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* f0 ~1 p6 t5 N 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 Y4 d8 J. ^+ q! p" h
5 r; z7 }/ x( O8 {If protective impedance is used, the current between the part and the supply source shall0 m4 X: n; z, @9 d6 A$ l
not exceed 2 mA for d.c., its peak value shall not exceed 0,7 mA for a.c. and: E/ F4 Z5 |5 k2 y# [* x! Z
– for voltages having a peak value over 42,4 V up to and including 450 V, the capacitance
7 w. M5 A" O$ B) W/ nshall not exceed 0,1 μF;
" E1 Y* b, h) S# _4 G. ?– for voltages having a peak value over 450 V up to and including 15 kV, the discharge shall
+ x7 }* B+ q8 L7 }4 }& `not exceed 45 μC;
2 z" M1 f. G( j, L7 ]! p3 l– for voltages having a peak value over 15 kV, the energy in the discharge shall not exceed( R0 @7 x; n ^9 P; d* A6 A2 I
350 mJ.4 t% X0 k3 p0 w9 d* I- I8 y. T* i
Compliance is checked by measurement, the appliance being supplied at rated voltage.0 Z3 b+ ~% ^/ L2 ]3 x' e: |& B
Voltages and currents are measured between the relevant parts and each pole of the supply
/ B* }2 k3 k& G" lsource. Discharges are measured immediately after the interruption of the supply. The* d7 P' g5 |/ F, V+ P
quantity of electricity and energy in the discharge is measured using a resistor having a: f% n: H5 G% Y3 [. A/ ?5 g5 D
nominal non-inductive resistance of 2 000 Ω .6 e# b: h4 k' G+ o# g( N
, T9 C$ U1 ]/ z- i( N
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发表于 2016-8-15 16:43
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