请教：一道初中物理竞赛题，看不懂解析，请高手帮忙，希望得到每个选项的详细解释，谢谢。
推荐回答（2个）
首先要搞清电路连接方式，它不是单纯的并联或串联，而是混联(既有串联又有并联)
R1与R4串联(在同一支路上)，再R3，R2并联(三条支路)，最后与R串联(R在干路上，
各电流，电压，电阻的关系：
有I=Ir=I1+I2+I3，I1=I4
U电源=Ur+U2，U2=U3=U1+U4
R总=Rr+R并，
1/R并=1/R2+1/R3+1/(R1+R4)
当R1断开，并联电路并联电阻R并反而增大，总电阻R总增大，总电流减小，R两端电压减小，U3增大，R3不变。所以I3(电流表示数)变大。
当R1断开，电压表测的U1。此时电压表相当于串联在电路中，如下图，由于电压表的电阻比R4大得多，所以它的示数接近=∪3。A错
当R2断开，并联减少一条支路，R并增大，R总增大，I减小，R两端电压减小，U不变，U3增大，I3电流表示数增大，U3增大，U1+U4也增大，这条支路的电流I1=I4增大，U1=I1R1，所以U1(电压表示数)增大。B错。
当R3断开时，电流表示数=0，与R2断开同理，电压表示数变大。C错。
当R4短路时，并联部分一条支路的电阻减小，R并也减小。R总也变小，U不变，I=Ir变大，Ur=IrR，变大，U3变小，所以I3(电流表示数)变小，电压表由测支路一部电压U1，U1+U4=U3，到变成测支路总电压∪1*=U3*，所以变大。D对
这解析真nb，等于说因为D是正确的所以选D。。。。
而且不对吧，r1断路，电压表不等于电源电压啊。。
电压表视为断路，电流表视为导线，此图就简化为一个R2，R3，R1+R4三者并联（可将他们视为一个整体），再和R串联的电路。无论R123哪一个断路了，结果都显然是导致这个整体的电阻变大。而R分得的电压与这个整体分的电压加起来就是电源电压，因此就会导致这个整体的电压变大，断路前，电压表测R1，R23断路，电压表仍测R1，显然R1分的电压变大。R1断，R4就流不过去了，此时R4不分压，即下图BC电压为0，又AC电压等于AB+BC，所以AC=AB，即电压表测AC电压，由之前分析，显然这个电压是增大的。
R4应该是最难分析的。若R4短，即原来的R1+R4变为R1，这条通路电阻变小，可以得出：（1）形成的整体分压变小，（2）全电路电阻也变小，总电流变大。而总电流等于R2支路+R3支路+R1支路。对R2支路，电阻不变，电压变小，故电流变小，R3同理，所以流过R1电流只能变大，所以R1的电压变大。
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