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DEVESHkishore Group TitleBest ResponseYou've already chosen the best response.1
i have given a possible solution of ur problem to Keni_jk plz look at it
 one year ago

DEVESHkishore Group TitleBest ResponseYou've already chosen the best response.1
dw:1369544893178:dw solve this u will get solution of 1st part
 one year ago

DEVESHkishore Group TitleBest ResponseYou've already chosen the best response.1
dw:1369544973454:dw solve box . get answere. now solve with resistor in series
 one year ago

prateekbansal Group TitleBest ResponseYou've already chosen the best response.0
tankzz helped a lot
 one year ago

DEVESHkishore Group TitleBest ResponseYou've already chosen the best response.1
welcome :)
 one year ago

KenLJW Group TitleBest ResponseYou've already chosen the best response.0
1K1K = 1K(11. 1K + 1K = 1K(1+1) So I'll only use 1's (1+1)1 +1 = 5/2 (1+11)1 = 3/5 I had to try different arrangements until I came upon this. 11 = 1/2 21 = 2/3 general rule 1a = a/(1+a)
 one year ago

KenLJW Group TitleBest ResponseYou've already chosen the best response.0
should be 5/3 This principle of factoring resistance can be used on all passive networks aRa, LL/R, CRC You need to make appropriate change in driving function, whether you preserving loop currents or nodal voltages This is used in filter design
 one year ago
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