Question

13£®ÈçͼËùʾ£¬ÀíÏë±äѹÆ÷µÄÔ­ÏßȦ½Óu=3880$\sqrt{2}$sinl00¦Ðt£¨V£©µÄ½»±äµçѹ£¬¸±ÏßȦͨ¹ýµ¼Ïߣ¨Ã¿¸ùµ¼ÏߵĵÈЧµç×è¾ùΪR₀=1¦¸£©£¬¶ÔÁ½¸ö¾ùΪ¡°380V£¬760W¡±µÄµç¶¯»úM₁¡¢M₂¹©µç£¬M₁¡¢M₂¾ùÕý³£¹¤×÷£¬ÓÉ´Ë¿ÉÖª£¨¡¡¡¡£©

• A£® ±äѹÆ÷µÄÊäÈ빦ÂÊΪ1520 W
• B£® Ô­¸±ÏßȦµÄÔÑÊý±ÈΪ10$\sqrt{2}$£º1
• C£® Ô­ÏßȦËù½ÓµçÁ÷±íAµÄ¶ÁÊýΪ0.4A
• D£® ÈôM₁¡¢M₂µÄµÈЧÄÚ×è¾ùΪ10¦¸£¬Ôòÿ¸öµç¶¯»úÊä³öµÄ»úе¹¦ÂʾùΪ720W

Answer 1

·ÖÎö ½»Á÷µçÁ÷±íµÄʾÊýΪÓÐЧֵ£¬Óɹ«Ê½P=UIÖªµç·ҪÕý³£¹¤×÷£¬µç¶¯»úÖеçÁ÷Ϊ2A£¬¶øµç¶¯»úÊÇ·Ç´¿µç×裮¸ù¾ÝÊäÈ빦ÂʵÈÓÚÊä³ö¹¦ÂÊ£¬Ô­¸±ÏßȦÔÑÊýÓëµçѹ³ÉÕý±È£¬µçÁ÷ÓëÔÑÊý³É·´±ÈÁÐʽÇó½â£®

½â´ð ½â£ºA¡¢¸ù¾ÝÌâÒ⣬µç¶¯»ú¹¤×÷£¬¶î¶¨µçÁ÷${I}_{M}^{\;}=\frac{P}{U}=\frac{760}{380}A=2A$£¬¸±ÏßȦÖеĵçÁ÷${I}_{2}^{\;}=2{I}_{M}^{\;}=4A$£¬µ¼ÏßÉÏÏûºÄµÄµç¹¦ÂÊ$2{I}_{\;}^{2}{R}_{0}^{\;}$=$2¡Á{4}_{\;}^{2}¡Á1=32W$£¬Êä³ö¹¦ÂÊ${P}_{2}^{\;}=760¡Á2+32=1552W$£¬ÀíÏë±äѹÆ÷ÊäÈ빦ÂʵÈÓÚÊä³ö¹¦ÂÊ£¬ËùÒÔÊäÈ빦ÂÊ1552W£¬¹ÊA´íÎó£»
B¡¢¸±ÏßȦÖеçÏßÉϵçѹ$2{I}_{2}^{\;}{R}_{0}^{\;}=2¡Á4¡Á1=8V$£¬¸±ÏßȦÁ½¶ËµÄµçѹ£º${U}_{2}^{\;}=380+8=388V$£¬Ô­ÏßȦµçѹµÄÓÐЧֵ${U}_{1}^{\;}=\frac{3880\sqrt{2}}{\sqrt{2}}=388V$£¬Ô­¸±ÏßȦµÄÔÑÊýÖ®±È$\frac{{n}_{1}^{\;}}{{n}_{2}^{\;}}=\frac{{U}_{1}^{\;}}{{U}_{2}^{\;}}=\frac{3880}{388}=\frac{10}{1}$£¬¹ÊB´íÎó£»
C¡¢Ô­¸±ÏßȦµçÁ÷ÓëÔÑÊý³É·´±È£¬$\frac{{I}_{1}^{\;}}{{I}_{2}^{\;}}=\frac{{n}_{2}^{\;}}{{n}_{1}^{\;}}$£¬$\frac{{I}_{1}^{\;}}{4}=\frac{1}{10}$£¬½âµÃ${I}_{1}^{\;}=0.4A$£¬¹ÊCÕýÈ·£»
D¡¢Ã¿¸öµç¶¯»úÏûºÄµÄÈȹ¦ÂÊ${P}_{r}^{\;}={I}_{M}^{2}r={2}_{\;}^{2}¡Á10=40W$£¬Êä³öµÄ»úе¹¦ÂÊ$P={P}_{µç}^{\;}-{P}_{r}^{\;}=760-40=720$£¬¹ÊDÕýÈ·£»
¹ÊÑ¡£ºCD

µãÆÀ ±¾Ìâ¹Ø¼üÊÇÃ÷È·±äѹÆ÷µÄ±äѹ±È¹«Ê½ºÍ±äÁ÷±È¹«Ê½£¬ÖªµÀµç¶¯»úÊÇ·Ç´¿µç×èµç·£¬Êä³ö¹¦ÂÊΪP_³ö=UI-I²r£®