ÌâÄ¿ÄÚÈÝ
£¨12·Ö£©Í¼¼×ΪijͬѧÑо¿×Ô¸ÐÏÖÏóµÄʵÑéµç·ͼ£¬ÓõçÁ÷´«¸ÐÆ÷ÏÔʾÆ÷¸÷ʱ¿Ìͨ¹ýÏßȦLµÄµçÁ÷¡£µç·ÖеçµÆµÄµç×èR1=6.0¦¸£¬¶¨Öµµç×èR=2.0¦¸£¬AB¼äµçѹU=6.0 V¡£¿ª¹ØSÔÀ´±ÕºÏ£¬µç·´¦ÓÚÎȶ¨×´Ì¬£¬ÔÚt1=1.0¡Á103 sʱ¿Ì¶Ï¿ª¹ØS£¬´Ëʱ¿ÌÇ°ºóµçÁ÷´«¸ÐÆ÷ÏÔʾµÄµçÁ÷Ëæʱ¼ä±ä»¯µÄͼÏßÈçͼÒÒËùʾ¡£
£¨1£©Çó³öÏßȦLµÄÖ±Á÷µç×èRL£»
£¨2£©ÔÚͼ¼×ÖÐÓüýÍ·±ê³ö¶Ï¿ª¿ª¹Øºóͨ¹ýµçµÆµÄµçÁ÷·½Ïò£»
£¨3£©ÔÚt2=1.6¡Á103 sʱ¿ÌÏßȦLÖеĸÐÓ¦µç¶¯ÊƵĴóСÊǶàÉÙ£¿
½âÎö£º
£¨1£©ÓÉͼ¿ÉÖª£¬Áãʱ¿Ìͨ¹ýµç¸ÐÏßȦLµÄµçÁ÷Ϊ£ºI0£½1.5A£¬
ÓÉÅ·Ä·¶¨ÂÉ£º
½âµÃ£º
£¨2£©R1ÖеçÁ÷·½ÏòÏò×󣨻òÄæʱÕë·½Ïò£©
£¨3£©ÓÉͼ¿ÉÖª£¬ÔÚt2=1.6¡Á103 sʱ¿Ìµç¸ÐÏßȦLµÄµçÁ÷I=0.2A¡£ÏßȦ´ËʱÏ൱ÓÚÒ»¸öµçÔ´£¬Óɱպϵç·µÄÅ·Ä·¶¨ÂɵãºE = I (RL+R+R1)
´úÈëÊý¾Ý½âÖ®µÃ£ºE£½2.0V
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿