(1)当只闭合开关S1时,只有R1接入电路此时U1=U=12V∴R1= U1 I1 = 12V 1A =12Ω(2)当开关S1S2同时闭合时,R1与R2并联接入电路I2=I-I1=1.5A-1A=0.5A∴P2=U2×I2=12V×0.5A=6W(3)当开关S1S2同时闭合时,R1与R2并联接入电路Q=W=UIt=12V×1.5A×10S=180J答:(1)R1的阻值为12Ω;(2)R2消耗的电功率为6W;(3)电路中产生的热量为180J.