电路计算,如图期待详细解决。A点与B点电阻是多少?
3个回答
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直接根据基尔霍夫定律列方程吧。令通过电阻Ri的电流为Ii,电流I3的正方向定义为从上到下,其他电流正方向定义为从左到右。设AB间电压为U。
U = I1R1 + I4R4
U = I2R2 + I5R5
I1R1 + I3R3 = I2R2
I3 = I1 - I4
I3 = I5 - I2
5个方程正好解出5个未知电流
I1 = U(R2R3+R2R4+R2R5+R3R5)/(R1 R2 R3+R1 R2 R4+R1 R2 R5+R1 R3 R5+R1 R4 R5+R2 R3 R4+R2 R4 R5+R3 R4 R5)
I2 = U(R1R3+R1R4+R1 R5+R3 R4)/(R1 R2 R3+R1 R2 R4+R1 R2 R5+R1 R3 R5+R1 R4 R5+R2 R3 R4+R2 R4 R5+R3 R4 R5)
I3 = U(R2R4-R1R5)/(R1 R2 R3+R1 R2 R4+R1 R2 R5+R1 R3 R5+R1 R4 R5+R2 R3 R4+R2 R4 R5+R3 R4 R5)
I4 = U(R1R5+R2 R3+R2 R5+R3 R5)/(R1 R2 R3+R1 R2 R4+R1 R2 R5+R1 R3 R5+R1 R4 R5+R2 R3 R4+R2 R4 R5+R3 R4 R5)
I5 = U(R1 R3+R1 R4+R2 R4+R3 R4)/(R1 R2 R3+R1 R2 R4+R1 R2 R5+R1 R3 R5+R1 R4 R5+R2 R3 R4+R2 R4 R5+R3 R4 R5)
AB两点间电阻为
R = U / (I1+I2)
= (R1R2R3+R1R2R4+R1R2R5+R1R3R5+R1R4R5+R2R3R4+R2R4R5+R3R4R5) / (R2R3+R2R4+R2R5+R3R5+R1R3+R1R4+R1R5+R3R4)
= 816/55
U = I1R1 + I4R4
U = I2R2 + I5R5
I1R1 + I3R3 = I2R2
I3 = I1 - I4
I3 = I5 - I2
5个方程正好解出5个未知电流
I1 = U(R2R3+R2R4+R2R5+R3R5)/(R1 R2 R3+R1 R2 R4+R1 R2 R5+R1 R3 R5+R1 R4 R5+R2 R3 R4+R2 R4 R5+R3 R4 R5)
I2 = U(R1R3+R1R4+R1 R5+R3 R4)/(R1 R2 R3+R1 R2 R4+R1 R2 R5+R1 R3 R5+R1 R4 R5+R2 R3 R4+R2 R4 R5+R3 R4 R5)
I3 = U(R2R4-R1R5)/(R1 R2 R3+R1 R2 R4+R1 R2 R5+R1 R3 R5+R1 R4 R5+R2 R3 R4+R2 R4 R5+R3 R4 R5)
I4 = U(R1R5+R2 R3+R2 R5+R3 R5)/(R1 R2 R3+R1 R2 R4+R1 R2 R5+R1 R3 R5+R1 R4 R5+R2 R3 R4+R2 R4 R5+R3 R4 R5)
I5 = U(R1 R3+R1 R4+R2 R4+R3 R4)/(R1 R2 R3+R1 R2 R4+R1 R2 R5+R1 R3 R5+R1 R4 R5+R2 R3 R4+R2 R4 R5+R3 R4 R5)
AB两点间电阻为
R = U / (I1+I2)
= (R1R2R3+R1R2R4+R1R2R5+R1R3R5+R1R4R5+R2R3R4+R2R4R5+R3R4R5) / (R2R3+R2R4+R2R5+R3R5+R1R3+R1R4+R1R5+R3R4)
= 816/55
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