CBSE BOARD X, asked by rukumanikumaran, 9 months ago

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Answered by amitnrw
4

22  Ω   is the Equivalent resistance between A & B

Explanation:

Taking 1st one

10 Ω   & 20 Ω   in Series    together  are in Parallel with  10 Ω &  10 Ω   in Series    together  and this resultant is in series with 10 Ω

Solving one be one

10 Ω   & 20 Ω   in Series   => 10 + 20  = 30 Ω    

10 Ω   & 10 Ω   in Series   => 10 + 10  = 20 Ω    

30 Ω  ║  20 Ω    

=> 1/R  = 1/30  + 1/20

=> 1/R  = (2 + 3)/60

=> 1/R = 5/60

=> 1/R = 1/12

=> R = 12 Ω  

10 Ω  &  12 Ω    in Series

Equivalent  resistance between A & B   = 10 + 12  = 22  Ω  

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Answered by Anonymous
8

Answer:

Explanation:

Taking 1st one

10 Ω   & 20 Ω   in Series    together  are in Parallel with  10 Ω &  10 Ω   in Series    together  and this resultant is in series with 10 Ω

Solving one be one

10 Ω   & 20 Ω   in Series   => 10 + 20  = 30 Ω    

10 Ω   & 10 Ω   in Series   => 10 + 10  = 20 Ω    

30 Ω  ║  20 Ω    

=> 1/R  = 1/30  + 1/20

=> 1/R  = (2 + 3)/60

=> 1/R = 5/60

=> 1/R = 1/12

=> R = 12 Ω  

10 Ω  &  12 Ω    in Series

Equivalent  resistance between A & B   = 10 + 12  = 22  Ω  

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