Question

11£®Êµ¼ÊµÄµçÔ´¶¼ÓÐÒ»¶¨µÄµç×裬Èç¸Éµç³Ø£¬ÎÒÃÇÐèÒªÓÃËüµÄµçѹ U ºÍµç×èrÁ½¸öÎïÀíÁ¿À´ÃèÊöËü£®ÄÚ×è»áÏûºÄµçÄܲ¢½«Æäת»¯ÎªÄÚÄÜ£®Êµ¼Ê¼ÆËã¹ý³ÌÖУ¬¿ÉÒÔ°ÑËü¿´³ÉÊÇÓÉÒ»¸öµçѹΪ U¡¢µç×èΪ 0 µÄÀíÏëµçÔ´ÓëÒ»¸öµç×èֵΪrµÄµç×è´®Áª¶ø³É£¬Èçͼ¼×Ëùʾ£ºÈçͼËùʾ£¬½«µç×èΪR½ÓÔÚµçѹΪU£¬ÄÚµç×è´óСΪrµÄµçÔ´Á½¶Ë£¬µ±¿ª¹Ø±ÕºÏºó£º
£¨1£©Ö¤Ã÷£ºµç·µÄ×ܵç×èR_×Ü=R+r
£¨2£©ÇóµçÔ´µÄÊä³ö¹¦ÂÊ£¨¼´µç×èRµÄµç¹¦ÂÊ£©
£¨3£©µ±µç×èRÂú×ãʲôÌõ¼þʱ£¬µçÔ´µÄÊä³ö¹¦ÂÊ×î´ó£¿×î´óÖµÊǶàÉÙ£¿

Answer 1

·ÖÎö £¨1£©¸ù¾Ýµç×èµÄ´®ÁªÇó½â£»
£¨2£©ÓÉÅ·Ä·¶¨ÂÉÇó³öµç·µÄµçÁ÷£¬¸ù¾Ýµç¹¦ÂÊP=I²RÇóµç¹¦ÂÊ£»
£¨3£©µç×èRµÄµç¹¦ÂÊP=I²R=$£¨\frac{U}{R+r}£©^{2}R$£¬ÓÉÊýѧ֪ʶÇó½â£®

½â´ð ½â£º£¨1£©Êµ¼ÊµÄµçÔ´¶¼ÓÐÒ»¶¨µÄµç×裬¿ÉÒÔ°ÑËü¿´³ÉÊÇÓÉÒ»¸öµçѹΪU¡¢µç×èΪ0 µÄÀíÏëµçÔ´ÓëÒ»¸öµç×èֵΪrµÄµç×è´®Áª¶ø³É£¬½«µç×èΪR½ÓÔÚµçѹΪU£¬ÄÚµç×è´óСΪrµÄµçÔ´Á½¶Ë£¬¸ù¾Ýµç×èµÄ´®Áª¹æÂÉ£¬
µç·µÄ×ܵç×èR_×Ü=R+r£»
£¨2£©¸ù¾ÝÅ·Ä·¶¨ÂÉ£¬µç·ÖеĵçÁ÷I=$\frac{U}{{R}_{×Ü}}=\frac{U}{R+r}$£¬µçÔ´µÄÊä³ö¹¦ÂÊ£¬¼´µç×èRµÄµç¹¦ÂÊP=I²R=$£¨\frac{U}{R+r}£©^{2}R$=$\frac{{U}^{2}R}{£¨R+r£©^{2}}$£»
£¨3£©µçÔ´µÄÊä³ö¹¦ÂÊ£º
P=$\frac{{U}^{2}R}{£¨R+r£©^{2}}$=$\frac{{U}^{2}}{£¨r+R£©^{2}\frac{1}{R}}$=$\frac{{U}^{2}}{\frac{{r}^{2}+2Rr{+R}^{2}}{R}}$=$\frac{{U}^{2}}{\frac{{r}^{2}-2Rr{+R}^{2}+4Rr}{R}}$=$\frac{{U}^{2}}{\frac{£¨r-R£©^{2}+4Rr}{R}}$=$\frac{{U}^{2}}{\frac{£¨r-R£©^{2}}{R}+4r}$£¬
¿ÉÖªµ±R=rʱ£¬·Öĸ×îС£¬P×î´ó£®
×î´ó¹¦ÂÊP_´ó=$\frac{{U}^{2}}{4R}$£®
´ð£º£¨1£©Ö¤Ã÷ÈçÉÏ£»
£¨2£©µçÔ´µÄÊä³ö¹¦ÂÊΪ$\frac{{U}^{2}R}{£¨R+r£©^{2}}$£»
£¨3£©µ±µç×èR=rʱ£¬µçÔ´µÄÊä³ö¹¦ÂÊ×î´ó£»×î´óֵΪ$\frac{{U}^{2}}{4R}$£®

µãÆÀ ±¾Ì⿼²é´®Áªµç·µÄ¹æÂɼ°Å·Ä·¶¨ÂÉ¡¢µç¹¦Âʹ«Ê½ºÍÊýѧ֪ʶµÄÔËÓã¬ÌåÏÖÁËÓë¸ßÖÐÎïÀí֪ʶµÄÏνӣ¬ÄѶȽϴó£®