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电磁感应中求解做功和电热能是一种常见题型,其解题途径往往有两条:一是用 Q=I~2Rt 直接求解;二是运用能的转化和守恒定律进行计算.需要注意的是由于中学数学知识的限制,公式 Q=I~2Rt 不能随便乱用,当导体运动情况不稳定,电流变化无规律时无法直接求解,即使电流变化有规律,也只能讨论两类情况:其一是电流 I 恒定或以方波规律变化;二是电流 I 以正弦规律变化,此时可根据电流热效应利用有效值计算焦耳热、求解外力所做的功.近年来高考试题、各地模拟试题中频频出现巧借正弦交流电设计的求解功和热的问题,学生常因不能很好地理解和应用有效值而落入陷阱.为提高学生的甄别能力,避免解题时出错.本文将几种常见的巧借正弦交流电的有效值求解电热和电功的情形分类总结如下:
Solving electromagnetic induction work and electric energy is a common problem, the solution to the problem there are often two: First, with Q = I ~ 2Rt direct solution; the second is the use of energy conversion and conservation laws to be calculated. Is due to the limitation of high school mathematics knowledge. The formula Q = I ~ 2Rt can not be arbitrarily used indiscriminately. When the motion of the conductor is unstable and the current changes irregularly, it can not be solved directly. Even if the current changes regularly, only two situations can be discussed: Is the current I constant or square wave law changes; the second is the current I sinusoidal law, this time according to the current thermal effect of the effective value of the Joule heat calculation to solve the power of the work done in recent years entrance examination questions around the simulation test frequently There is the problem of solving the power and the problem of using the sine wave alternating current design, the students often fall into the trap because they can not understand and apply the effective value well.To improve the ability of the students to discriminate and avoid the mistakes in solving the problem, Coincidentally, the effective value of sinusoidal AC electric heating and electric power to solve the situation are summarized as follows: