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 DSH 534
; r' R; v( k' \) k2 R4 s) S8 R3 { | Figure 21, presence of resistor of 45 kO
6 V2 V1 q, l0 b* L9 }8 F6 Y! Z) }( k | 19
0 w8 Y- s' U$ r2 | | 60601-1(ed.2);am1;am2) S+ C' D1 @$ ~4 b4 j/ e) w
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' h' m9 U* g/ X: D) r' RStandard(s)- (year and edition):9 @2 V2 y P* r, n3 l
IEC 60601-1:1988 Ed.2
1 M6 J# M$ l4 m0 b: _Am1+Am2
, T4 t% E$ z- G3 g/ Q5 X. |Sub clause(s):
m m( t6 M7 J$ \# L& a- s& \: k193 C) N+ P* Y" n6 I
Sheet n°: 5349 g# Q D6 c: E$ m8 X" }
Subject:7 \; L8 W) w' m: o
Figure 21, presence of resistor* \- H2 Q1 x- d M; |7 o
of 45 kO- F$ v% f* A. i0 l* W) ^
Key words: Decision taken at the 40th
$ t! \0 n" H) R2 A' M1 Umeeting 2003
$ ~- D2 A5 [% B3 S3 y0 sQuestion:
/ u" K5 y7 y6 s8 ]Should the 45 kO resistor be used since the current is limited to 5 mA and the limit is 5 mA?6 l" u2 x( [: `1 G9 o0 q. I
Decision:
# g' c- q. }5 @3 G y1 fFor the 1st edition, use the method of 2nd edition (use any resistance).5 m" a. y6 y, J9 Q) f
For 2nd edition, instead of using any resistance, you may also use alteration of the voltage or fuses.2 s' B& Z6 f9 A$ A+ @8 t8 f
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