(1)RL=UL2PL=(3V)21.5W=6Ω;
当开关S1、S2都断开时,灯L与R1串联;
则I1=IL=I=0.3A;
R1=U1I1=U-UL′I1=U-ILRLI1=3V-1.8V0.3A=4Ω;
PL′=IL2RL=(0.3A)2×6Ω=0.54W;
答:R1的电阻为4Ω;灯泡消耗的电功率为0.54W.
(2)当开关S1、S2都闭合时,R1短路,灯L与R2并联;
UL=U2=U=3V;
I′=IL′+I2=ULRL+U2R2=3V6Ω+3V3Ω=1.5A.
答:电路中的总电流为1.5A.