图1是某“汽车超载记录器”原理图。控制电路中，电源电压U=6V，压敏电阻的阻值随压力变化关系如图2所示，继电器线圈电阻R0=10Ω。当通过线圈的电流达到0.2A以上时，继电器的衔铁被吸下，工作电路中的电控照相机工作，拍摄超载车辆。
（1）质量超过多少吨的车辆经过压敏电阻时才会被拍照？
（2）闭合开关S，控制电路每分钟至少消耗电能多少焦？
（3）请指出汽车超载的危害（一种即可）。
解：
（1）对临界条件电流I0＝0.2A有：R＝
=
=30Ω
R敏＝R－R0＝30Ω－10Ω＝20Ω. 由图可查得此时F0＝2×105N，
得：m0＝
=
=2×104kg＝20t ；
（2）要控制电路电功率小，则需电路电阻大，由图可得压敏电阻最大阻值为80Ω，有：R′max＝R0＋Rmax＝10Ω＋80Ω＝90Ω.
Pmin＝
=
=0.4W，Wmin＝Pmin×60s＝0.4W×60s＝24J；
（3）不容易刹车，易造成交通事故．
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