Question

6£®Ä³ÎïÀí̽¾¿Ð¡×éÉè¼ÆÁËÈçͼ1ËùʾµÄµç·²âÁ¿Ò»¿éµç¶¯³µÐîµç³ØµÄµç¶¯ÊƺÍÄÚ×裮ÒÑÖª¸Ãµç¶¯³µÐîµç³ØµÄµç¶¯ÊÆԼΪ12V£¬¶¨Öµµç×èR₀=5¦¸£¬R₁=20¦¸£¬µçÁ÷±íÁ½¸öÁ¿³Ì·Ö±ðΪ0.6A£¨ÄÚ×èΪ0.5¦¸£©ºÍ3A£¨ÄÚ×èΪ0.1¦¸£©£®
£¨1£©Çë¸ù¾Ýͼ1Ëùʾµç·ͼ½«Í¼2µÄʵÑéÆ÷²ÄÓñʻ­´úÌæµ¼ÏßÁ¬½Ó³ÉʵÑéµç·£®
£¨2£©¸Ä±äµç×èÏäµÄ×èÖµR£¬µÃµ½¶ÔÓ¦µÄµçÁ÷±í¶ÁÊýI£®Èç¹ûʵÑé²âµÃ¶à×飨R£¬I£©ÊµÑéÊý¾Ý£¬ÓÃͼÏó·¨´¦ÀíʵÑéÊý¾Ý£¬ÒªÊ¹µÃµ½µÄͼÏóΪֱÏߣ¬Ôò×Ý¡¢ºá×ø±ê±íʾµÄÎïÀíÁ¿·Ö±ðÊÇD£®
A£®I¡¢R      B.$\frac{1}{I}$¡¢$\frac{1}{R}$   C.$\frac{1}{I}$¡¢$\frac{1}{R+{R}_{1}}$    D.$\frac{1}{I}$¡¢$\frac{1}{R+{R}_{0}}$
£¨3£©ÈôÕýÈ·Ñ¡Ôñ×ø±êºó£¬µÃµ½µÄͼÏóбÂÊΪk£¬ÔÚ×ÝÖáµÄ½Ø¾àΪb£¬ÔòµçÔ´µç¶¯ÊÆΪE=$\frac{{£¨{R}_{1}+{r}_{A}£©}^{2}}{b{R}_{1}+b{r}_{A}-k}$£¬ÄÚ×èΪr=$\frac{k£¨{R}_{1}+{r}_{A}£©}{b{R}_{1}+b{r}_{A}-k}$£®£¨µçÁ÷±íÄÚ×èÓÃr_A±íʾ£¬¶¨Öµµç×è·Ö±ðÓÃR₀ºÍR₁±íʾ£©

Answer 1

·ÖÎö £¨1£©Ïȸù¾ÝµçÔ´µç¶¯Êƺ͵緽ṹÇó³öµçÁ÷±íµçÁ÷×î´óÖµ£¬Ñ¡ÔñÁ¿³Ì£¬ÔÙ¸ù¾Ýµç·ͼÁ¬½ÓʵÎïͼ£»
£¨2£©¸ù¾Ý±ÕºÏµç·ŷķ¶¨ÂÉÁÐʽ·ÖÎö¼´¿É£»
£¨3£©¸ù¾ÝͼÏóµÄбÂÊÓë½Ø¾à½áºÏ±í´ïʽ·ÖÎö¼´¿É£®

½â´ð ½â£º£¨1£©µçÔ´µçѹΪ12V£¬R₁=20¦¸£¬Ôòͨ¹ýµçÁ÷±íµÄµçÁ÷×î´ó${I}_{max}£¼\frac{E}{{R}_{1}}=\frac{12}{20}=0.6A$£¬ÔòÁ¿³ÌÑ¡Ôñ0-0.6A£¬
¸ù¾Ýµç·ͼÁ¬½ÓʵÎïͼ£¬ÈçͼËùʾ£º

£¨2£©¸ù¾Ý±ÕºÏµç·ŷķ¶¨Âɵãº
$E=I{£¨R}_{1}+{r}_{A}£©+[\frac{I{£¨R}_{1}+{r}_{A}£©}{{R}_{0}+R}+I]r$
ÕûÀíµÃ£º$\frac{1}{I}=\frac{{£¨R}_{1}+{r}_{A}£©r}{E}•\frac{1}{£¨{R}_{0}+R£©}+\frac{r+{R}_{1}+{r}_{A}}{E}$£¬ÒªÊ¹µÃµ½µÄͼÏóΪֱÏߣ¬Ôò×Ý¡¢ºá×ø±ê±íʾµÄÎïÀíÁ¿·Ö±ðÊÇ$\frac{1}{I}$¡¢$\frac{1}{R+{R}_{0}}$£¬¹ÊDÕýÈ·£¬ABC´íÎó£®
¹ÊÑ¡£ºD£»
£¨3£©¸ù¾Ý$\frac{1}{I}=\frac{{£¨R}_{1}+{r}_{A}£©r}{E}•\frac{1}{£¨{R}_{0}+R£©}+\frac{r+{R}_{1}+{r}_{A}}{E}$£¬½áºÏͼÏó¿ÉÖª£¬k=$\frac{{£¨R}_{1}+{r}_{A}£©r}{E}$£¬b=$\frac{r+{R}_{1}+{r}_{A}}{E}$£¬
½âµÃ£ºr=$\frac{k£¨{R}_{1}+{r}_{A}£©}{b{R}_{1}+b{r}_{A}-k}$£¬E=$\frac{£¨{R}_{1}+{r}_{A}£©^{2}}{b{R}_{1}+b{r}_{A}-k}$
¹Ê´ð°¸Îª£º£¨1£©ÈçͼËùʾ£»£¨2£©D£»£¨3£©$\frac{{£¨{R}_{1}+{r}_{A}£©}^{2}}{b{R}_{1}+b{r}_{A}-k}$£»$\frac{k£¨{R}_{1}+{r}_{A}£©}{b{R}_{1}+b{r}_{A}-k}$£®

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éÁ˱պϵç·ŷķ¶¨ÂɵÄÖ±½ÓÓ¦Óã¬×¢ÒâµçÁ÷±íµÄÄÚ×è²»ÄܺöÂÔ£¬ÄܽáºÏÊýѧ֪ʶÇó½â£¬ÄѶÈÊÊÖУ®