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- Z6 o% G8 \% y 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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2 y, P0 e1 A# p9 x7 e5 ^4 @If protective impedance is used, the current between the part and the supply source shall
, ]1 J$ n2 ~) l- Nnot exceed 2 mA for d.c., its peak value shall not exceed 0,7 mA for a.c. and# w; D% g" u+ y: j; P" S
– for voltages having a peak value over 42,4 V up to and including 450 V, the capacitance" P2 t/ K' y2 O$ k: _
shall not exceed 0,1 μF;
1 r: ^" M: J& w* y* I7 ~– for voltages having a peak value over 450 V up to and including 15 kV, the discharge shall
' F1 B# P1 \9 a' ]; J+ @. `not exceed 45 μC;/ h6 A0 g& D" J# W( X
– for voltages having a peak value over 15 kV, the energy in the discharge shall not exceed5 I5 G0 d7 K# S6 i( D% z* r! _
350 mJ.
- x1 D) j; D; C0 sCompliance is checked by measurement, the appliance being supplied at rated voltage.
: v P5 D+ b2 R% ]0 \5 e4 ~8 T1 LVoltages and currents are measured between the relevant parts and each pole of the supply: I5 k# J1 U0 i9 T- u# B
source. Discharges are measured immediately after the interruption of the supply. The* ~: ]9 @! m- ^
quantity of electricity and energy in the discharge is measured using a resistor having a
2 ]- q3 @1 Z6 x0 B& P' w% l) a% hnominal non-inductive resistance of 2 000 Ω .
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