A 5 ampere current is distibuted in three branches. The ratio of the lengths of the wires in the three branches is 1:2:3.Determine the magnitude of current in each branch. The material and the cross sectional area of each wire are the same.

Explanation

Resistance of the wire is proportional to its length as the material and the cross sectional area are the same.So we take the resistances as x,2x and 3x and the equivalent resistance as R, we have,
`1/R=1/x+1/(2x)+1/(3x)=(6+3+2)/(6x)=11/(6x)or,R=(6x)/11`
so, potential difference between the two terminals
`V=IR=I.(6x)/11`
As the potential difference across each wire is the same, the current in the first branch
`=V/x=I.6/11=5times6/11=30/11A`
Current in the second branch
`=V/(2x)=I.3/11=5times3/11=15/11A`
current in the third branch
`=V/(3x)=I.2/11=5times2/11=10/11A`