一道电学难题( • ̀ω•́ )✧题目:
若图中甲乙均为电压表
甲:5v
乙:3v
若甲乙均为电流表
甲:0.9a
乙:0.5a
求R1,R2,R3
ããå ææç½æ¯ä¸ªçµåãçµæµæµéçæ¯åªäºçµé»çï¼
ãã1ãU1=5Vï¼R2串èR3两端çµåï¼U2=3Vï¼æµéR1串èR2两端çµåã
ãã2ãI1=0.9Aï¼æµéçæ¯R1并èR2两个çµé»ççµæµï¼I2=0.5Aï¼æµéR2并èR3æµè¿ççµæµã
æ¥çµå表æ¶ï¼ä¸ä¸ªçµé»ä¸²èï¼è®¾çµæºçµå为Eï¼ä¸²èçµæµä¸ºIãåï¼
ããIÃï¼R1+R2+R3ï¼=Eï¼ä¸ï¼IÃï¼R1+R2ï¼=3ï¼IÃï¼R2+R3ï¼=5ã
ãã两å¼ç¸æ¯ï¼ï¼R1+R2ï¼/ï¼R2+R3ï¼=3/5ã......â
ããæ以ï¼E=IÃï¼R1+R2ï¼+IÃï¼R2+R3ï¼-IR2=3+5-IR2=8-IR2ã
ããç±I=3/ï¼R1+R2ï¼ï¼å¾å°ï¼E=8-3R2/ï¼R1+R2ï¼=ï¼8R1+5R2ï¼/ï¼R1+R2ï¼ã......â '
ããç±I=5/ï¼R2+R3ï¼ï¼å¾å°ï¼E=8-5R2/ï¼R2+R3ï¼=ï¼3R2+8R3ï¼/ï¼R2+R3ï¼ã......â¡'
æ¥çµæµè¡¨æ¶ï¼ä¸ä¸ªçµé»å¹¶èï¼çµæºçµåEä¸åã
ããE/R1+E/R2=0.9ï¼E/R2+E/R3=0.5ã
ããå¾å°ï¼Eï¼R1+R2ï¼/ï¼R1R2ï¼=0.9ï¼Eï¼R2+R3ï¼/ï¼R2R3ï¼=0.5ã...............â¡
ãã两å¼ç¸æ¯ï¼[ï¼R1+R2ï¼/ï¼R2+R3ï¼]ÃR3/R1=9/5ã......â¢
ããâ ä»£å ¥â¢ï¼3R3/5R1=9/5ï¼å¾å°R3=3R1ã
ããå°â 'ãâ¡'ä»£å ¥â¡ä¸ç两个å¼åï¼
ããï¼8R1+5R2ï¼/ï¼R1R2ï¼=0.9ã
ããï¼3R2+8R3ï¼/ï¼R2R3ï¼=0.5ã
ããæ´çï¼å¾ï¼
ãã5/R1+8/R2=0.9ã
ãã8/R2+3/R3=0.5ã
ãã两å¼ç¸åï¼5/R1-3/R3=0.4ï¼èR3=3R1ï¼è§£å¾ï¼R1=10ï¼Î©ï¼ï¼R3=30Ωã
ããR2=20ï¼Î©ï¼ã
ããå¦å¤è¿å¯æ±å¾ï¼E=ï¼8Ã10+5Ã20ï¼/ï¼10+20ï¼=6ï¼Vï¼ãæE=ï¼3Ã20+8Ã30ï¼/ï¼20+30ï¼=6ï¼Vï¼ã
ãã串èçµè·¯ççµæµï¼I=3/ï¼10+20ï¼=0.1ï¼Aï¼æI=5/ï¼20+30ï¼=0.1ï¼Aï¼ã
都是电流表时为并联电路,则电压相等:0.9*R1R2/(R1+R2)=0.5*R2R3/(R2+R3)
又因为总电压等于8-U2=8-R2*5/(R2+R3)=0.9*R1R2/(R1+R2)
解三方程组得:R1=10
R2=20
R3=30