两个阻值均为100欧的电阻并联,如果通过它们的总电流是3安,通电10秒它们共同产生的热量是多少？ 用三个阻值为100欧的电阻,通过串联并联可以得到几种不同的阻...
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3\uff0e\u8ba1\u7b97\u516c\u5f0f\uff1a
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W=13\u8f6c\u00f7\uff081950\u8f6c\uff0f\u5343\u74e6\u65f6\uff09=13\u00f7\u30141950/(1000\u00d73600)\u3015=24000(\u7126) (1)
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=0.435\uff08\u5343\u74e6\uff09 (3)
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W\u603b=P\u5176t\u5176+P\u51b0t\u51b0
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R=32.2\u6b27 121\u79d2\u6240\u505a\u7684\u7535\u529f
W=1.5\u00d7105\u7126 \u5b9e\u9645\u529f\u7387
P\u2019=1240\u74e6
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\u2234\u5e94\u4e32\u4e00\u4e2a\u963b\u503c\u4e3a60.5\u6b27\uff0c\u5141\u8bb8\u901a\u8fc7\u6700\u5927\u7535\u6d41\u4e3a3.64\u5b89\u7684\u6ed1\u52a8\u53d8\u963b\u5668
\u603b\u5171\u80fd\u5f97\u5230\u56db\u79cd\u4e0d\u540c\u963b\u503c\u7684\u7535\u963b\u3002
1\u3001\u4e09\u4e2a\u7535\u963b\u4e32\u8054\uff0c\u603b\u7535\u963b\u4e3a
R\u603b=3r=300\u03a9
2\u3001\u4e09\u4e2a\u7535\u963b\u5e76\u8054\uff0c\u603b\u7535\u963b\u4e3a
R\u5e76=r/3=100\u03a9/3
3\u3001\u4e24\u4e2a\u5148\u5e76\u8054\u518d\u548c\u53e6\u4e00\u4e2a\u4e32\u8054\uff0c\u603b\u7535\u963b\u4e3a
R=50\u03a9+100\u03a9=150\u03a9
4\u3001\u4e24\u4e2a\u5148\u4e32\u518d\u548c\u53e6\u4e00\u4e2a\u5e76\u8054\uff0c\u603b\u7535\u963b\u4e3a
R=200\u03a9/3
\u5982\u679c\u4e09\u4e2a\u7535\u963b\u963b\u503c\u5404\u4e0d\u76f8\u540c\uff0c\u540e\u4e24\u79cd\u7535\u963b\u4f4d\u7f6e\u4e0d\u540c\uff0c\u603b\u7535\u963b\u4e5f\u4e0d\u540c\u3002\u79cd\u7c7b\u5c31\u66f4\u591a\u3002
总电流3安，则每个电阻的电流为1.5安。
再有利用热量公式：Q1=I^2*R*T=I*I*R*T=1.5*1.5*100*10=2250J
共同产生的热量为：Q=2Q1=4500J
总电阻为50欧，功率为150W，这样热量为1500J
Q=I^2Rt=(3^2)*(100/2)*10=4500J
4500
绛旓細鐏场L1 L2骞惰仈鐨勯樆鍊锛歊鐏紳R1脳R2/(R1锛婻2)锛100脳100/(100锛100)锛50(惟)鐏场涓ょ鐢靛帇涓鐢甸樆涓ょ鐢靛帇鐨勬瘮鍊硷細50/20锛1/4 姹傜伅娉″苟鑱旂數娴佷笌鐢甸樆鐢垫祦姣斿硷細涓茶仈鐢佃矾澶勫鐩哥瓑锛屾墍浠ワ紝姣斿硷細1/1 姣忎釜鐏场鐢垫祦(1/2)涓庣數闃荤數娴佹瘮鍊硷細0.5/1 锛1/2 ...
绛旓細鎬鐢甸樆鍊=(1/0.1+1/100)鐨勫掓暟=100/1001= ---鑷繁绠楀惂.100闄や互1001
绛旓細涓ゅ彧闃诲间负100娆х殑鐢甸樆涓茶仈锛屽叾闃诲间负200娆э紝闃诲艰闄愬埗鍦200娆с備覆鑱旂數璺紝鐢甸樆鍊肩浉鍔犮
绛旓細鍥犱负鏄骞惰仈鐢甸樆锛屾牴鎹叕寮忓苟鑱旂數璺鐨勭數闃鐨勫掓暟绛変簬鎵鏈夋敮璺數闃荤殑鍊掓暟鐨勫拰銆傛墍浠ユ墍姹傜數闃诲氨鏄1/锛1/100+1/100锛=50惟銆
绛旓細1 娆у涓2娆у 骞惰仈鐢甸樆 鏄紙1*2锛/锛1+2锛=2/3娆у 涓や釜100娆у鐨勫苟鑱旂數闃鏄紙100*100锛/锛100+100锛=50娆у銆備篃鍙互鐢ㄦ洿绠鍗曠殑100/2=50璁＄畻锛堣繖鏍风殑鏂规硶浠呴檺浜闃诲奸兘鐩哥瓑鐨勭數闃诲苟鑱銆傦級
绛旓細涓涓畾鍊肩數闃诲湪鐢佃矾涓秷鑰楃殑鍔熺巼鍙栧喅浜庢祦杩囪繖涓數闃荤殑鐢垫祦锛屼篃鍙互璇村彇鍐充簬杩欎釜鐢甸樆涓ょ鐨勭數鍘嬨傚鏋滃彇涓や釜鐩稿悓闃诲涓100娆х殑鐢甸樆锛屽皢鍏朵覆鑱斿悗鎺ラ氫竴涓數鍘嬩负100V鐨勭數婧 锛屾鏃剁數娴佷负100/200=0.5 A, 涓涓數闃荤殑鍔熺巼涓0.5X0.5X100=25W銆傚悓鏍涓や釜鐢甸樆锛骞惰仈鍦100V鐢垫簮涓婏紝涓や釜鐢甸樆鍒嗗埆閫氳繃1A鐢垫祦...
绛旓細鍥犱负骞惰仈鐢佃矾鐨勬荤數娴佺瓑浜庡悇涓敮璺數娴佺殑鍜,骞惰仈鐢佃矾鐨勬荤數鍘嬬瓑浜庡悇涓敮璺殑鐢靛帇銆傛墍浠/R1+U/R2=U/R鍗1/R1+1/R2=1/R鎵浠1/1000+1/1000=1/500鎵浠ユ荤數娴侊紳500娆
绛旓細涓や釜100鐨勫苟鑱锛鐢甸樆涓 锛100+100锛/100*100锛=50 娆у 50鐨勫啀涓244 鐨勫苟鑱 = 锛50+244锛/50*244 绾︾瓑浜 41.6 娆у
绛旓細1/10+1/100=11/100锛岋紙鍗佸垎涔嬩竴鍔犱竴鐧惧垎涔嬩竴绛変簬浠ョ櫨鍒嗕箣鍗佷竴锛夛紝100/11=9.091娆у銆傝浣忥細涓や釜锛堟垨澶氫釜锛鐢甸樆骞惰仈锛屾闃诲浼氬皬浜庡叾涓樆鍊兼渶灏忕殑閭ｄ釜鐢甸樆锛屽惁鍒欏氨鏄绠楅敊璇
绛旓細R=100+50=150 100Ω涓50Ω鐨勭數闃涓茶仈鍚,鍏舵诲间负150Ω
