解:(1)S1、S2断开时,R1、R3串联
R串= R1+R3=8Ω+12Ω=20Ω
I串=U/ R串=6V/20Ω=0.3A
I3= I串=0.3A
U3=I3R3=0.3A×12Ω=3.6V
(2)P1=U1I串= I2串R1=(0.3A)2 ×8Ω=0.72W
(3)S1、S2闭合时,R1、R2并联,电压表被短路,故电压表示数为0
I1=U/ R1=6V/8Ω=0.75A
I2=U/R2=6V/4Ω=1.5A
电流表示数:I=I1+I2=0.75A+1.5A=2.25A
© 2017-2019 超级试练试题库,All Rights Reserved.