设ï¼ååç¯ççµååå«ä¸ºU1ãU2ï¼çµæµåå«ä¸ºI1ãI2ï¼çµåçåå«ä¸ºP1ãP2ã
解ï¼
ï¼1ï¼åé»å¨ççµåä¸ç¯ççµåä¹åä¸åï¼çäºçµæºçµåï¼ï¼åé»å¨ççµååå°1Vï¼åç¯ççµåå¢å¤§1Vã
对äºç¯ï¼U1=I1Rç¯ï¼U2=I2Rç¯
U2-U1=ï¼I2 - I1)Rç¯
Rç¯=(U2-U1)/(I2 - I1) = 1V/0.2A = 5Ω
I2 - I1=0.2
I2=0.2 + I1
ç¯çµåçåå P2-P1=I2å¹³æ¹Rç¯-I1å¹³æ¹Rç¯
=(I2å¹³æ¹ - I1å¹³æ¹)Rç¯
=ï¼»(0.2 + I1) å¹³æ¹ - I1å¹³æ¹ï¼½Rç¯
=2I1+0.2
å 为 P2-P1=0.4ï¼æ以 2I1+0.2=0.4
å¾ ç¯åæ¥ççµæµ I1=0.1(A)
ç¯åæ¥ççµæµ I2=0.2+ I1 = 0.2+0.1=0.3(A)
ç¯åæ¥ççµå U2=I2Rç¯= 0.3Ã5 =1.5(V)
(2) ç¯åæ¥ççµåç P2=(9/25)Pé¢
æ ¹æ®P=UU/R,
P2÷Pé¢=(U2å¹³æ¹/Rç¯)÷Ué¢å¹³æ¹/Rç¯)
= U2å¹³æ¹Ã·Ué¢å¹³æ¹
(9/25)Pé¢Ã·Pé¢=1.5å¹³æ¹Ã·Ué¢å¹³æ¹
ç¯çé¢å®çµå Ué¢=2.5(V).
ç¯çé¢å®çµæµ Ié¢=Ué¢/Rç¯=2.5V/5Ω=0.5A (å°äº0.6Aï¼çµæµè¡¨è½å®å
¨å·¥ä½ãå°äº1Aï¼åé»å¨è½å®å
¨å·¥ä½ãï¼
å½åé»å¨ä¸¤ç«¯çµåçäº3V(åé»å¨è½å®å
¨å·¥ä½)ï¼ç¯ä¸¤ç«¯çµåçäº2.5Vï¼ç¯æ£å¸¸åå
) æ¶ï¼
çµæºçµåçæ大å¼U=2.5V+3V=5.5V
ï¼æ³¨ï¼åé»å¨è¿å
¥çµè·¯ä¸çæå°çµé»ä¸º Ré»ï¼=3V/0.5A=6Ωï¼è¥åé»å¨çé»å¼å°äº6Ωï¼ç¯ä¼ç§ãï¼
(3) çµè·¯ä¸çæ大çµæµä¸º0.5A(çäºç¯çé¢å®çµæµ)ï¼
çµè·¯æ¶èçæ大åç P=UI=4VÃ0.5A=2W
çµè·¯ä¸çæå°çµæµ Iï¼=U/(Rç¯+Råé»ï¼)=4V/(5Ω+20Ω)=0.16A
çµè·¯æ¶èçæå°åç Pï¼=UIï¼=4VÃ0.16A=0.64W
çï¼
ï¼1ï¼ç¯æ³¡ççµé»æ¯5Ωãæ»çPæ»å¨åç¯æ³¡ä¸¤ç«¯çå®é
çµåæ¯1.5Vã
ï¼2ï¼ä¸ºä¿è¯çµè·¯ä¸åå
件çè½å®å
¨ä½¿ç¨ï¼çµæºçµåçæ大å¼ä¸º5.5Vã
ï¼3ï¼è¥ä½¿ç¨çµå为4Vççµæºï¼åæ´ä¸ªçµè·¯æ¶èçµåççæ大å¼ä¸º2Wï¼
æå°å¼ä¸º0.64W ã
温馨提示:答案为网友推荐,仅供参考