I?EO?2
RO?RRREORO题1-14图(a)E5?3?IS2IS1题1-14图I2VI2V5?6?3?RO6?题1-14图(b)
将R?4?,I?1A代入上式 1?则等效电源电动势为
EO?1?(2?4)?2?8V 则当R?12?时,电流I为 I?EO?2 2?4EO?28?23??A
RO?R2?1271-15 如题1-15图所示为某电路的部分电路,试求A、B、C各点的电位及电阻R。
A1?6A7A5?R3?A1?6AI5?5?IRRUA3?I3?URUCCI10?C
B10?5V50V题1-15图10?5V7AB2?2?50VUB题1.15图(a)解:将题1-15图的是个节点看成一个虚的大节点,如题1-15图(a)所示,根据基尔霍夫电流定律可得
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6?7?I3? I3??13A 由欧姆定理得
I10??5?0.5A 10由基尔霍夫电流定律可知 6?I5??I10?
故 I5??6?I10??6?0.5?5.5A 又 I5??IR?I3?
故 IR?I5??I3??5.5?13??7.5A 由基尔霍夫电压定律可列方程为
I5??5?UR?5??50
UR??50?5?5.5?5??72.5V 则 6?1?5?50?UA?0 UA?6?1?5?50??39V
7?2?50?UB?0 UB?7?2?50??36V
I3??3?UC?UR?0
UC?UR?I3??3??72.5?13?3??111.5V R?UR?72.529???9.67? IR?7.531-16 题1-16图为某一复杂电路的一部分,已知I1?2A,I2?2A,I5?1A,E3?3V,
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E4?4V,E5?6V,R1?2?,R2?3?,R3?4?,R4?5?,R5?6?,求电压UAF和C、D两点的电位VC及VD。 解:由基尔霍夫电流定律得
I3?I1?I2?2?2?4A I4?I3?I5?4?1?3A
A R1I1BR2CI2R3I3CR2E3R5DR4I5I4E5题1-16图E4FAR1I1BI2R3I3VAVCE3VDDR4R5I5I4E5E4F
VF题1-16图(a)则对于题1-16图(a)所示,由基尔霍夫电压定律可得 I1R1?I3R3?I5R5?VA?E3
VA?I1R1?I3R3?I5R5?E3?E5?2?2?4?4?1?6?3?6?17V VF?I5R5?I4R4??E4?E5
VF??E4?E5?I4R4?I5R5??4?6?3?5?1?6??19V UAF?VA?VF?17?(?19)?36V I2R2?I3R3?I5R5?VC?E3?E5
VC?I2R2?I3R3?I5R5?E3?E5?2?3?4?4?1?6?3?6?19V I5R5?VD?E5
VD?I5R5?E5?1?6?6?0V
1-17 电路如题1-17图所示,试求A、B、C、D各点的电位VA、VB、VC、VD、电压UAB及
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恒流源端电压US。
解:题1-17图可还原成题1-17图(a),由基尔霍夫定律 5I?10I?10?20 I? -20V
题1-17图30?2A 15c5?10?abUab1?3?d5?I10?ABUab1?I1?USC3?DI3?2AI4?4?4?US10V2A20V4?4?10V题1-17图(a)由于A、B两点是断开的,故I1??0,因此I3??2A,I4?? VA??I?10?10??2?10?10??10V
1I3??1A,则所求的量为 2 VB?1?I1??VC?VC?3I3??4I4??3?2?4?1?10V VC?US?10V
VD?4?I4??4?1?4V
UAB?VA?VB??10?10??20V
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