论文部分内容阅读
高中物理甲种本第二册第七章中有一道习题:现有电动势为1.5伏,内电阻为1欧的电池若干.每个电池允许输出的电流为0.05安,又有不同阻值的电阻可作分压电阻.试设计一种电路,使额定电压为6伏,额定电流为0.1安的用电器正常工作.画出电路图,并标明分压电阻的值. 《标准化训练与教学》一书对该题作出的解答是: 如图1的电路可用.用电器的电阻为R=U/I=6/0.1欧=60欧. 共用12节电池,6个串联一组,然后两组再并联,电路中再串联R.若用电器正常工作,则 _总=I(R_总+T_总), 6×1.5=0.1(60+R+6×1/2), 即 R=27欧. 乍一看,上述解法设计出来的电路,由6个电池串联一组再两组并联成混联电池组向用电器供电,用
There is an exercise in the seventh chapter of the second volume of high school physics. There is a number of batteries with an electromotive force of 1.5 volts and an internal resistance of 1 ohm. The allowable output current of each battery is 0.05 ampere and there are resistances of different resistances. Can be used as a voltage divider resistor. Try to design a circuit, so that the rated voltage of 6 volts, rated current of 0.1 A normal electrical appliances. Draw a circuit diagram, and marked the value of the voltage divider resistance. “Standardization of training and teaching” a book The answer to this question is: The circuit shown in Figure 1 is available. The resistance of the appliance is R = U/I = 6/0.1 ohm = 60 ohms. There are 12 batteries in common, 6 in series, and then the two groups are connected in parallel. R. If the electrical equipment works normally, then _ total = I (R_ total + T_ total), 6 × 1.5 = 0.1 (60 + R + 6 × 1/2), ie R = 27 euro At first glance, the circuit designed by the above solution consists of six batteries connected in series and two sets of parallel connected battery packs.