Question

20£®ÔÚÈçͼËùʾµÄµç·ÖУ¬Á½Æ½ÐÐÕý¶Ô½ðÊô°åA¡¢Bˮƽ·ÅÖã¬Á½°å¼äµÄ¾àÀëd=4.0cm£®µçÔ´µç¶¯ÊÆE=400V£¬ÄÚµç×èr=20¦¸£¬µç×èR₁=1980¦¸£®±ÕºÏ¿ª¹ØS£¬´ýµç·Îȶ¨ºó£¬´ø¸ºµçµÄСÇò£¨¿ÉÊÓΪÖʵ㣩ǡºÃÔÚÁ½¼«°å¼ä¾²Ö¹£®ÈôСÇòËù´øµçºÉÁ¿q=1.0¡Á10^-6C£¬ÖÊÁ¿m=5.0¡Á10^-4kg£¬²»¿¼ÂÇ¿ÕÆø×èÁ¦£¬ºöÂÔСÇò¶Ôµç·µÄÓ°Ï죬ȡg=10m/s²£®Çó£º
£¨1£©A¡¢BÁ½½ðÊô°å¼äµÄµçѹµÄ´óСU£»
£¨2£©»¬¶¯±ä×èÆ÷ÏûºÄµÄµç¹¦ÂÊP_»¬£»
£¨3£©µçÔ´µÄЧÂʦǣ®

Answer 1

·ÖÎö £¨1£©Ð¡ÇòÔڵ糡Á¦ºÍÖØÁ¦µÄ×÷ÓÃÏ´¦ÓÚƽºâ״̬£¬¸ù¾ÝqE=mg£¬ÓëE=$\frac{U}{d}$¼´¿ÉÇó½âU£®
£¨2£©±ä×èÆ÷Á½¶ËµÄµçѹµÈÓÚU£¬ÓÉI=$\frac{E}{{R}_{1}+{R}_{»¬}+r}$£¬U=IR_»¬£¬
Ïà½áºÏÇó³öR_»¬£¬»¬¶¯±ä×èÆ÷ÏûºÄµÄµç¹¦ÂÊ P_»¬=$\frac{{U}^{2}}{{R}_{»¬}}$£®
£¨3£©µçÔ´µÄЧÂʦÇ=$\frac{{P}_{³ö}}{{P}_{×Ü}}$=$\frac{{I}_{1}£¨{R}_{1}+{R}_{»¬}£©}{{I}_{2}£¨{R}_{1}+{R}_{»¬}+r£©}$£®

½â´ð ½â£º£¨1£©Ð¡ÇòÔڵ糡Á¦ºÍÖØÁ¦µÄ×÷ÓÃÏ´¦ÓÚƽºâ״̬£¬
ÉèA¡¢BÁ½¼«°å¼äµçѹΪU£¬
¸ù¾ÝqE=mg£¬ÓëE=$\frac{U}{d}$
½âµÃU=$\frac{mgd}{q}$=$\frac{5¡Á1{0}^{-4}¡Á10¡Á0.04}{1¡Á1{0}^{-6}}$=200V
£¨2£©Éè´Ëʱ»¬¶¯±ä×èÆ÷½ÓÈëµç·Öеĵç×èֵΪR_»¬£¬¸ù¾Ý±ÕºÏµç ·ŷķ¶¨ÂÉ¿ÉÖª£¬µç·ÖеĵçÁ÷
µÃ I=$\frac{E}{{R}_{1}+{R}_{»¬}+r}$£¬
¸ù¾Ý²¿·Öµç·ŷķ¶¨ÂÉ¿ÉÖª   U=IR_»¬
½âµÃ    R_»¬=2.0¡Á10³¦¸
»¬¶¯±ä×èÆ÷ÏûºÄµÄµç¹¦ÂÊ  P_»¬=$\frac{{U}^{2}}{{R}_{»¬}}$=20W
 £¨3£©µçÔ´µÄЧÂʦÇ=$\frac{{P}_{³ö}}{{P}_{×Ü}}$¡Á100%=$\frac{{I}_{1}£¨{R}_{1}+{R}_{»¬}£©}{{I}_{2}£¨{R}_{1}+{R}_{»¬}+r£©}$¡Á100%=99.5%£®
´ð£º
£¨1£©A¡¢BÁ½½ðÊô°å¼äµÄµçѹµÄ´óСUÊÇ200V£»
£¨2£©»¬¶¯±ä×èÆ÷ÏûºÄµÄµç¹¦ÂÊP_»¬ÊÇ20W£®
£¨3£©µçÔ´µÄЧÂʦÇÊÇ99.5%£®

µãÆÀ ±¾Ìâµç³¡Óëµç·µÄ×ÛºÏÓ¦Óã¬Ð¡ÇòÔڵ糡ÖÐ×öÔȼõËÙÔ˶¯£¬Óɶ¯Äܶ¨ÀíÇóµçѹ£®¸ù¾Ýµç·µÄ½á¹¹£¬ÓÉÅ·Ä·¶¨ÂÉÇó±ä×èÆ÷½ÓÈëµç·µÄµç×裮