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1 ?0 @/ J6 b1 y0 E
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？4 g7 N! L6 M1 j2 k; Z% z( f. |) y
2 L9 m) B% ]' a4 \1 g5 P
If protective impedance is used, the current between the part and the supply source shall, q! z9 V" U. X4 Y
not exceed 2 mA for d.c., its peak value shall not exceed 0,7 mA for a.c. and) ]" W& }$ p- y7 `. U, O: {
– for voltages having a peak value over 42,4 V up to and including 450 V, the capacitance; \3 z" \0 b+ e5 z2 O- E/ c/ J
shall not exceed 0,1 μF;1 b! m9 v) d7 P h2 ]5 Q
– for voltages having a peak value over 450 V up to and including 15 kV, the discharge shall
' v: D3 f* {3 @not exceed 45 μC;! G9 p4 }- f- w j- m
– for voltages having a peak value over 15 kV, the energy in the discharge shall not exceed
}' T T3 U* y2 c* W# M Y* |350 mJ.( f& h/ K9 H; N) s
Compliance is checked by measurement, the appliance being supplied at rated voltage.
) s2 w; X f; e$ r7 |3 l; N5 W; EVoltages and currents are measured between the relevant parts and each pole of the supply& v4 v: N9 P' n, K
source. Discharges are measured immediately after the interruption of the supply. The% N# `: O) e5 l& @
quantity of electricity and energy in the discharge is measured using a resistor having a
3 c* d9 }8 T7 M! O( J/ @) wnominal non-inductive resistance of 2 000 Ω .; K/ g3 a4 e! y; V
) A" k* t* i2 O/ q. j' N+ P |
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发表于 2016-8-15 16:43
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