ÌâÄ¿ÄÚÈÝ
£¨2010?³¯Ñô£©Ð¡»ªÔÚ¡°¶àÖÖ·½·¨²âµç×衱µÄ¿ÎÍâ̽¾¿ÊµÑéÖУ¬Éè¼Æ³öÈçͼ¼×ËùʾµÄÔÀíͼ£¬R1=20¦¸£¬R2=10¦¸£¬¾ùΪÒÑÖª×èÖµµÄ¶¨Öµµç×裮RxΪ´ý²âµç×裮
£¨1£©Ð¡»ª½øÐÐÁËÁ½²½ÊµÑé²Ù×÷£¬²¢»ñÈ¡Ïà¹ØÊý¾ÝÈç±íÖÐËùʾ£¬Ôò¿ÉÒÔÍƶÏRx=
£¨2£©Ç밴С»ªÉè¼ÆµÄÔÀíͼ£¬ÒԱʻÏß´úÌæµ¼Ïß½«Í¼ÒÒʵÎïÁ¬½ÓÆðÀ´£®
£¨3£©Èç¹ûС»ªÉè¼ÆºÃÔÀíͼºó£¬Î´ÕÒµ½µçѹ±í£¬¶øÆäËûÆ÷²Ä¾ùÒÑÕÒµ½£¬Ö®ºó£¬ÀÏʦ¸øС»ªÒ»Ö»µçÁ÷±í£®Ç뽫µçÁ÷±í½ÓÈëÊʵ±µÄλÖã¨ÊµÑé¹ý³ÌÖв»µÃ¸ü»»ÆäλÖã©£¬²¢½«Ðµç·ͼ»ÔÚ¡°Í¼±û¡±´¦·½¿òÖУ®ÈôʵÑé²Ù×÷ÈÔÈç±íÖÐËùÊö£¬µÃ³öµçÁ÷±í¶ÁÊýΪI1¡¢I2£¬ÔòRx=
£¨ÓÃI1¡¢I2¡¢R1¡¢R2±íʾ£©£®
£¨4£©»ØÒäƽʱµÄѧϰÌåÑ飬µç³Ø±íÃæµÄ·âËÜÖ½ÉÏ£¬ÍùÍù±êÓУ¨Èç1.5V£©µçѹֵ£¬ÊµÑéÖÐ
ʵÑé²Ù×÷ | µç±í¶ÁÊý | ||
Sl¡¢S2±ÕºÏ£¬S3¶Ï¿ª | U1=1V | ||
Sl¡¢S3±ÕºÏ£¬S2¶Ï¿ª | U2=2V |
40¦¸
40¦¸
£®£¨2£©Ç밴С»ªÉè¼ÆµÄÔÀíͼ£¬ÒԱʻÏß´úÌæµ¼Ïß½«Í¼ÒÒʵÎïÁ¬½ÓÆðÀ´£®
£¨3£©Èç¹ûС»ªÉè¼ÆºÃÔÀíͼºó£¬Î´ÕÒµ½µçѹ±í£¬¶øÆäËûÆ÷²Ä¾ùÒÑÕÒµ½£¬Ö®ºó£¬ÀÏʦ¸øС»ªÒ»Ö»µçÁ÷±í£®Ç뽫µçÁ÷±í½ÓÈëÊʵ±µÄλÖã¨ÊµÑé¹ý³ÌÖв»µÃ¸ü»»ÆäλÖã©£¬²¢½«Ðµç·ͼ»ÔÚ¡°Í¼±û¡±´¦·½¿òÖУ®ÈôʵÑé²Ù×÷ÈÔÈç±íÖÐËùÊö£¬µÃ³öµçÁ÷±í¶ÁÊýΪI1¡¢I2£¬ÔòRx=
I1(R1+R2)-I2R1 |
I2 |
I1(R1+R2)-I2R1 |
I2 |
£¨4£©»ØÒäƽʱµÄѧϰÌåÑ飬µç³Ø±íÃæµÄ·âËÜÖ½ÉÏ£¬ÍùÍù±êÓУ¨Èç1.5V£©µçѹֵ£¬ÊµÑéÖÐ
²»ÄÜ
²»ÄÜ
£¨Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±£©½«¸ÃµçѹֵÈ϶¨ÎªËüʵ¼Ê¸øµç·ÌṩµÄµçѹ£¬ÒòΪʵ¼ÊµçѹÍùÍùÓë±êʶµçѹֵ²»Ò»ÖÂ
ʵ¼ÊµçѹÍùÍùÓë±êʶµçѹֵ²»Ò»ÖÂ
£®·ÖÎö£º£¨1£©¸ù¾Ý¿ª¹ØµÄ¿ªºÏÇé¿ö£¬·ÖÎö³öµç·µÄÁ¬½Ó·½Ê½£¬²¢¸ù¾Ý·½¸ñÖиø³öµÄÌõ¼þÁгöÏàÓ¦µÄ¹Øϵʽ£¬È»ºó¸ù¾Ý¹Øϵʽ±ã¿ÉÇó³öËùÇóµÄÁ¿£®
£¨2£©°´ÕÕµç·ͼ½«¸÷¸öµçѧԪ¼þÁ¬½ÓÆðÀ´£¬×¢ÒâµçÁ÷´Óµçѹ±íµÄÕý½ÓÏßÖùÁ÷È룬´Ó¸º½ÓÏßÖùÁ÷³ö£¬µçÔ´µçѹΪ3V£¬ËùÒÔµçѹ±í¿ÉÑ¡Ôñ0¡«3VµÄÁ¿³Ì£®
£¨3£©¸ù¾Ý¿ª¹ØµÄ¿ªºÏÇé¿ö·ÖÎöµç·µÄÁ¬½ÓÇé¿ö£¬²¢¸ù¾Ý¸ø³öµÄÒÑÖªÊý¾ÝÁгö±í´ïʽ£¬ÁªÁ¢±í´ïʽ±ã¿ÉµÃ³öRxµÄ±í´ïʽ£®
£¨4£©ËäÈ»ÔÚµç³Ø±íÃæµÄ·âËÜÖ½ÉϱêÓÐ1.5VµÄ×ÖÑù£¬µ«Êµ¼ÊµçѹÍùÍùÓëËù±êµçѹ²»Í¬£¬ËùÒÔ²»Äܽ«¸ÃµçѹֵÈ϶¨ÎªËüʵ¼Ê¸øµç·ÌṩµÄµçѹ£®
£¨2£©°´ÕÕµç·ͼ½«¸÷¸öµçѧԪ¼þÁ¬½ÓÆðÀ´£¬×¢ÒâµçÁ÷´Óµçѹ±íµÄÕý½ÓÏßÖùÁ÷È룬´Ó¸º½ÓÏßÖùÁ÷³ö£¬µçÔ´µçѹΪ3V£¬ËùÒÔµçѹ±í¿ÉÑ¡Ôñ0¡«3VµÄÁ¿³Ì£®
£¨3£©¸ù¾Ý¿ª¹ØµÄ¿ªºÏÇé¿ö·ÖÎöµç·µÄÁ¬½ÓÇé¿ö£¬²¢¸ù¾Ý¸ø³öµÄÒÑÖªÊý¾ÝÁгö±í´ïʽ£¬ÁªÁ¢±í´ïʽ±ã¿ÉµÃ³öRxµÄ±í´ïʽ£®
£¨4£©ËäÈ»ÔÚµç³Ø±íÃæµÄ·âËÜÖ½ÉϱêÓÐ1.5VµÄ×ÖÑù£¬µ«Êµ¼ÊµçѹÍùÍùÓëËù±êµçѹ²»Í¬£¬ËùÒÔ²»Äܽ«¸ÃµçѹֵÈ϶¨ÎªËüʵ¼Ê¸øµç·ÌṩµÄµçѹ£®
½â´ð£º½â£º£¨1£©ÓÉʵÑé±í¸ñÖª£¬µ±Sl¡¢S2±ÕºÏ£¬S3¶Ï¿ªÊ±£¬R1ÓëR2´®Áª£¬µçѹ±íµÄʾÊýΪ1V£¬ÔòR2Á½¶ËµÄµçѹΪ1V£®
Ôò
?R2=1V ¢Ù
µ±Sl¡¢S3±ÕºÏ£¬S2¶Ï¿ªÊ±£¬R1ÓëR3´®Áª£¬µçѹ±íʾÊýΪ2V£¬ËùÒÔ´ËʱR3Á½¶ËµÄµçѹΪ2V£®
Ôò
?Rx=2V ¢Ú
µÃ£¬
=
¢Û
½«R1=20¦¸£¬R2=10¦¸´úÈë¢ÛµÃ£¬Rx=40¦¸£¬
£¨2£©°´µç·ͼÒÀ´ÎÁ¬½Ó¸÷¸öµç·Ԫ¼þ£¬ÔÚ½øÐÐÁ¬½ÓʱעÒâµçѹ±íµÄÕý¸º½ÓÏßÖùºÍÁ¿³ÌµÄÑ¡Ôñ£®
£¨3£©ÒòΪҪ·Ö±ð²â³öÁ½´Îеç·ÖеĵçÁ÷£¬ËùÒÔµçÁ÷±íÓ¦´®ÁªÔڸɷÖУ®
¸ù¾Ý±íÊö£ºI1=
¢Ü
I2=
¢Ý
ÁªÁ¢¢Ü¢ÝµÃ£¬Rx=
£¨4£©ËäÈ»ÔÚµç³Ø±íÃæµÄ·âËÜÖ½ÉϱêÓÐ1.5VµÄ×ÖÑù£¬µ«Êµ¼ÊµçѹÍùÍùÓëËù±êµçѹ²»Í¬£¬ËùÒÔ²»Äܽ«¸ÃµçѹֵÈ϶¨ÎªËüʵ¼Ê¸øµç·ÌṩµÄµçѹ£®
¹Ê´ð°¸Îª£º£¨1£©40¦¸£»£¨2£©Èçͼ£»£¨3£©Èçͼ£»
£»£¨4£©²»ÄÜ£»Êµ¼ÊµçѹÍùÍùÓë±êʶµçѹֵ²»Ò»Ö£®
Ôò
U |
R1+R2 |
µ±Sl¡¢S3±ÕºÏ£¬S2¶Ï¿ªÊ±£¬R1ÓëR3´®Áª£¬µçѹ±íʾÊýΪ2V£¬ËùÒÔ´ËʱR3Á½¶ËµÄµçѹΪ2V£®
Ôò
U |
R1+Rx |
¢Ù |
¢Ú |
(R1+Rx)?R2 |
(R1+R2)?Rx |
1 |
2 |
½«R1=20¦¸£¬R2=10¦¸´úÈë¢ÛµÃ£¬Rx=40¦¸£¬
£¨2£©°´µç·ͼÒÀ´ÎÁ¬½Ó¸÷¸öµç·Ԫ¼þ£¬ÔÚ½øÐÐÁ¬½ÓʱעÒâµçѹ±íµÄÕý¸º½ÓÏßÖùºÍÁ¿³ÌµÄÑ¡Ôñ£®
£¨3£©ÒòΪҪ·Ö±ð²â³öÁ½´Îеç·ÖеĵçÁ÷£¬ËùÒÔµçÁ÷±íÓ¦´®ÁªÔڸɷÖУ®
¸ù¾Ý±íÊö£ºI1=
U |
R1+R2 |
I2=
U |
R1+Rx |
ÁªÁ¢¢Ü¢ÝµÃ£¬Rx=
I1(R1+R2)-I2R1 |
I2 |
£¨4£©ËäÈ»ÔÚµç³Ø±íÃæµÄ·âËÜÖ½ÉϱêÓÐ1.5VµÄ×ÖÑù£¬µ«Êµ¼ÊµçѹÍùÍùÓëËù±êµçѹ²»Í¬£¬ËùÒÔ²»Äܽ«¸ÃµçѹֵÈ϶¨ÎªËüʵ¼Ê¸øµç·ÌṩµÄµçѹ£®
¹Ê´ð°¸Îª£º£¨1£©40¦¸£»£¨2£©Èçͼ£»£¨3£©Èçͼ£»
I1(R1+R2)-I2R1 |
I2 |
µãÆÀ£º´ËÌ⿼²éÁ˶¯Ì¬µç·µÄ·ÖÎö£¬Òª¸ù¾Ý¿ª¹ØµÄ¿ªºÏÇé¿öµÃ³öµç·µÄÁ¬½Ó£¬È»ºó¸ù¾ÝÅ·Ä·¶¨ÂɵÄ֪ʶÁгöÏàÓ¦µÄ±í´ïʽ£¬½«¸÷±í´ïʽÁªÁ¢µÃ³ö½á¹û£®Í¬Ê±¿¼²éÁËʵÎïµç·µÄÁ¬½ÓºÍµç·ͼµÄ»·¨£¬ÔÚ°´µç·ͼÁ¬½ÓʵÎïʱ£¬Òª×¢ÒâÔª¼þµÄÁ¬½Ó˳Ðò¼°µçѹ±íºÍµçÁ÷±íµÄÕý¸º½ÓÏßÖù¡¢Á¿³ÌµÄ½Ó·¨£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿