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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?
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' u, v Y2 k8 f% V) dIf protective impedance is used, the current between the part and the supply source shall9 Y5 f9 q9 B: Z% q; V3 M- g' _
not exceed 2 mA for d.c., its peak value shall not exceed 0,7 mA for a.c. and
. @2 K) {# c8 t' U– for voltages having a peak value over 42,4 V up to and including 450 V, the capacitance
+ E: X, G4 J& X( Tshall not exceed 0,1 μF;
* H) R% g- {' c: h0 _# J. M– for voltages having a peak value over 450 V up to and including 15 kV, the discharge shall8 I2 A; K0 u7 c N
not exceed 45 μC;
. y' p+ P# m" K# {1 N) z) L– for voltages having a peak value over 15 kV, the energy in the discharge shall not exceed
+ Z9 _ q/ T! J6 y2 Y$ \350 mJ.% z% a$ G- g8 r1 P5 b
Compliance is checked by measurement, the appliance being supplied at rated voltage.% A( _, V0 }& O6 y( o
Voltages and currents are measured between the relevant parts and each pole of the supply
E$ e3 l9 t5 C+ o& k2 wsource. Discharges are measured immediately after the interruption of the supply. The. F7 N0 ?+ A& e, y5 w- Q
quantity of electricity and energy in the discharge is measured using a resistor having a$ X r- M: C# E1 D
nominal non-inductive resistance of 2 000 Ω .
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