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) o* m8 x& x+ }/ O, \: B 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?
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If protective impedance is used, the current between the part and the supply source shall+ l3 @2 p& [1 t1 L: _( g- N, ^
not exceed 2 mA for d.c., its peak value shall not exceed 0,7 mA for a.c. and
, U* k) T! c$ P$ s# O& `– for voltages having a peak value over 42,4 V up to and including 450 V, the capacitance
' u6 N& Z' g0 K- \! K" S3 K6 n1 |" pshall not exceed 0,1 μF;0 m- E# j" X* q M; e0 H
– for voltages having a peak value over 450 V up to and including 15 kV, the discharge shall9 m p' e4 _ b6 N* T5 M) c/ A
not exceed 45 μC;
0 h" Q' x# V, J+ l! q" B– for voltages having a peak value over 15 kV, the energy in the discharge shall not exceed. k c( Z; h8 Q6 y
350 mJ.
) G [+ P a0 B, m% H5 z, V1 m) tCompliance is checked by measurement, the appliance being supplied at rated voltage.
4 I# |0 s- {7 U; r. I5 X+ H3 dVoltages and currents are measured between the relevant parts and each pole of the supply/ {, I$ z0 g h) Y
source. Discharges are measured immediately after the interruption of the supply. The
) @- W! a+ r ^- t2 h) n1 _6 Rquantity of electricity and energy in the discharge is measured using a resistor having a
: b+ _2 S; X4 jnominal non-inductive resistance of 2 000 Ω .
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