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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?( y( A: v; ~7 t; D$ c- g
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If protective impedance is used, the current between the part and the supply source shall" b$ ^* A4 g5 r0 H; x) u( V
not exceed 2 mA for d.c., its peak value shall not exceed 0,7 mA for a.c. and( g' V* V0 l* H% g$ f; O1 ^ z
– for voltages having a peak value over 42,4 V up to and including 450 V, the capacitance
5 P$ Y1 |: D+ U; m# sshall not exceed 0,1 μF;& T* p" u7 `6 ~6 |
– for voltages having a peak value over 450 V up to and including 15 kV, the discharge shall
% @0 O! {: O3 R& p' l# |not exceed 45 μC;0 `. u M2 u }3 e! h
– for voltages having a peak value over 15 kV, the energy in the discharge shall not exceed9 r6 A1 A/ F# h, G5 i n
350 mJ.
' F7 w$ ?3 N6 f3 ]; hCompliance is checked by measurement, the appliance being supplied at rated voltage.
& k1 H8 I% h6 \# G! qVoltages and currents are measured between the relevant parts and each pole of the supply9 a) u0 K0 n2 j+ c1 r& c
source. Discharges are measured immediately after the interruption of the supply. The/ t6 z4 d: h" c0 F& I1 h2 {
quantity of electricity and energy in the discharge is measured using a resistor having a
8 I& W1 O& d; |5 ~* Z, w$ @2 Unominal non-inductive resistance of 2 000 Ω .
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