并联电路中电阻上乘下加的公式是怎么推出来的？ 求并联电路中总电阻的公式R总=R1XR2/R1 R2是怎么推...
\u7535\u963b\u7684\u5e76\u8054\uff0c\u516c\u5f0f\u662f\u600e\u4e48\u63a8\u7b97\u7684\uff1f\u7535\u963bR1\u3001R2\u3001R3\u2026\u2026Rn\u5e76\u8054\uff0c
\u7535\u538bU1=U2=\u2026\u2026=Un
\u5e72\u8def\u7535\u6d41\uff1aIn=I1+I2+\u2026\u2026+In\uff08\u5e76\u8054\u7535\u8def\uff1a\u5e76\u8054\u7684\u5404\u652f\u8def\u7535\u538b\u76f8\u7b49\uff0c\u5e72\u8def\u7535\u6d41\u7b49\u4e8e\u5404\u4e2a\u652f\u8def\u548c\uff09
\u7531 \u4e8eI=U/R\uff0c\u4ee3\u5165\uff0c
In=U/R1+U/R2+U/R3......+U/Rn
In=U*\uff081/R1+1/R2+1/R3+......+1/Rn\uff09
\u6574\u7406\u4e0a\u5f0f
U/In=1/\uff081/R1+1/R2+1/R3+......+1/Rn\uff09
\u56e0\u4e3aU/In=R\u603b\uff0c\u6240\u4ee5
R\u603b=1/\uff081/R1+1/R2+1/R3+......+1/Rn\uff09
\u5e76\u8054\u7535\u8def\u7684\u963b\u503c\u4e3aR1\u3001R2\uff0c\u5e76\u8054\u540e\u7684\u7535\u963b\u4e3aR\uff0c\u7535\u8def\u4e24\u7aef\u6709\u7535\u4f4d\u538bU\uff0c\u6839\u636e\u6b27\u59c6\u5b9a\u5f8b\u6211\u4eec\u6709I=U/R\u3001I1=U/R1\u3001I2=U/R2\uff0c\u7531\u4e8e\u5e76\u8054\u5206\u6d41\uff0cI=I1+I2\u3001\u6545\u6709U/R=U/R1+U/R2\uff0c\u53731/R=1/R1+1/R2\uff0c\u5316\u7b80\u7684\u8bdd\u5c31\u662fR=R1R2/R1+R2.\u3002\u4e32\u8054\u7684\u8bdd\u5c31\u662fR=R1+R2\u3002
在并联电路中I＝I1+I2I=U／R
U/R=U1/R1+U2/R2
又U=U1=U2
1／R=1/R1+1/R2=(R1+R2)/R1R2
所以R=R1R2/(R1+R2)
两个电阻R1和R2并联后的电阻为1/[(1/R1)+(1/R2)],通分1/[(R1+R2)/(R1*R2)]=(R1*R2)/(R1+R2)
利用通分公式加出来的
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绛旓細鍦骞惰仈鐢佃矾涓 鎬鐢甸樆鐨璁＄畻鍏紡鏄1/R=1/R1+1/R2+1/R3+鈥1/Rn 澶氫釜鐢甸樆骞惰仈鐩稿綋浜庢荤殑妯埅闈㈢Н鍙樼矖锛屾墍浠ユ荤數闃讳細鍙樺皬 鎺ㄥ杩囩▼锛氣埖I=I1+I2 U=U1+U2 I=U/R 鈭碔1=U1/R2 I2=U2/R2 U/R=U1/R1+U2/R2锛堝苟鑱旂數璺腑鎬荤數娴佺瓑浜庡悇娈电數娴佷箣鍜岋級鈭靛苟鑱 鈭碪=U1=U2 鎵浠1/R=1/R1+1...
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