论文部分内容阅读
对《高级中学课本·物理(必修)》第二册38页的例题:有一盏弧光灯,额定电压U_1=40V,正常工作时通过的电流I=0.5A,应该怎样把它连入U=220V的照明电流中?现提两点个人看法.1.弧光灯是一种气体电光源,按工作电源可分为交流供电型和直流供电型.一般只在直流供电的电路中,才采用串入电阻的办法维持弧光灯的正常工作;在交流供电型电路中一般采用串入扼流圈的办法限制其工作电流,以保证其正常工作.而220V照明用电显然是交流电源(仅从这一点来讲,将本题作为恒定电流的例题似乎就不太妥当),从技术上讲应接入扼流圈来限制工作电流而又不致引起过大的功率损失.在题设条
On the 38th page of the book “High school textbooks and physics (required)” in Book 2 of Volume 2: There is an arc lamp with a rated voltage U_1=40V. The current passed during normal operation is I=0.5A. How should it be connected to U=220V? The lighting current in the current? Now mention two personal views. 1. Arc lamp is a gas electric light source, according to the power supply can be divided into AC power supply type and DC power supply type. Generally only in the DC power supply circuit, only use the string into The method of resistance maintains the normal operation of the arc lamp; in the AC power supply type circuit, the method of limiting the operating current by using a choke coil is generally used to ensure its normal operation. The 220V lighting power is obviously an AC power supply (only from this point In terms of the example of the constant current, it seems that this question is not appropriate. Technically, it should be connected to the choke coil to limit the operating current without causing excessive power loss.