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% u0 W% \; Y; f0 t. e# D% q0 ]
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？1 n4 |( q3 k: E8 T1 k) \- a" f
. U! l h, i0 |2 u% }1 }
If protective impedance is used, the current between the part and the supply source shall, R/ M1 i) a: {
not exceed 2 mA for d.c., its peak value shall not exceed 0,7 mA for a.c. and
. R+ M) \: i1 O– for voltages having a peak value over 42,4 V up to and including 450 V, the capacitance: f+ R! T" f, b$ }2 X. @) u
shall not exceed 0,1 μF;
9 E: b X& b; S: Q5 a– for voltages having a peak value over 450 V up to and including 15 kV, the discharge shall# L. V' k4 m' E
not exceed 45 μC;
0 y& {$ z B( U% o– for voltages having a peak value over 15 kV, the energy in the discharge shall not exceed b' d7 W6 b0 ^$ A
350 mJ.7 r1 u9 m* A) \6 Y9 o: R
Compliance is checked by measurement, the appliance being supplied at rated voltage.4 P* E! X1 g3 O1 \( n
Voltages and currents are measured between the relevant parts and each pole of the supply. Y6 Z7 ?! A5 E9 |, f
source. Discharges are measured immediately after the interruption of the supply. The
. h' Z5 k" w3 q5 D3 nquantity of electricity and energy in the discharge is measured using a resistor having a
! x! f6 A- \+ H. }4 P+ l5 d, Tnominal non-inductive resistance of 2 000 Ω .# S5 C# B$ o+ [/ S4 h
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发表于 2016-8-15 16:43
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