CBSE Class 12-science Answered

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Asked by Varsneya Srinivas | 21 Feb, 2018, 08:57: PM

Expert Answer

We know the resistance of a conductor is directly proportional to length and inversly proportional to Area of cross section.

Lengths of wires are in the ratio 2 : 3 : 4 , Radii of wires in the ratio 3 : 4 : 5 hence Areas of Cross section 9 : 16 : 25

Hence resistances of wires are in the ratio (2/9) : (3/16) : (4/25)

we can assume the three resistances R₁ = (2/9)R , R₂ = (3/16)R and R₃ = (4/25)R

Let V be the potenial difference across the parallel combination of three resistances R₁, R₂ and R₃ .

Let i₁ be current through resitance R₁. Let i₂ be current through resitance R₂. Let i₃ be current through resitance R₃.

begin mathsize 12px style i space equals i subscript 1 plus i subscript 2 plus i subscript 3 i subscript 1 R subscript 1 space equals space i subscript 2 R subscript 2 space equals space i subscript 3 R subscript 3 space equals space V i subscript 1 over i equals space fraction numerator begin display style bevelled V over R subscript 1 end style over denominator space open parentheses bevelled fraction numerator space V over denominator R subscript 1 end fraction space close parentheses space plus space open parentheses space bevelled V over R subscript 2 space close parentheses plus open parentheses space bevelled V over R subscript 3 space close parentheses end fraction space equals space fraction numerator R subscript 2 space cross times space R subscript 3 over denominator R subscript 1 space cross times space R subscript 2 space plus space R subscript 2 space cross times space R subscript 3 space plus space R subscript 3 space cross times space R subscript 1 end fraction s i m i l a r l y space i subscript 2 over i equals space fraction numerator R subscript 3 space cross times space R subscript 1 over denominator R subscript 1 space cross times space R subscript 2 space plus space R subscript 2 space cross times space R subscript 3 space plus space R subscript 3 space cross times space R subscript 1 end fraction i subscript 3 over i equals space space fraction numerator R subscript 1 space cross times space R subscript 2 over denominator R subscript 1 space cross times space R subscript 2 space plus space R subscript 2 space cross times space R subscript 3 space plus space R subscript 3 space cross times space R subscript 1 end fraction end style

by substituting the values for R₁, R₂ and R₃ and from the known current i.e. 5A, we get i₁ = 1.40 A , i₂ = 1.66 A and i₃ = 1.94 A

Answered by Thiyagarajan K | 22 Feb, 2018, 11:47: AM

CBSE Class 12-science Answered

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CBSE Class 12-science Answered

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