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CBSE Class 10 Answered

DETERMINE THE EQUIVALENT RESISTANCE ACROS AB IN THE ABOVE CIRCUIT?
Asked by abinash.gupta003 | 28 May, 2017, 05:38: PM
answered-by-expert Expert Answer
begin mathsize 12px style The space resistors space are space named space straight R subscript 1 space to space straight R subscript 6 space as space shown space in space the space figure space above. Consider space parallel space combination space of space straight R subscript 3 space and space straight R subscript 4 colon straight R subscript 34 equals fraction numerator straight R subscript 3 cross times straight R subscript 4 over denominator straight R subscript 3 plus straight R subscript 4 end fraction equals fraction numerator 8 cross times 8 over denominator 8 plus 8 end fraction equals 64 over 16 equals 4 space straight capital omega Now comma space straight R subscript 5 space and space straight R subscript 6 space are space in space series space and space that space series space combination space is space in space parallel space with space straight R subscript 34. therefore straight R subscript 56 equals straight R subscript 5 plus straight R subscript 6 equals 4 plus 8 equals 12 space straight capital omega therefore straight R subscript straight P equals fraction numerator straight R subscript 34 cross times straight R subscript 56 over denominator straight R subscript 34 plus straight R subscript 56 end fraction equals fraction numerator 4 cross times 12 over denominator 4 plus 12 end fraction equals 48 over 16 equals 3 space straight capital omega Now comma space this space parallel space equivalent space resistance space straight R subscript straight P space is space in space series space with space straight R subscript 2. therefore straight R subscript straight P 2 end subscript equals straight R subscript straight P plus straight R subscript 2 equals 3 plus 3 equals 6 space straight capital omega This space series space combination space is space finally space connected space in space parallel space with space straight R 1 space to space give space the space equivalent space resistance space of space the space circuit. therefore straight R subscript eq equals fraction numerator straight R subscript straight P 2 end subscript cross times straight R subscript 1 over denominator straight R subscript straight P 2 end subscript plus straight R subscript 1 end fraction equals fraction numerator 6 cross times 20 over denominator 6 plus 20 end fraction equals 120 over 26 equals 60 over 13 equals 4.62 space straight capital omega end style
Answered by Romal Bhansali | 29 May, 2017, 09:43: AM

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