ÌâÄ¿ÄÚÈÝ
ÔÚ¡°²âÁ¿¶¨Öµµç×è×èÖµ¡±ºÍ¡°²âÁ¿Ð¡µÆÅݵ繦ÂÊ¡±µÄʵÑéÖУº

£¨1£©Ð¡Ã÷ÕýÈ·Á¬½ÓºÃµç·ºó£¬±ÕºÏ¿ª¹Ø£¬Òƶ¯»¬¶¯±ä×èÆ÷µÄ»¬Æ¬µ½Ä³Ò»Î»ÖÃʱ£¬¹Û²ìµ½µçÁ÷±íʾÊýΪ0.2A£¬µçѹ±íʾÊýÈçͼ¼×Ëùʾ£¬Ôò¶¨Öµµç×è×èֵΪ ¦¸£»
£¨2£©Ð¡Ã÷½ö°ÑÕâÒ»´ÎʵÑé²âµÃµÄ×èÖµ×÷Ϊ×îºóµÄ½á¹û£¬ÄãÈÏΪºÏÊÊÂð?
ÀíÓÉÊÇ £»
£¨3£©²â³öµç×èºó£¬Ð¡Ã÷½«µç·Öеֵ͍µç×è»»³ÉÁËÒ»¸ö¶î¶¨µçѹΪ2.5VµÄСµÆÅÝ£¬²âÁ¿Ð¡µÆÅݵĵ繦ÂÊ£¬Í¼ÒÒÊÇСÃ÷ÖØÐÂÁ¬½ÓµÄ²»ÍêÕûµÄµç·£¬ÇëÄãÓñʴúÌæµ¼Ïß½«ÊµÎïµç·Á¬½ÓÍêÕû£¨ÒªÇ󣺻¬Æ¬PÏòÓÒÒÆ¶¯Ê±£¬Ð¡µÆÅݱäÁÁ£©£»
£¨4£©±ÕºÏ¿ª¹Øºó£¬µ÷½Ú»¬¶¯±ä×èÆ÷»¬Æ¬P£¬Ê¹µçѹ±íʾÊýΪ Vʱ£¬Ð¡µÆÅÝÕý³£·¢¹â£¬Èç¹û´ËʱµçÁ÷±íʾÊýÈçͼ±ûËùʾ£¬ÔòСµÆÅݵĶ¹¦ÂÊÊÇ W£»

£¨5£©ÁíһʵÑéС×é°´ÕýÈ··½·¨Íê³ÉʵÑéºó£¬·Ö±ð»æÖÆÁ˶¨Öµµç×èºÍСµÆÅݵÄU - IͼÏñ£¬È綡ͼËùʾ£¬Ôò±íʾ¡°¶¨Öµµç×衱µÄU - IͼÏñµÄÊÇ (Ìî¡°¼×¡±»ò¡°ÒÒ¡±)£¬ÕâÑùÅжϵÄÒÀ¾ÝÊÇ £®
£¨1£©Ð¡Ã÷ÕýÈ·Á¬½ÓºÃµç·ºó£¬±ÕºÏ¿ª¹Ø£¬Òƶ¯»¬¶¯±ä×èÆ÷µÄ»¬Æ¬µ½Ä³Ò»Î»ÖÃʱ£¬¹Û²ìµ½µçÁ÷±íʾÊýΪ0.2A£¬µçѹ±íʾÊýÈçͼ¼×Ëùʾ£¬Ôò¶¨Öµµç×è×èֵΪ ¦¸£»
£¨2£©Ð¡Ã÷½ö°ÑÕâÒ»´ÎʵÑé²âµÃµÄ×èÖµ×÷Ϊ×îºóµÄ½á¹û£¬ÄãÈÏΪºÏÊÊÂð?
ÀíÓÉÊÇ £»
£¨3£©²â³öµç×èºó£¬Ð¡Ã÷½«µç·Öеֵ͍µç×è»»³ÉÁËÒ»¸ö¶î¶¨µçѹΪ2.5VµÄСµÆÅÝ£¬²âÁ¿Ð¡µÆÅݵĵ繦ÂÊ£¬Í¼ÒÒÊÇСÃ÷ÖØÐÂÁ¬½ÓµÄ²»ÍêÕûµÄµç·£¬ÇëÄãÓñʴúÌæµ¼Ïß½«ÊµÎïµç·Á¬½ÓÍêÕû£¨ÒªÇ󣺻¬Æ¬PÏòÓÒÒÆ¶¯Ê±£¬Ð¡µÆÅݱäÁÁ£©£»
£¨4£©±ÕºÏ¿ª¹Øºó£¬µ÷½Ú»¬¶¯±ä×èÆ÷»¬Æ¬P£¬Ê¹µçѹ±íʾÊýΪ Vʱ£¬Ð¡µÆÅÝÕý³£·¢¹â£¬Èç¹û´ËʱµçÁ÷±íʾÊýÈçͼ±ûËùʾ£¬ÔòСµÆÅݵĶ¹¦ÂÊÊÇ W£»
£¨5£©ÁíһʵÑéС×é°´ÕýÈ··½·¨Íê³ÉʵÑéºó£¬·Ö±ð»æÖÆÁ˶¨Öµµç×èºÍСµÆÅݵÄU - IͼÏñ£¬È綡ͼËùʾ£¬Ôò±íʾ¡°¶¨Öµµç×衱µÄU - IͼÏñµÄÊÇ (Ìî¡°¼×¡±»ò¡°ÒÒ¡±)£¬ÕâÑùÅжϵÄÒÀ¾ÝÊÇ £®
£¨1£©10 £¨2£©²»ºÏÊÊ£¬Ã»ÓнøÐжà´Î²âÁ¿£¬ÊµÑéÎó²î´ó£¨3£©¼ûͼ

£¨4£©2.5£¬ 0.75 £¨5£©¼× (1·Ö) ¶¨Öµµç×è×èÖµÒ»¶¨£¬µçÁ÷Óëµçѹ³ÉÕý±È(»òµÆÅݵĵÆË¿µç×èÊÜζÈÓ°Ï죬µçÁ÷Óëµçѹ²»³ÉÕý±È)(2·Ö)
£¨4£©2.5£¬ 0.75 £¨5£©¼× (1·Ö) ¶¨Öµµç×è×èÖµÒ»¶¨£¬µçÁ÷Óëµçѹ³ÉÕý±È(»òµÆÅݵĵÆË¿µç×èÊÜζÈÓ°Ï죬µçÁ÷Óëµçѹ²»³ÉÕý±È)(2·Ö)
£¨1£©µçѹ±íÑ¡Ôñ0¡«3VÁ¿³Ì£¬Ã¿Ò»¸ö´ó¸ñ´ú±í1V£¬Ã¿Ò»¸öС¸ñ´ú±í0.1V£¬µçѹΪ2V£¬R="U" /I ="2V" /0.2A =10¦¸£®
£¨2£©Ö»°ÑÒ»´ÎʵÑé²âµÃµÄ×èÖµ×÷Ϊ×îºó½á¹ûÊDz»ºÏÊʵģ¬ÒòΪ¶¨Öµµç×èµÄ×èÖµÊDz»±äµÄ£¬Òª½øÐжà´Î²âÁ¿Ç󯽾ùÖµÀ´¼õСÎó²î£®
£¨3£©»¬Æ¬ÓÒÒÆ£¬µÆÅݱäÁÁ£¬µç·µçÁ÷±ä´ó£¬µç·µç×è±äС£¬»¬¶¯±ä×èÆ÷Òª½ÓÈëÓÒ°ë¶Î£¬»¬¶¯±ä×èÆ÷Òª½ÓB½ÓÏßÖù£¬°Ñ»¬¶¯±ä×èÆ÷´®ÁªÔÚµç·ÖУ®Èçͼ£®

£¨4£©µçѹ±í²âµÆÅÝÁ½¶Ëµçѹ£¬µçѹ±íʾÊýΪ2.5VʱµÆÅÝÕý³£¹¤×÷£®
µçÁ÷±íÑ¡Ôñ0¡«0.6AÁ¿³Ì£¬Ã¿Ò»¸ö´ó¸ñ´ú±í0.2A£¬Ã¿Ò»¸öС¸ñ´ú±í0.02A£¬µçÁ÷Ϊ0.3A£¬
P=U'I'=2.5V¡Á0.3A=0.75W£®
£¨5£©¼×ͼ²»±äµÄ£¬µçÁ÷¸úµçѹ³ÉÕý±È£¬¼×ÊǶ¨Öµµç×裮
ÒÒͼµÄU /I µÄ±ÈÖµÊÇÖð½¥Ôö´ó£¬ÒÒͼ±íʾµÆÅݵç×裮
£¨2£©Ö»°ÑÒ»´ÎʵÑé²âµÃµÄ×èÖµ×÷Ϊ×îºó½á¹ûÊDz»ºÏÊʵģ¬ÒòΪ¶¨Öµµç×èµÄ×èÖµÊDz»±äµÄ£¬Òª½øÐжà´Î²âÁ¿Ç󯽾ùÖµÀ´¼õСÎó²î£®
£¨3£©»¬Æ¬ÓÒÒÆ£¬µÆÅݱäÁÁ£¬µç·µçÁ÷±ä´ó£¬µç·µç×è±äС£¬»¬¶¯±ä×èÆ÷Òª½ÓÈëÓÒ°ë¶Î£¬»¬¶¯±ä×èÆ÷Òª½ÓB½ÓÏßÖù£¬°Ñ»¬¶¯±ä×èÆ÷´®ÁªÔÚµç·ÖУ®Èçͼ£®
£¨4£©µçѹ±í²âµÆÅÝÁ½¶Ëµçѹ£¬µçѹ±íʾÊýΪ2.5VʱµÆÅÝÕý³£¹¤×÷£®
µçÁ÷±íÑ¡Ôñ0¡«0.6AÁ¿³Ì£¬Ã¿Ò»¸ö´ó¸ñ´ú±í0.2A£¬Ã¿Ò»¸öС¸ñ´ú±í0.02A£¬µçÁ÷Ϊ0.3A£¬
P=U'I'=2.5V¡Á0.3A=0.75W£®
£¨5£©¼×ͼ²»±äµÄ£¬µçÁ÷¸úµçѹ³ÉÕý±È£¬¼×ÊǶ¨Öµµç×裮
ÒÒͼµÄU /I µÄ±ÈÖµÊÇÖð½¥Ôö´ó£¬ÒÒͼ±íʾµÆÅݵç×裮
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿