【答案】:
(1)①1 3 ②400 Ω (2)C
【解析】:
(1)①1 3 ②设电源电压为$U$,当温度为$42.0^\circ{C}$时,$R_{t1}=400\ \Omega$,$U_{1}=3\ V$,此时电路中电流$I_{1}=\frac{U_{1}}{R_{1}}$,电源电压$U=I_{1}(R_{t1}+R_{1})=\frac{U_{1}}{R_{1}}(R_{t1}+R_{1})$。当温度为$32.0^\circ{C}$时,$R_{t2}=600\ \Omega$,$U_{2}=2.4\ V$,此时电路中电流$I_{2}=\frac{U_{2}}{R_{1}}$,电源电压$U=I_{2}(R_{t2}+R_{1})=\frac{U_{2}}{R_{1}}(R_{t2}+R_{1})$。联立可得$\frac{3}{R_{1}}(400 + R_{1})=\frac{2.4}{R_{1}}(600 + R_{1})$,解得$R_{1}=400\ \Omega$。
(2)C