ÌâÄ¿ÄÚÈÝ

ÈçͼËùʾµÄµç·ÊÇÑо¿µçÁ÷¸úµçѹ¡¢µç×è¹ØÏµµÄµç·ͼ£®Í¼Öиø³öÁ˸ÃʵÑéËùÓõÄʵÑéÔª¼þ£®ÏÖ½«¸ÃʵÑéµÄÊý¾Ý£¬·Ö±ðÔÚ±í1¡¢±í2ÖÐÁгö£®
±í1µç×èRx=
15
15
¦¸£¬±í2µçѹU=
2
2
V


£¨1£©°´µç·ͼ˳Ðò£¬Óñʻ­Ïß±íʾµ¼Ïߣ¬½«Í¼ÒÒÖеÄʵÎïÁ¬½ÓÆðÀ´£®
£¨2£©·ÖÎö±í1µÄÊý¾Ý¿ÉµÃ½áÂÛ£º
±£³Öµç×è²»±äʱ£¬µçÁ÷¸úµçѹ³ÉÕý±È
±£³Öµç×è²»±äʱ£¬µçÁ÷¸úµçѹ³ÉÕý±È
£¬·ÖÎö±í2µÄÊý¾Ý¿ÉµÃ½áÂÛ£º
±£³Öµçѹ²»±äʱ£¬µçÁ÷¸úµç×è³É·´±È
±£³Öµçѹ²»±äʱ£¬µçÁ÷¸úµç×è³É·´±È

£¨3£©±í1ÖУ¬Rx=
15
15
¦¸
·ÖÎö£º´ËʵÑéÊÇÑо¿µ¼ÌåÖеĵçÁ÷Óëµçѹ¡¢µç×èµÄ¹ØÏµ£®ÓÉÓÚµçÁ÷¸úµçѹºÍµç×è¾ùÓйأ¬Òò´ËÔÚ¡°Ì½¾¿µç×èÉϵĵçÁ÷¸úµçѹµÄ¹ØÏµ¡±ÊµÑéÖУ¬Ó¦¿ØÖƵ¼ÌåµÄµç×è´óС²»±ä£®ÔÚ·ÖÎöµçÁ÷¸úµç×èµÄ¹ØÏµÊ±Ó¦¿ØÖƵçѹ²»±ä£®
»¬¶¯±ä×èÆ÷ÔÚ´ËʵÑéÖеÄ×÷Ó㺾ÍÊǸıäµç·ÖеĵçÁ÷£¬µ÷½Ú²¿·Öµç·Á½¶ËµÄµçѹ£®
½â´ð£º½â£º£¨1£©ÊµÎïÁ¬½ÓÈçÏÂͼËùʾ£®

£¨2£©Óɱí1µÄÊý¾ÝµÄÊý¾Ý¿ÉÖª£¬ÔÚµç×è²»±äʱ£¬µçÁ÷ËæµçѹµÄÔö´ó¶øÔö´ó£¬ÇÒÔö´óµÄ±¶ÊýÏàͬ£¬ËùÒԿɵÃʵÑé½áÂÛΪ£º±£³Öµç×è²»±äʱ£¬µçÁ÷¸úµçѹ³ÉÕý±È£»
Óɱí2µÄÊý¾ÝµÄÊý¾Ý¿ÉÖª£¬ÔÚµçѹ²»±äʱ£¬µçÁ÷Ëæµç×èµÄÔö´ó¶ø¼õС£¬ÇÒÔö´óµÄ±¶Êý»¥Îªµ¹Êý£¬ËùÒԿɵÃʵÑé½áÂÛΪ£º
±£³Öµçѹ²»±äʱ£¬µçÁ÷¸úµç×è³É·´±È£®
£¨3£©Óɱí1Êý¾Ý¿ÉÖªµ±µçѹµÈÓÚ1.5Vʱ£¬Í¨¹ýµ¼ÌåµÄµçÁ÷Ϊ0.1A£¬Óɹ«Ê½¿ÉµÃ£ºR=
U
I
=
1.5V
0.1A
=15¦¸
£®
¹Ê´ð°¸Îª£º£¨1£©ÊµÎïÁ¬½ÓÈçÉÏͼËùʾ£®
£¨2£©±£³Öµç×è²»±äʱ£¬µçÁ÷¸úµçѹ³ÉÕý±È£»±£³Öµçѹ²»±äʱ£¬µçÁ÷¸úµç×è³É·´±È£®
£¨3£©15£®
µãÆÀ£º±¾ÌâÖ÷Òª¿¼²éÓ¦ÓÿØÖƱäÁ¿·¨·ÖÎöʵÑé½áÂÛºÍÀûÓÃÅ·Ä·¶¨ÂɼÆËãµç×è´óСµÄÄÜÁ¦£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø