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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？
) b2 \. l. f) g1 h9 o( l8 d1 N/ m$ X7 d6 C6 W2 f7 R
If protective impedance is used, the current between the part and the supply source shall
. a, B) Q$ }3 Rnot exceed 2 mA for d.c., its peak value shall not exceed 0,7 mA for a.c. and: E5 i+ W/ e3 c% Z- D
– for voltages having a peak value over 42,4 V up to and including 450 V, the capacitance/ C% d- r" S0 o% |
shall not exceed 0,1 μF;
5 Z; p( |; b1 v1 b% R9 {1 m– for voltages having a peak value over 450 V up to and including 15 kV, the discharge shall
/ H' ^* ?: X* {3 T: n! _not exceed 45 μC;
. z; ~6 r3 _) r. V. p– for voltages having a peak value over 15 kV, the energy in the discharge shall not exceed# q% T$ K4 _2 t9 ^/ A/ W
350 mJ.
' E0 w7 i( q5 e# m' N& ?+ CCompliance is checked by measurement, the appliance being supplied at rated voltage.
# [, V( [0 I+ {' [+ LVoltages and currents are measured between the relevant parts and each pole of the supply
$ X1 \+ `' N$ ~, Asource. Discharges are measured immediately after the interruption of the supply. The
- Z$ t. r* M3 {9 Y" O1 {; P+ v% ]quantity of electricity and energy in the discharge is measured using a resistor having a
8 S5 f$ d5 w7 v( l) @3 ^nominal non-inductive resistance of 2 000 Ω .; N+ i Y5 P7 l+ f
8 m% N; e) k, `9 r6 ~1 @7 ]5 I
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发表于 2016-8-15 16:43
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