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Asked by Varsneya Srinivas 21st February 2018, 8:57 PM
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.

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 Expert 22nd February 2018, 11:47 AM
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