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?
来源：学生作业帮助网 编辑：六六作业网 时间：2024/12/25 21:31:42
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?
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?
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
等效电路图如下：
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,
R...
全部展开
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.
收起
假设总电阻为E X为R1,R2，R3，R4组合以后总电阻
E=R+X
由于此图连接方式电压表相当于断路，而电流表为短路
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