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25 Feb 2007 19:11:31 IST

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Q:- Establish the current distribuition over 2 resisitances R1 & R2 connected in parallel corresponds to the minimum thermal power generated in the circuit

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vir sharma

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Joined: 17 Feb 2007

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25 Feb 2007 19:25:57 IST

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Hi i think the sol. is

Let R1 & R2 be the 2 resistances . thermal power generated in the circuit is

P =i^2( R1R2/R1+R2)
in parallel combination,thermal power will be minimum if:

i^2 (R1R2 / R1+R2) < i^2 (R1 + R2).......(res. in series)
or
i^2 [R1R2/R1+R2 - (R1+R2)] < 0............(1)

= i^2 [R1R2 - (R1+R2)^2/ R1+R2]

= i^2 [-(R1^2 + R2^2 + R1R2] / R1+ R2

since it is less than zero hence =n (1) satisfied

hope my ans is correct plz give me the hat

Himanshu

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Joined: 19 Feb 2007

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25 Feb 2007 19:28:21 IST

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thanks vir u r cool

Himanshu

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Joined: 19 Feb 2007

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25 Feb 2007 19:29:04 IST

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now i 'i give u a desrving salute

vir sharma

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Joined: 17 Feb 2007

Posts: 162

25 Feb 2007 19:29:58 IST

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hey thanks

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