如图所示电路中R1=R2=R3=R4=3Ω,R=1Ω,电源电压为12V,则电路总电阻为???

总电阻=R+R1+(R2||R3)=1+3+1.5=5.5Ω 等效悄羡电路图如碧衡下悔运做:
如图所示电路中R1=R2=R3=R4=3Ω,R=1Ω,电源电压为12V,则电路总电阻为???

总电阻=R+R1+(R2||R3)=1+3+1.5=5.5Ω

等效悄羡电路图如碧衡下悔运做:



Solution: R2, R3, R4 after the parallel connection of resistance R123 = 3 Omega 3 = 1 omega,

R = R1+R123+R = 3 Omega + 1 Omega + 1 omega = 5 ohm,

下面是其他几小题
I = UR = 12V5 omega = 2.4A,

R2, R3, R4 and R1 parallel and series voltage:

U = I (R1+R123) = 2.4A x (omega 3 Omega + 1) = 9.6V,

R2, R3, R4 parallel, R2 = R3 = R4,

R2, R3 parallel current

I = 23I = 23 x 2.4A = 1.6A.

解:假设总电阻为E X为R1,R2,R3,R4组合以后总电阻

E=R+X
由于此图连接扒野毕方式电压表相春芹当于断路,而电流表为短脊察路
分析为,R2,R3,R4并联 再与R1串联
电阻X=3+1/(1/3+1/3+1/3)=4 Ω
E=3+4=7Ω

5欧姆

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