Question

4£®Ä³Í¬Ñ§Éè¼ÆÁËÈçͼ¼×ËùʾµÄµç·À´²âÁ¿µçÔ´µç¶¯ÊÆE¼°ÄÚ×èrºÍR₁µÄ×èÖµ£¬ÊµÑéÆ÷²ÄÓУº´ý²âµçÔ´£¨µç¶¯ÊÆE£¬ÄÚ×èr£©£¬´ý²âµç×èR₁£¬µçѹ±íV£¨Á¿³ÌΪ3V£¬ÄÚ×èºÜ´ó£©£¬µç×èÏäR£¨0¡«99.99¦¸£©£¬µ¥µ¶µ¥ÖÀ¿ª¹ØS₁£¬µ¥µ¶Ë«ÖÀ¿ª¹ØS₂£¬µ¼ÏßÈô¸É£®

£¨1£©ÏȲâµç×èR₁µÄ×èÖµ£®
A£®±ÕºÏS₁£¬½«S₂Çл»µ½a£¬µ÷½Úµç×èÏ䣬¶Á³öÆäʾÊýR₀ºÍ¶ÔÓ¦µÄµçѹ±íʾÊýU₁£»
B£®±£³Öµç×èÏäʾÊý²»±ä£¬½«S₂Çл»µ½b£¬¶Á³öµçѹ±íµÄʾÊýU₂£»
C£®Ôòµç×èR₁µÄ±í´ïʽΪR₁=$\frac{{U}_{2}-{U}_{1}}{{U}_{1}}$R₀£»
£¨2£©¸ÃͬѧÒѾ­²âµÃµç×èR₁=3.0¦¸£¬¼ÌÐø²âµçÔ´µç¶¯ÊÆEºÍÄÚ×èr£¬Æä×ö·¨ÊÇ£º±ÕºÏS₁£¬½«S₂Çл»µ½a£¬¶à´Îµ÷½Úµç×èÏ䣬¶Á³ö¶à×éµç×èÏäʾÊýRºÍ¶ÔÓ¦µÄµçѹ±íʾÊýU£¬ÓɲâµÃµÄÊý¾Ý£¬»æ³öÁËÈçͼÒÒËùʾµÄ$\frac{1}{U}$-$\frac{1}{R}$ͼÏߣ¬ÔòµçÔ´µç¶¯ÊÆE=2.0V£¬µç×èr=1.0¦¸£»
£¨3£©ÀûÓÃÈçͼ£¨2£©µÄ·½·¨Ëù²âµÃµÄµç¶¯ÊÆÓëÕæʵֵ±È½Ï£ºE_²âСÓÚE_Õ棨ѡÌî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±¡¢¡°µÈÓÚ¡±£©

Answer 1

·ÖÎö £¨1£©ÀûÓÃÌâÄ¿Öиø³öµÄµç·¼°²½ÖèµÄÌáʾ£¬Í¬Ê±½áºÏ±ÕºÏµç·µÄÅ·Ä·¶¨ÂɿɵóöʵÑéµÄ²½Ö裻
£¨2£©ÔÙÓÉÔ­Àí¿ÉµÃ³öʵÑéÖÐÊý¾Ý´¦ÀíµÄ·½·¨¼°¹«Ê½£¬½áºÏͼÏó¿ÉÇóµÃµç¶¯ÊƺÍÄÚµç×裮
£¨3£©¸ù¾Ý±ÕºÏµç·ŷķ¶¨ÂɽøÐзÖÎö£¬Ã÷È·µçѹ±íÄÚ×èµÄÓ°Ïì´Ó¶ø·ÖÎöʵÑéÖгöÏÖµÄÎó²îÇé¿ö£®

½â´ð ½â £¨1£©ÓÉÅ·Ä·¶¨ÂÉÓУºU₂=I£¨R₀+R₁£©£¬
U₁=IR₀£¬
ÁªÁ¢½âµÃ£ºR₁=$\frac{{U}_{2}-{U}_{1}}{{U}_{1}}$R₀£»
£¨2£©¸ù¾ÝE=U+$\frac{U}{R}$£¨R₁+r£©£¬±äÐεãº
$\frac{1}{U}$=$\frac{1}{E}$+$\frac{1}{R}$•$\frac{R_{1}+r}{E}$
¶Ô±ÈͼÏóÓУº$\frac{1}{E}$=0.5£¬µÃ£ºE=2 V£»
$\frac{R_{1}+r}{E}$=$\frac{1.5-0.5}{0.5-9}$=2.0£¬
ÒÑÖª£ºR₁=3.0¦¸£¬
½âµÃ£ºr=1.0¦¸£®
£¨3£©±¾·½·¨ÖÐÀûÓÃÅ·Ä·¶¨ÂÉÇó½â¸É·µçÁ÷£¬ÓÉÓÚµçѹ±íµÄ·ÖÁ÷×÷Óã¬Ê¹²âÁ¿³öµÄµçÁ÷ֵƫС£»Èô¿¼Âǵçѹ±íÄÚ×裬Ôò±í´ïʽӦΪ£º
E=U+£¨$\frac{U}{R}$+$\frac{U}{{R}_{V}}$£©£¨R₁+r£©£¬
±äÐκóӦΪ£º
$\frac{1}{U}$=$\frac{1}{E}$+$\frac{1}{R_{V}}$$\frac{R_{1}+r}{E}$+$\frac{1}{R}$$\frac{R_{1}+r}{E}$
¹ÊʵÑéֵӦΪ£º$\frac{1}{E}$+$\frac{1}{R_{V}}$$\frac{R_{1}+r}{E}$=0.5
ËùÒÔ˵µç¶¯ÊƵIJâÁ¿ÖµÆ«Ð¡£»
¹Ê´ð°¸Îª£º£¨1£©$\frac{{U}_{2}-{U}_{1}}{{U}_{1}}$R₀£»£¨2£©2.0£»1.0£»£¨3£©Æ«Ð¡£®

µãÆÀ ±¾Ì⿼²éÁËʵÑéµÄʵÎïͼÁ¬½Ó¡¢ÇóµçÔ´µç¶¯ÊÆÓëÄÚ×èµÄÊý¾Ý·ÖÎöºÍÎó²î·ÖÎö£»Ó¦ÓÃͼÏó·¨´¦ÀíʵÑéÊý¾ÝÊdz£Óõķ½·¨£¬Òª×¢Òâ±ä»¯¹«Ê½µÃ³öÏßÐÔ¹ØϵÔÙ½øÐзÖÎöÇó½â£®