CL.8.1.4 OF IEC 60335-1 电量的计算
求解：标准中有说电量的计算是通过2000 Ω电阻，测出电压/时间曲线，求出电路二端的放电电量。好久没用高数了，对于此电压/时间曲线，如何计算？积分公式是怎样的？
If protective impedance is used, the current between the part and the supply source shall
not exceed 2 mA for d.c., its peak value shall not exceed 0,7 mA for a.c. and
– for voltages having a peak value over 42,4 V up to and including 450 V, the capacitance
shall not exceed 0,1 μF;
– for voltages having a peak value over 450 V up to and including 15 kV, the discharge shall
not exceed 45 μC;
– for voltages having a peak value over 15 kV, the energy in the discharge shall not exceed
350 mJ.
Compliance is checked by measurement, the appliance being supplied at rated voltage.
Voltages and currents are measured between the relevant parts and each pole of the supply
source. Discharges are measured immediately after the interruption of the supply. The
quantity of electricity and energy in the discharge is measured using a resistor having a
nominal non-inductive resistance of 2 000 Ω .
有的示波器好像是带这个功能的， 本帖最后由 vcgtrf 于 2016-8-16 13:39 编辑
直接用示波器的都是，积分出来的值再除以2000就是要的值了 查DLM 示波器，只有此项功能：通过计算焦耳积分(I2t) 测量浪涌电流，是用它除以2000？依据是什么？ 直接用示波器 本帖最后由 vcgtrf 于 2016-8-17 10:23 编辑
积分出来是mVs的话，除以2000欧姆电阻得到电量，电流积分那个就不知道了 示波器的确是个好东西 测量时，终止电压怎么设定？ 直接用示波器测出来的，自己算是算不了的。另外注意这个测试是断电之后才开始测量喔！ 哪位能否分享一个示波器测试的截图，并文字润墨以点缀:(
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