电工学郭木森第三章题答案

1P172/1：如图3-6-1所示电路，E1、E2、E3、R1、R2、R3都是已知量，试用叠加原理求通过电阻R的电流。1R1E2R2E3R3ER解：各电源单独作用时，电路图如下：1R1E2R3RR图3-6-1(a)(b)1R2E2R3RR(c)1R3E2R3RRIIIIRRRRRRRRREI32322211////////323121321321RRRRRRRRRRRRRRE2RRRRRRRRREI31313122////////323121321312RRRRRRRRRRRRRRERRRRRRRRREI21212133////////323121321213RRRRRRRRRRRRRREIIII323121321213312321RRRRRRRRRRRRRRERRERRE3P172/2：如图3-6-2所示电路，Z1=j0.5Ω,IS1=1A,Z2=2Ω,IS2=j2A，电流表内阻不计，试求其读数。+1SI2SI1Z2ZA1SI1Z2ZA2SI1Z2ZA解：电路改画成每个电流源单独作用时情形。SISI04.1406.2905.015.025.02111jjZZZIISS0.236A0.059A96.75243.0j04.1406.2905.045.02222122jjZZZIISS1.884A471.0A96.7592.1jSSIII236.0059.0j12.253.0j884.1471.0jA96.75075.2图3-6-24P173/3：如图3-6-3所示电路，将电压源等效变换为电流源，设E=10V，。2,2V,906jZRE图3-6-3(a)(b)(c)(d)REZERERZEA5RA03ZA03RE906不能等效变换5P173/4：如图3-6-3所示电路，将电流源等效变换为电压源，设IS=3A，。S5.0,2A,03SYRISIR图3-6-4(a)(b)SIY(c)SIR(d)SIRYREEV621YSIEV621Y不能等效变换6P173/5：用戴维宁定理求解第一章第七题。求I。105.252014V5.12EIAB解：205205.125.25.2105.12aboUV5.7105.2620//55.2//10abRA375.01465.7IabR14aboUAB7P173/7：用戴维宁定理和叠加原理求解第二章第14题。求0I解：戴维宁定理，a、b间断开，如图1R1ECX1R2RLX4RabI1E2E4432LL11aboRRREjXjXREU1E1R2R4R2ELXab06645550602aboEjjU时，问E2等于多少？66490545520602E26109054552060EV4571.702E81RCX1R2RLX4RabI1E2E1E1R2R4R2ELXab解：叠加定理：0IbaUUV4542.4290525060LL11ajXjXREUV4542.42baUUV4571.701064542.42)(4342RRRUEb91RCX2RLX1I1E2E3E已知：31R42R4CX4LXV01001EV901002EV01003E求：?1I解：用简化分析方法1、叠加法：C1CC1C1L11)(jXRjXjXRjXRjXEI1ICXLX1R16904100)(90CLC1L1C1XXjXRXjRXEA25A9025jP173/8：用简化分析方法求解第二章第15题。102ECX2RLX1R1IL1LL1L1C21jXRjXjXRjXRjXEILCL1C1L2XXjXRXjRjXEA1802516904901003E2RCXLX1R1I0,12IE短路后)25(25111jIIIA1352252525jA13535.3511P173/9求图3-6-5所示电路中1、2两点间等效电压源的电动势及内阻。已知V,5.2E。1800,400,200mA,5321SRRRI解：用戴维宁定理：E1R2R3RSI122213S12RRRERIU4004002005.218001053V3.733.19334002004002001800//21312RRRRV3.733.19331212用诺顿定理求：YI和Sd解：电路改画成如图E1R2R3RSI121RE2R3R12SdI1R2RSI3R12SdI322321Sd//RRRRRREI18004004001800//4002005.2A000862.0mA862.033.133400200400200//21RRmA655.4180033.133180033.13333.1335//////21321SSdRRRRRIImA79.3862.0655.4SdSdSdIIIS000517.033.19331//11321RRRRY12SdIY7.33V000517.079.3oeE13P174/10求图3-6-5所示电路中1、2两点间等效电激流及内电导。已知E1R2RSI12V,5.2E。1800,200mA,531SRRI解：（1）用电源变换方法求1、2之间的短路电流利用上题解得的电压和内阻，进行等效变换。V3.733.1933123.8mAA0038.033.19333.712OCsdRUIS00052.033.1933111212RY另解：（1）用诺顿定理求1、2之间的短路电流先将电路变换：求1、2之间的短路电流E1R2R3R3RsdI1R2R3RsdIsdsdsdIIISI1212SIY1432232321sdRRRRRRRREI1800400400180040018004002005.2116000010007200004400004005.221321Ssd//////RRRRRIImA655.4180033.1331800//33.1335sdImA79.3862.0655.4sdsdsdIII3211//11RRRY000555.0//0025.0005.00.000517S//0.0005550075.033.133400200400200//21RR220040022007200002005.2862mA.0A000862.015已知：,2000OR,104RV,5.2EmA,5SI,4002R,18003R。100LR求：负载上的电流。解：用戴维宁定理：2213S12RRRERIU18001053V3.721312//RRRRV3.733.1933124004002005.233.19334002004002001800A00359.010033.19333.7LI3.6mA,2001R2E1ROR3RSI14R2RLRLR16P175/15在图3-6-11所示电路中，已知：V,10EV,)302sin(100fte50KHz,fF,01CH,01L,0100321RRR求三个电阻的电压瞬时表达式。E1R2RCL3Re解：1C11R1RjXREU10500021001000100030100jV18.30100183.3100030105j10500021000100030100102L2R2jERjXREU82.893141603010105V)82.592sin(318.010V82.59318.010ftV)302sin(100103RftuV)18.302sin(1001RftuV)82.592sin(318.0102RftuE1R2R3R1ELXCX17P175/16在图3-6-12所示电路中，已知：A,)628sin5314sin10(Stti,10R求（1）电流的有效值；（2）电阻R吸收的平均功率。SiiR18P176/17在图3-6-13所示电路中，已知：。8,5,511RXR问R0等于多少欧时，通过它的电流对电压0IU有2有相位角？1X1RabR1X1RR0RU解：运用戴维宁定理，求R0两端的开路电压及从a、b端看进去的等效内阻。以U为参考相量1X1RRR1X1Rab111111abk)(jXRRjXRUjXRRRUUUjjjXRRUjXRR51353)(1111UU067.80418.004.2193.1303.5983.5558)55(82)(21111abjjjXRRjXRRZ96.2312.804.2193.1345251619戴维宁等效电路如图：abZababkUoIURRZUI00ababko96.2312.8067.80418.0URj03.342.7067.80418.00I与U的相位差为22ZZ于是分母的阻抗)90067.80(3.342.70ZjR933.9Z]172.0985.0[jZ)172.0(3.3jZj186.19172.03.3Z898.18985.0186.1942.70R5.11478.1142.7989.180R,5.110时当R20差的电流与端电压的相位通过R20P176/18如图3-6-14所示电路，在什么条件下，输入端电流和电压波形同相？IULC1R2R解：输入端电压、电流同相须电路为纯电阻的性质。即：阻抗角为0CjRCjRLjRLjRZ112211CjRCRjLjRLjR12211CjRCjRCjRCRjLjRLjRLjRLjR1112222111121222222222121211CRCRjCRLRLjRLR2222212212222222212111CRLRLRCRjCRCRLjRLR221222222221211LRCRjCRCRLjRLRCRLjCRRjCLRjLRjR22222212212221