ÌâÄ¿ÄÚÈÝ

ijµçÁ÷±íµÄÁ¿³ÌΪ50mA£¬ÄÚ×èΪ50¦¸£¬Æä±íÃæµÄ¿Ì¶ÈÅÌÒÑË𻵣¬ÒªÖØÐÂͨ¹ý²âÁ¿À´¿Ì»­³ö¿Ì¶ÈÖµ£¬ÓÐÏÂÁÐÆ÷²Ä£º

 A£®´ý²âµçÁ÷±í

 B£®9VÖ±Á÷µçÔ´

 C£®¡°0-10¦¸£¬1A¡±»¬¶¯±ä×èÆ÷

 D.¡°0-100¦¸£¬50mA¡±»¬¶¯±ä×èÆ÷

 E£®¡°0.6A£¬0.5¦¸¡±±ê×¼µçÁ÷±í

 F£®¡°3A£¬0.01¦¸¡±±ê×¼µçÁ÷±í

 G£®5¦¸±ê×¼µç×è

 F£®20¦¸±ê×¼µç×è

 I. ¿ª¹Ø¼°µ¼ÏßÈô¸É

(1)ӦѡÓõÄÆ÷²Ä                  (ÌîÑ¡ÓÃÆ÷²ÄÐòºÅ)¡£

(2)»­³öʵÑéµç·ͼ

(3)´ý¿Ì¶ÈµçÁ÷±íµÄµçÁ÷±í¿Ì¶ÈÖµx =        £®ËµÃ÷ʽÖи÷ÎïÀíÁ¿µÄ±íʾÒâÒå                        ¡£

 


£¨1£©ABCEGI£¨2·Ö£©£¨2£©µç·ͼÈçͼ¡£

£¨3£©              £¨2·Ö£©

ʽÖÐIΪE±íʾÊý£¬ÎªE±íÄÚ×裬Ϊ5¦¸±ê×¼µç×裬RΪ´ý¿Ì»­±íÄÚ×è

½âÎö£ºÖØÐÂͨ¹ý²âÁ¿À´¿Ì»­³ö¿Ì¶ÈÖµ£¬ÐèÒª²âÁ¿µçÁ÷±íÖеçÁ÷¡£ÓÉÓÚ±ê×¼µçÁ÷±íÁ¿³ÌÔ¶´óÓÚ´ý²âµçÁ÷±í£¬ÇÒ´ý²âµçÁ÷±íÄÚ×èÒÑÖª£¬ËùÒÔÉè¼Æ³É¡°0.6A£¬0.5¦¸¡±±ê×¼µçÁ÷±íÓë5¦¸±ê×¼µç×è´®ÁªºóÓë´ý²âµçÁ÷±í²¢Áªµç·£¬»¬¶¯±ä×èÆ÷²ÉÓ÷Öѹʽ½Ó·¨¡£ÓÉIxR=I(+)½âµÃ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

I£®ºÍ´òµã¼ÆʱÆ÷Ò»Ñù£¬¹âµç¼ÆʱÆ÷Ò²ÊÇÒ»ÖÖÑо¿ÎïÌåÔ˶¯Çé¿öµÄ³£ÓüÆʱÒÇÆ÷£¬Æä½á¹¹Èçͼ¼×Ëùʾ£¬a¡¢b·Ö±ðÊǹâµçÃŵļ¤¹â·¢ÉäºÍ½ÓÊÕ×°Ö㬵±ÓÐÎïÌå´Óa¡¢b¼äͨ¹ýʱ£¬¹âµç¼ÆʱÆ÷¾Í¿ÉÒÔÏÔʾÎïÌåµÄµ²¹âʱ¼ä£®ÏÖÀûÓÃÈçͼÒÒËùʾµÄ×°ÖòâÁ¿»¬¿éºÍ³¤1 m×óÓÒµÄľ°å¼äµÄ¶¯Ä¦²ÁÒòÊý£¬Í¼ÖÐMNÊÇˮƽ×ÀÃ棬QÊÇľ°åÓë×ÀÃæµÄ½Ó´¥µã£¬1ºÍ2Êǹ̶¨ÔÚľ°åÉÏÊʵ±Î»ÖõÄÁ½¸ö¹âµçÃÅ£¬ÓëÖ®Á¬½ÓµÄÁ½¸ö¹âµç¼ÆʱÆ÷ûÓл­³ö£®´ËÍ⣬ÔÚľ°å¶¥¶ËµÄPµã»¹Ðü¹Ò×ÅÒ»¸öǦ´¸£®Èû¬¿éd´Óľ°åµÄ¶¥¶Ë»¬Ï£¬¹âµçÃÅl£¬2¸÷×ÔÁ¬½ÓµÄ¼ÆʱÆ÷ÏÔʾµÄµ²¹âʱ¼ä·Ö±ðΪ2.5102sºÍ1.0102s£¬Ð¡»¬¿édµÄ¿í¶ÈΪ0.5cm£®

(1)»¬¿éͨ¹ý¹âµçÃÅ1µÄËÙ¶Èv1=      m/s£¬»¬¿éͨ¹ý¹âµçÃÅ2µÄËÙ¶Èv2=     m/s¡£

(2)ÒÑÖªµ±µØµÄÖØÁ¦¼ÓËÙ¶ÈΪg£¬v1¡¢v2ºÍÁ½¸ö¹âµçÃÅÖ®¼äµÄ¾àÀëLÒÑÖª£¬ÓÃÌṩµÄÃ׳߲â³öPµãµ½×ÀÃæ¸ß¶Èh£»ÖØ´¸ÔÚ×ÀÃæÉÏËùÖ¸µÄµãÓëQµãµÄ¾àÀëa£¬¶¯Ä¦²ÁÒòÊýµÄ±í´ïʽ

=         (ÓÃ×Öĸ±íʾ)¡£

II£®Ä³µçÁ÷±í   µÄÁ¿³ÌΪI0=50mA£¬ÄÚ×èΪr0=50£¬Æä±íÅ̶̿ÈÏßÒÑÄ£ºý²»Ç壬ҪÖØÐÂͨ¹ý²âÁ¿À´¿Ì»­³ö´ÓÁãµ½Âú¿Ì¶ÈµÄ¿Ì¶ÈÖµ£¬ÓÐÏÂÁÐÆ÷²Ä£º   

  A£®´ý²âµçÁ÷±í                     B£®6 VÖ±Á÷µçÔ´E

  C£®¡°0~10£¬l A¡±»¬¶¯±ä×èÆ÷R1    D£®¡°0~100£¬50 mA¡±»¬¶¯±ä×èÆ÷R2

  E£®¡°0.6 A£¬0.5¡±±ê×¼µçÁ÷±í        F£®¡°3 A£¬0.01¡±±ê×¼µçÁ÷±í      

G£®5¶¨Öµµç×èR3                  H£®20¶¨Öµµç×èR4       

I£®¿ª¹Ø¼°µ¼ÏßÈô¸É

  (1)ӦѡÓõÄÆ÷²ÄÓР                     (Ö»ÐèÌîдËùÑ¡Æ÷²ÄÐòºÅ)

  (2)ÔÚÓÒ²àÐéÏß¿òÄÚ»­³öʵÑéµç·ͼ£®

 (3)´ý²âµçÁ÷±íµÄµçÁ÷¿Ì¶ÈÖµµÄ±í´ïʽI=   £¬Ê½Öи÷ÎïÀíÁ¿Ëù±íʾµÄÒâÒå·Ö±ðΪ    £®

 

I£®ºÍ´òµã¼ÆʱÆ÷Ò»Ñù£¬¹âµç¼ÆʱÆ÷Ò²ÊÇÒ»ÖÖÑо¿ÎïÌåÔ˶¯Çé¿öµÄ³£ÓüÆʱÒÇÆ÷£¬Æä½á¹¹Èçͼ¼×Ëùʾ£¬a¡¢b·Ö±ðÊǹâµçÃŵļ¤¹â·¢ÉäºÍ½ÓÊÕ×°Ö㬵±ÓÐÎïÌå´Óa¡¢b¼äͨ¹ýʱ£¬¹âµç¼ÆʱÆ÷¾Í¿ÉÒÔÏÔʾÎïÌåµÄµ²¹âʱ¼ä£®ÏÖÀûÓÃÈçͼÒÒËùʾµÄ×°ÖòâÁ¿»¬¿éºÍ³¤1 m×óÓÒµÄľ°å¼äµÄ¶¯Ä¦²ÁÒòÊý£¬Í¼ÖÐMNÊÇˮƽ×ÀÃ棬QÊÇľ°åÓë×ÀÃæµÄ½Ó´¥µã£¬1ºÍ2Êǹ̶¨ÔÚľ°åÉÏÊʵ±Î»ÖõÄÁ½¸ö¹âµçÃÅ£¬ÓëÖ®Á¬½ÓµÄÁ½¸ö¹âµç¼ÆʱÆ÷ûÓл­³ö£®´ËÍ⣬ÔÚľ°å¶¥¶ËµÄPµã»¹Ðü¹Ò×ÅÒ»¸öǦ´¸£®Èû¬¿éd´Óľ°åµÄ¶¥¶Ë»¬Ï£¬¹âµçÃÅl£¬2¸÷×ÔÁ¬½ÓµÄ¼ÆʱÆ÷ÏÔʾµÄµ²¹âʱ¼ä·Ö±ðΪ2.5102sºÍ1.0102s£¬Ð¡»¬¿édµÄ¿í¶ÈΪ0.5cm£®

(1)»¬¿éͨ¹ý¹âµçÃÅ1µÄËÙ¶Èv1=       m/s£¬»¬¿éͨ¹ý¹âµçÃÅ2µÄËÙ¶Èv2=      m/s¡£

(2)ÒÑÖªµ±µØµÄÖØÁ¦¼ÓËÙ¶ÈΪg£¬v1¡¢v2ºÍÁ½¸ö¹âµçÃÅÖ®¼äµÄ¾àÀëLÒÑÖª£¬ÓÃÌṩµÄÃ׳߲â³öPµãµ½×ÀÃæ¸ß¶Èh£»ÖØ´¸ÔÚ×ÀÃæÉÏËùÖ¸µÄµãÓëQµãµÄ¾àÀëa£¬¶¯Ä¦²ÁÒòÊýµÄ±í´ïʽ

=          (ÓÃ×Öĸ±íʾ)¡£

II£®Ä³µçÁ÷±í   µÄÁ¿³ÌΪI0=50mA£¬ÄÚ×èΪr0=50£¬Æä±íÅ̶̿ÈÏßÒÑÄ£ºý²»Ç壬ҪÖØÐÂͨ¹ý²âÁ¿À´¿Ì»­³ö´ÓÁãµ½Âú¿Ì¶ÈµÄ¿Ì¶ÈÖµ£¬ÓÐÏÂÁÐÆ÷²Ä£º   

  A£®´ý²âµçÁ÷±í                      B£®6 VÖ±Á÷µçÔ´E

  C£®¡°0~10£¬l A¡±»¬¶¯±ä×èÆ÷R1     D£®¡°0~100£¬50 mA¡±»¬¶¯±ä×èÆ÷R2

  E£®¡°0.6 A£¬0.5¡±±ê×¼µçÁ÷±í        F£®¡°3 A£¬0.01¡±±ê×¼µçÁ÷±í      

G£®5¶¨Öµµç×èR3                   H£®20¶¨Öµµç×èR4       

I£®¿ª¹Ø¼°µ¼ÏßÈô¸É

  (1)ӦѡÓõÄÆ÷²ÄÓР                      (Ö»ÐèÌîдËùÑ¡Æ÷²ÄÐòºÅ)

  (2)ÔÚÓÒ²àÐéÏß¿òÄÚ»­³öʵÑéµç·ͼ£®

 (3)´ý²âµçÁ÷±íµÄµçÁ÷¿Ì¶ÈÖµµÄ±í´ïʽI=   £¬Ê½Öи÷ÎïÀíÁ¿Ëù±íʾµÄÒâÒå·Ö±ðΪ     £®

 

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

¾«Ó¢¼Ò½ÌÍø