题目

如图所示电路，电阻R1＝10Ω，若断开S2、闭合S1，电流表的示数I1＝0.30A；若闭合S1、S2，电流表的示数I2＝0.50A。求   （1）电源两端的电压；   （2）电阻R2的阻值；   （3）电阻R2通电1min产生的热量。 答案：解：（1）由欧姆定律，电源两端电压 Ｕ＝I1R1＝3V       （2）通过电阻R2的电流I＝I2－I1=0.2A          由欧姆定律 R2＝＝15Ω________ I haven’t had any success. ________, I’ll keep trying.A.So far; ButB.Now; HoweverC.So far; HoweverD.Until now; But