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在图1所示的电路中,a、b两点,是那点的电势高,在有些参考书和杂志上,都是给出一个答案:“a点的电势高于b点的电势”。然而这个答案是不全面的,值得商榷。我们把图1改画成图2的等效图。从图2可以明显看出:流过电压表中的电流方向,是取决于I_1、I_2迭加的结果。那么a、b两点到底是那点的电势高呢? 我们利用电势升降方程来讨论。在图1中有: U_b+ε_1-I_1r_1=U_a∴U_(ab)=ε_1-I_1r_1=ε_1-ε_1+ε_2/r_1+r_2+R r_1 =ε_1r_1+ε_1r_2+ε_1R-ε_1r_1-ε_2r_1/r_1+r_2+R U_(ab)=ε_1(r_2+R)-ε_2r_1/r_1+r_2+R从上式可知: 当ε_1(r_2+R)>ε_2r_1时,U_(ab)>0,这时a点的电势高于b点的电势;
In the circuit shown in Fig. 1, a and b are two points. The potential at that point is high. In some reference books and magazines, an answer is given: “The potential at point a is higher than the potential at point b.” ". However, this answer is not comprehensive and it is debatable. We will change Figure 1 to the equivalent of Figure 2. It can be clearly seen from Figure 2 that the direction of the current flowing through the voltmeter depends on the superposition of I_1 and I_2. So what is the point a and b at which point the potential is high? We use the potential equation to discuss. In FIG. 1, there are: U_b+ε_1-I_1r_1=U_a∴U_(ab)=ε_1-I_1r_1=ε_1-ε_1+ε_2/r_1+r_2+R r_1 =ε_1r_1+ε_1r_2+ε_1R-ε_1r_1-ε_2r_1/r_1+r_2+ R U_(ab)=ε_1(r_2+R)-ε_2r_1/r_1+r_2+R It can be known from the above equation that when ε_1(r_2+R)>ε_2r_1, U_(ab)>0, the potential at point a is high. The potential at point b;