(1)由欧姆定律得:
电源电压U=U1=I1R1=0.6A×3Ω=1.8V;(2)通过R2的电流;
I2=UR2=1.8V6Ω=0.3A;(3)则干路电流I=I1+I2=0.6A+0.3A=0.9A; 电流表A的示数是0.9A.
故选C.来源:中国机电网
(1)由欧姆定律得:
电源电压U=U1=I1R1=0.6A×3Ω=1.8V;(2)通过R2的电流;
I2=UR2=1.8V6Ω=0.3A;(3)则干路电流I=I1+I2=0.6A+0.3A=0.9A; 电流表A的示数是0.9A.
故选C.