ÌâÄ¿ÄÚÈÝ

12£®ÊµÑéÊÒÀïÓÐÁ½Ö»·Ö±ð±êÓÐΪ6V 3WºÍ6V 6WµÄСµÆÅÝL1ºÍL2£¬½«ËüÃÇÁ¬½Óµ½ÈçͼËùʾµÄµçÔ´µçѹ¿Éµ÷µÄµç·ÖУ¬±ÕºÏ¿ª¹ØS1¡¢S2¡¢S3£¬»¬Æ¬PÖÃÓÚ×îÓÒ¶Ëʱ£¬µ÷½ÚµçÔ´µçѹʹСµÆÅÝL1Õý³£·¢¹â£¬µçÁ÷±íµÄʾÊýΪ0.8A£®µçÁ÷±íµçѹ±í¾ùʹÓôóÁ¿³Ì£¨0-3A£¬0-15V£©£¬²»¼ÆÎ¶ȶԵç×èÓ°Ï죮Çó£º
£¨1£©»¬¶¯±ä×èÆ÷µÄ×î´óµç×裮
£¨2£©Ö»±ÕºÏ¿ª¹ØS2ʱ£¬µ÷½ÚµçÔ´µçѹ£¬±£Ö¤µÆÅݰ²È«Ç°ÌáÏ£¬µçÔ´µçѹµÄ¿Éµ÷µÄ×î´óÖµ£®
£¨3£©Ö»±ÕºÏ¿ª¹ØS1ʱ£¬µç·°²È«Ç°ÌáÏ£¬µ÷½ÚµçÔ´µçѹºÍ»¬¶¯±ä×èÆ÷×èÖµ£¬µ±µçÔ´µçѹµ÷µ½×î´óʱ£¬´Ëʱ»¬¶¯±ä×èÆ÷µÄ¹¦ÂÊΪ¶àÉÙ£¿

·ÖÎö £¨1£©±ÕºÏ¿ª¹ØS1¡¢S2¡¢S3£¬»¬Æ¬PÖÃÓÚ×îÓÒ¶Ëʱ£¬»¬¶¯±ä×èÆ÷µÄ×î´ó×èÖµÓëµÆÅÝL1²¢Áª£¬µçÁ÷±í²â¸É·µçÁ÷£¬¸ù¾Ý¶î¶¨µçѹϵÆÅÝÕý³£·¢¹âºÍ²¢Áªµç·µÄµçÑ¹ÌØµã¿ÉÖªµçÔ´µÄµçѹ£¬¸ù¾ÝP=UIÇó³öͨ¹ýL1µÄµçÁ÷£¬¸ù¾Ý²¢Áªµç·µÄµçÁ÷ÌØµãÇó³öͨ¹ý»¬¶¯±ä×èÆ÷µÄµçÁ÷£¬ÔÙ¸ù¾ÝÅ·Ä·¶¨ÂÉÇó³ö»¬¶¯±ä×èÆ÷µÄ×î´ó×èÖµ£»
£¨2£©Ö»±ÕºÏ¿ª¹ØS2ʱ£¬L1ÓëL2´®Áª£¬¸ù¾ÝP=$\frac{{U}^{2}}{R}$Çó³öL2µÄ×èÖµ£¬¸ù¾ÝÅ·Ä·¶¨ÂÉÇó³öL2Õý³£·¢¹âʱµÄµçÁ÷£¬±È½ÏÁ½µÆÅÝÕý³£·¢¹âʱµÄµçÁ÷¿ÉÖªµç·ÖеÄ×î´óµçÁ÷£¬ÔÙ¸ù¾ÝÅ·Ä·¶¨ÂÉÇó³öL2Á½¶ËµÄµçѹ£¬ÀûÓô®Áªµç·µÄµçÑ¹ÌØµãÇó³öµçÔ´µçѹµÄ¿Éµ÷µÄ×î´óÖµ£»
£¨3£©Ö»±ÕºÏ¿ª¹ØS1ʱ£¬L2Ó뻬¶¯±ä×èÆ÷´®Áª£¬µçÁ÷±í²âµç·ÖеĵçÁ÷£¬µçѹ±í²â»¬¶¯±ä×èÆ÷Á½¶ËµÄµçѹ£¬µ±L2Õý³£·¢¹âÇÒµçѹ±íµÄʾÊý×î´óʱµçÔ´µÄµçѹ×î´ó£¬¸ù¾ÝP=UIÇó³ö´Ëʱ»¬¶¯±ä×èÆ÷µÄ¹¦ÂÊ£®

½â´ð ½â£º£¨1£©±ÕºÏ¿ª¹ØS1¡¢S2¡¢S3£¬»¬Æ¬PÖÃÓÚ×îÓÒ¶Ëʱ£¬»¬¶¯±ä×èÆ÷µÄ×î´ó×èÖµÓëµÆÅÝL1²¢Áª£¬µçÁ÷±í²â¸É·µçÁ÷£¬
Òò²¢Áªµç·Öи÷֧·Á½¶ËµÄµçѹÏàµÈ£¬ÇÒСµÆÅÝL1Õý³£·¢¹â£¬
ËùÒÔ£¬µçÔ´µÄµçѹU=U1=6V£¬
ÓÉP=UI¿ÉµÃ£¬Í¨¹ýL1µÄµçÁ÷£º
I1=$\frac{{P}_{1}}{{U}_{1}}$=$\frac{3W}{6V}$=0.5A£¬
Òò²¢Áªµç·ÖиÉ·µçÁ÷µÈÓÚ¸÷֧·µçÁ÷Ö®ºÍ£¬
ËùÒÔ£¬Í¨¹ý»¬¶¯±ä×èÆ÷µÄµçÁ÷£º
I»¬=I-I1=0.8A-0.5A=0.3A£¬
ÓÉI=$\frac{U}{R}$¿ÉµÃ£¬»¬¶¯±ä×èÆ÷µÄ×î´ó×èÖµ£º
R»¬=$\frac{U}{{I}_{»¬}}$=$\frac{6V}{0.3A}$=20¦¸£»
£¨2£©Ö»±ÕºÏ¿ª¹ØS2ʱ£¬L1ÓëL2´®Áª£¬
ÓÉP=$\frac{{U}^{2}}{R}$¿ÉµÃ£¬L2µÄ×èÖµ£º
R2=$\frac{{{U}_{2}}^{2}}{{P}_{2}}$=$\frac{£¨6V£©^{2}}{6W}$=6¦¸£¬
L2Õý³£·¢¹âʱµÄµçÁ÷£º
I2=$\frac{{U}_{2}}{{R}_{2}}$=$\frac{6V}{6¦¸}$=1A£¬
Òò´®Áªµç·Öи÷´¦µÄµçÁ÷ÏàµÈ£¬
ËùÒÔ£¬µç·ÖеÄ×î´óµçÁ÷I´ó=0.5A£¬
´ËʱL1Á½¶ËµÄµçѹU1=6V£¬L1Á½¶ËµÄµçѹU2¡ä=I´óR2=0.5A¡Á6¦¸=3V£¬
Òò´®Áªµç·ÖÐ×ܵçѹµÈÓÚ¸÷·Öµçѹ֮ºÍ£¬
ËùÒÔ£¬µçÔ´µçѹµÄ¿Éµ÷µÄ×î´óÖµ£º
U´ó=U1+U2¡ä=6V+3V=9V£»
£¨3£©Ö»±ÕºÏ¿ª¹ØS1ʱ£¬L2Ó뻬¶¯±ä×èÆ÷´®Áª£¬µçÁ÷±í²âµç·ÖеĵçÁ÷£¬µçѹ±í²â»¬¶¯±ä×èÆ÷Á½¶ËµÄµçѹ£¬
µ±L2Õý³£·¢¹âÇÒµçѹ±íµÄʾÊý×î´óʱµçÔ´µÄµçѹ×î´ó£¬
´Ëʱµç·ÖеĵçÁ÷I2=1A£¬»¬¶¯±ä×èÆ÷Á½¶ËµÄµçѹU»¬´ó=15V£¬
Ôò´Ëʱ»¬¶¯±ä×èÆ÷µÄ¹¦ÂÊ£º
P»¬=U»¬´óI2=15V¡Á1A=15W£®
´ð£º£¨1£©»¬¶¯±ä×èÆ÷µÄ×î´óµç×èΪ20¦¸£»
£¨2£©Ö»±ÕºÏ¿ª¹ØS2ʱ£¬µ÷½ÚµçÔ´µçѹ£¬±£Ö¤µÆÅݰ²È«Ç°ÌáÏ£¬µçÔ´µçѹµÄ¿Éµ÷µÄ×î´óֵΪ9V£»
£¨3£©Ö»±ÕºÏ¿ª¹ØS1ʱ£¬µç·°²È«Ç°ÌáÏ£¬µ÷½ÚµçÔ´µçѹºÍ»¬¶¯±ä×èÆ÷×èÖµ£¬µ±µçÔ´µçѹµ÷µ½×î´óʱ£¬´Ëʱ»¬¶¯±ä×èÆ÷µÄ¹¦ÂÊΪ15W£®

µãÆÀ ±¾Ì⿼²éÁË´®Áªµç·ºÍ²¢Áªµç·µÄÌØµãÒÔ¼°Å·Ä·¶¨ÂÉ¡¢µç¹¦Âʹ«Ê½µÄÓ¦Ó㬻áÈ·¶¨Á½µÆÅÝ´®ÁªÊ±µÄ×î´óµçÁ÷ºÍL2Ó뻬¶¯±ä×èÆ÷´®Áªµç·µçѹ×î´óʱµÄÁ¬½Ó·½Ê½Êǹؼü£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø