Question

11£®½»Á÷·¢µç»úµÄ·¢µçÔ­ÀíÊǾØÐÎÏßȦÔÚÔÈÇ¿´Å³¡ÖÐÈÆ´¹Ö±Óڴų¡µÄÖáOO¡äÔÈËÙת¶¯£®Ò»Ð¡ÐÍ·¢µç»úµÄÏßȦ¹²50ÔÑ£¬ÏßȦÃæ»ýS=0.5m²£¬ÏßȦת¶¯µÄ½ÇËٶȦØ=l00rad/sÏßȦÄÚ×è²»¼Æ£¬´Å³¡µÄ´Å¸ÐӦǿ¶ÈB=$\frac{\sqrt{2}}{10}$T£¬Ò»¸öÀíÏë±äѹÆ÷µÄÔ­ÏßȦ½Ó·¢µç»úµÄÊä³ö¶Ë£¬¸±ÏßȦÁ½¶Ë²¢Áª½ÓÈë¹æ¸ñΪ¡°220V 40W¡±µÄ£®µÆÅÝ10¸ö£¬µÆÅݾùÕý³£·¢¹â£®³ýµÆÅÝÍâÆäËüµç×è¾ù²»¼Æ£¬µçѹ±íµçÁ÷±í¾ùΪÀíÏë½»Á÷±í£¬µç·Èçͼ£®Çó£º
£¨1£©ÏßȦÓÉÖÐÐÔÃæλÖÃת¶¯¿ªÊ¼¼Æʱ£¬¸ÐÓ¦µç¶¯ÊƵÄ˲ʱ±í´ïʽ£»
£¨2£©µçÁ÷±íµÄʾÊý£»
£¨3£©·¢µç»úÊä³ö¹¦ÂÊ£®

Answer 1

·ÖÎö £¨1£©´ÓÏßȦ´¦ÓÚÖÐÐÔÃ濪ʼ¼Æʱ£¬ÏßȦÖиÐÓ¦µç¶¯ÊƵÄ˲ʱֵ±í´ïʽe=E_msin¦Øt£¬ÓÉE_m=NBS¦Ø£¬Çó³öE_m£®´Ó¶ø¼´¿ÉÇó½â£»
£¨2£©¸ù¾ÝµçѹÓëÔÑÊý³ÉÕý±ÈÇó³ö¸±ÏßȦÁ½¶ËµÄµçѹ£¬Çó³ö¸±ÏßȦµÄµçÁ÷£¬¸ù¾ÝµçÁ÷ÓëÔÑÊý³É·´±ÈÇóÔ­ÏßȦµÄµçÁ÷£¬¼´µçÁ÷±íµÄ¶ÁÊý£»
£¨3£©·¢µç»úµÄÊä³ö¹¦ÂʵÈÓÚÔ­ÏßȦµÄÊäÈ빦ÂÊ£¬¸ù¾ÝP=${U}_{1}^{\;}{I}_{1}^{\;}$¼´¿ÉÇó½â

½â´ð ½â£º£¨1£©ÏßȦת¶¯²úÉúµÄµç¶¯ÊÆ×î´óֵΪ
E_m=NBS¦Ø=250$\sqrt{2}$V    
¸ÐÓ¦µç¶¯ÊƵÄ˲ʱ±í´ïʽ
e=E_msin¦Øt=250$\sqrt{2}$sin100t£¨V£©  
£¨2£©µçѹʾÊýΪU₁=$\frac{{E}_{m}^{\;}}{\sqrt{2}}$=250 V
Ô­¡¢¸±ÏßȦÔÑÊý±È$\frac{{n}_{1}^{\;}}{{n}_{2}^{\;}}=\frac{{U}_{1}^{\;}}{{U}_{2}^{\;}}=\frac{25}{22}$
ͨ¹ýµÆÅݵĵçÁ÷I_R=$\frac{P}{U}$=$\frac{2}{11}$A
ͨ¹ý¸±ÏßȦµÄµçÁ÷I₂=$\frac{20}{11}$A      
µçÁ÷±íµÄʾÊý¡¡$\frac{{I}_{1}^{\;}}{{I}_{2}^{\;}}=\frac{{n}_{2}^{\;}}{{n}_{1}^{\;}}$
I₁=$\frac{8}{5}$A¡¡     
£¨3£©·¢µç»úÊä³ö¹¦ÂÊ
P₁=U₁I₁=400W   
´ð£º£¨1£©ÏßȦÓÉÖÐÐÔÃæλÖÃת¶¯¿ªÊ¼¼Æʱ£¬¸ÐÓ¦µç¶¯ÊƵÄ˲ʱ±í´ïʽ$e=250\sqrt{2}sin100t$£»
£¨2£©µçÁ÷±íµÄʾÊý$\frac{8}{5}A$£»
£¨3£©·¢µç»úÊä³ö¹¦ÂÊ400W£®

µãÆÀ ¿¼²é¶Ô½»Á÷·¢µç»úÔ­ÀíµÄÀí½âÄÜÁ¦£®¶ÔÓÚ½»Á÷µç±í£¬ÏÔʾµÄÊǽ»Á÷µçµÄÓÐЧֵ£®Ë²Ê±Öµ±í´ïʽҪעÒâ¼ÆʱÆðµã£¬²»Í¬µÄ¼ÆʱÆðµã±í´ïʽµÄ³õÏàλ²»Í¬£®