When the current is I2=0.50A, the power supply voltage U = I2R2+I2R0 = 0.50a× 7ω+0.50a× R0②.
① ② Available at the same time.
R0 = 5Ω，U = 6V
(2) when the temperature in the pot is 125, the current is I3=0.60A, and the thermistor rx = ui3-r0 = 6v0.60a-5ω =10ω-5ω = 5ω.
The power consumed at this time is px = i32rx = (0.60a) 2× 5ω =1.8w. 。
Answer: When the temperature in the pot is 125, the resistance RX of the thermistor is 5 Ω, and the power PX it consumes at this time is1.8w. 。