有一种电加热恒温箱，工作原理如图甲所示．控制电路由电压为U1=9V的电源、电磁继电器（线圈电阻不计）、 有一种电加热恒温箱，工作原理如图（甲）所示．控制电路由电压为...
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\uff081\uff09\u7535\u70ed\u4e1d\u5de5\u4f5c\u65f6\u7684\u529f\u7387\uff1aP=U2R2=(220V)244\u03a9=1100W\uff1b\uff082\uff09\u7535\u70ed\u4e1d\u6b63\u5e38\u5de5\u4f5c100s\u4ea7\u751f\u7684\u70ed\u91cf\uff1aQ=W=Pt=1100W\u00d7100s=1.1\u00d7105J\uff1b\uff083\uff09\u5982\u679c\u6052\u6e29\u7bb1\u7684\u6e29\u5ea6\u8bbe\u5b9a\u4e3a50\u2103\uff0c\u7531\u56fe\u4e59\u53ef\u77e5\uff0c\u70ed\u654f\u7535\u963b\u7684\u963b\u503cR1=100\u03a9\uff0c\u7531\u9898\u77e5\uff0c\u6b64\u65f6\u63a7\u5236\u7535\u8def\u7684\u7535\u6d41I=60mA=0.06A\uff0c\u2235I=U1R0+R1\uff0c\u5373\uff1a0.06A=9VR0+100\u03a9\uff0c\u2234\u7535\u963b\u7bb1R0\u5e94\u63a5\u5165\u7535\u8def\u7684\u963b\u503c\uff1aR0=50\u03a9\uff0e\u7b54\uff1a\uff081\uff09\u7535\u70ed\u4e1d\u6b63\u5e38\u5de5\u4f5c\u65f6\u7684\u529f\u7387\u4e3a1100W\uff1b\uff082\uff09\u7535\u70ed\u4e1d\u6b63\u5e38\u5de5\u4f5c100s\u4ea7\u751f\u7684\u70ed\u91cf\u4e3a1.1\u00d7105J\uff1b\uff083\uff09\u5982\u679c\u6052\u6e29\u7bb1\u7684\u6e29\u5ea6\u8bbe\u5b9a\u4e3a50\u2103\uff0c\u7535\u963b\u7bb1R0\u63a5\u5165\u7535\u8def\u7684\u963b\u503c\u4e3a50\u03a9\uff0e
（1）如图所示（2）产生的热量：
Q=
| U22 |
| R0 |
| (220V)2 |
| 44Ω |
（3）当电磁继电器线圈的电流达到60mA时，设此时热敏电阻的阻值为R01，
则R01=
| U1 |
| I |
| 9V |
| 60mA |
由图可知，热敏电阻为100Ω时对应的温度为50℃，
即恒温箱的温度将稳定在50℃．
（4）从图乙可知，要提高恒温箱的设定温度，就要减小热敏电阻的阻值，根据欧姆定律可知，热敏电阻两端的电压减小；
（5）从图乙可知，要提高恒温箱的设定温度，就要减小热敏电阻的阻值；但为了使衔铁吸合工作电流即线圈中的电流仍为60mA，根据欧姆定律就要增大电阻箱连入的电阻值，即R2的阻值必须调大．
故答案为：（1）如上图所示；
（2）电热丝R0正常工作100s产生的热量为1.1×105J；
（3）可使恒温箱的温度保持在50℃；
（4）变小；
（5）变大．
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