1.
因为串联电路电流相等,所以I1:I2=1:1;
因为R1=6,R2=24,所以P1:P2=I1^2R1/I2^R2=R1/R2=1/4
R总=R1+R2=30,I=P2/U2=0.25A,U总=IR总=0.25A*30=7.5V
2.
设原先电流为xA.
6/x=(6+2)/(x+0.1A)
x=0.3
P1=U1I1=6V*0.3A=1.8W,P2=U2I2=8V*0.4=3.2W
P2-P1=3.2W-1.8W=1.4W
所以选B
1.
因为串联电路电流相等,所以I1:I2=1:1;
因为R1=6,R2=24,所以P1:P2=I1^2R1/I2^R2=R1/R2=1/4
R总=R1+R2=30,I=P2/U2=0.25A,U总=IR总=0.25A*30=7.5V
2.
设原先电流为xA.
6/x=(6+2)/(x+0.1A)
x=0.3
P1=U1I1=6V*0.3A=1.8W,P2=U2I2=8V*0.4=3.2W
P2-P1=3.2W-1.8W=1.4W
所以选B