b) (c) (d) 5-2 已知一正弦电流的波形如题5-2图所示， (1) 试求此正弦电流的幅值、周期、频率、角频率和初相； (2) 写出此正弦电流的瞬时函数 表 关于同志近三年现实表现材料材料类招标技术评分表图表与交易pdf视力表打印pdf用图表说话 pdf 达式。 解：(1) 设 (2) 5-3 将下列各正弦量表示为有效值相量，并在同一复平面上绘出相量图。 解：(1) (2) 5-4 写出对应于下列各有效值相量的瞬时函数表达式，并在同一复平面上绘出相量图。 解： (1) (2) (3) 5-7 电路如题5-7图所示，已知 ， ， ， 。求电压 和 。 解：相量模型如图所示。 则：u1(t)=2.795cos(2t-63.4o) V ， u2(t)=1.25cos2t V 5-9 在题5-9图所示电路中，已知R1=10 (，XC=17.32 (， I1=5 A，U=120 V，UL=50 V，电压 与电流 同相。 求R、R2和XL。 解：由KVL，得： 由于电压 与电流 同相。 则：电压 应与电流 同相。 以电压 为参考相量，画相量图如图所示。 由KVL，得： 则： 由KVL，得： 则： 由相量图，得： 得： 则： 由KVL，得： ，且同相。 则：UR=U-URL=120-100=20 V 得： 5-10 在题5-10图所示电路中，iS=20cos100t mA，求电阻、受控源及独立源吸收的平均功率，并求电容的无功功率。 解：电路的相量模型如图所示。 设节点电压为 ，则节点电压方程为： 解方程，得： 则： 得： 5-13 在题5-13图所示电路中， ，求电路吸收的平均功率、视在功率及功率因数。 解：各支路电流的参考方向如图所示。 由KCL，得： 则： 5-15 在题5-15图所示电路中，已知 ， ，用节点法求各节点电压及流 过电容的电流 。 解：设节点电压为 、 ，则节点电压方程为： 即： 解方程，得： 则： 5-16 试求题5-16图所示电路的端口等效阻抗Z。 解：外加电源法，如图所示。 由KVL，得： 解方程，得： 则： 5-17用三只电流表测定一电容性负载功率的电路如题5-17图所示，设其中表 的读数为7 A，表 的读数为2 A，表 的读数为6 A，电源电压有效值为220 V，试画出电流、电压的相量图，并计算负载Z所吸收的平均功率及其功率因数。 解：电路的相量图如图所示。 由相量图，得： 即： 解方程，得： ， 即： 得： 则： 5-19 题5-19图表示在工频下测量线圈参数(R和L)的电路。测量时调节可变电阻使电压表(设内阻为无限大)的读数最小。若此时电源电压为100 V，R1为5 (，R2为15 (，R3为6.5 (。电压表读数为30 V，试求R和L之值。 解：电路的相量图如图所示，要使电压表的读数最小， 需要 与 正交，即向量垂直。 由KVL，得： 则： 由已知，Uab=30 V 根据向量图，得： 则： 根据相似三角形对应边成比例，得： 则： 根据相似三角形对应边成比例，得： 则： 5-20 在题5-20图所示电路中，已知 ，电流表 ， 的读数均为1A，设流过电感L2的电流初相位角为零度，求 。 解：电流参考方向如图所示，电路的相量图如图所示。 得： 则： 得： 则： 得： 5-22 在题5-22图所示正弦电路中，已知电压表、电流表、和功率表的读数分别为69.3V、3A和180W，C =139μF，电源频率f =50Hz，求电路中的R、L和整个电路的功率因数cos(。 解：由题意，得： 则： 设： 又： 则： (1)当 得： 则： 得：R=20 ( ，L=36.7 mH (2)当 得： 则： 得：R=6.66 ( ，L=36.7 mH u + (b) i 20mH uL + (d) iL 10mH uC + (c) iC 5μF u + (a) i 10μF � EMBED Equation.3 ��� � EMBED Equation.3 ��� � EMBED Equation.3 ��� � EMBED Equation.3 ��� � 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