Question

如图所示的电路中，电源电压为8V，R₂是阻值为40的定值电阻，压敏电阻R₁的阻值随压力的变化情况如图乙所示。当开关S闭合后，问：

（1）当压敏电阻R₁受到的压力从0增加到3N的过程中，电流表示数的变化是       （填“增大”“减少”）？

（2）当电流表示数为0.4A时，压敏电阻受到的压力为多少？

（3）当电路连续工作10mim，电阻R₂产生的热量为多少？

（4）若此热量的84%给20g的水加热，则水温升高多少？（水的比热容为）

Answer 1

（1）增大。········································································································ (2分)

（2）I₂=U/R₂=8V/40Ω=0.2A

I₁=I－I₂=0.4A－0.2A=0.2A···································································· (1分)

R₁= U/ I₁=8V/0.2A=40Ω 

F=2N··································································································· (1分)

（3）Q₂= I₂² R₂t=(0.2A)²×40Ω×10×60s=960J·················································· (2分)

（4）Q_吸＝cm△t·························································································· (1分)

△t =

=9.6℃······························································································ (1分)