电工基础的题目：`````````````````````````````````````````````````````````````````````````````习题:在题图中,已知E=220v,R1=25Ω,R2=55Ω,R3=30Ω.求：（1）开关S打开时电路中的电流及电阻上的电压.（2）开关S合上后
来源：学生作业帮助网 编辑：六六作业网 时间：2024/11/18 04:39:26
电工基础的题目：`````````````````````````````````````````````````````````````````````````````习题:在题图中,已知E=220v,R1=25Ω,R2=55Ω,R3=30Ω.求：（1）开关S打开时电路中的电流及电阻上的电压.（2）开关S合上后
电工基础的题目：`````````````````````````````````````````````````````````````````````````````
习题:
在题图中,已知E=220v,R1=25Ω,R2=55Ω,R3=30Ω.求：
（1）开关S打开时电路中的电流及电阻上的电压.
（2）开关S合上后,各电阻上的电压是增大还是减小,为什么?
电工基础的题目：`````````````````````````````````````````````````````````````````````````````习题:在题图中,已知E=220v,R1=25Ω,R2=55Ω,R3=30Ω.求：（1）开关S打开时电路中的电流及电阻上的电压.（2）开关S合上后
开关S打开时,R1、R2、R3处于串联状态,因此:
I=U/(R1+R2+R3)=220/(25+55+30)=2(A)
U1=I*R1=2*25=50(V)
U2=I*R2=2*55=110(V)
U3=I*R3=2*30=60(V)
开关S合上后:
R2处于短路状态,只有R1、R3处于串联状态,因此：
I=U/(R1+R3)=220/(25+30)=4(A)
U1=I*R1=4*25=100(V) 增大
U2=0(V)
U3=I*R3=4*30=120(V) 增大
开关S打开时:
I=U/(R1+R2+R3)=220/(25+55+30)=2(A)
U1=I*R1=2*25=50(V)
U2=I*R2=2*55=110(V)
U3=I*R3=2*30=60(V)
开关S合上后:
R2短路
I=U/(R1+R3)=220/(25+30)=4(A)
U1=I*R1=4*25=100(V) 增大
U2=0(V)
U3=I*R3=4*30=120(V) 增大