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CBSE Class 12-science Answered

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

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