An electric current is passed through a circuit containing two wires of the same material connected in parallel. If the lengths and radii of the wire are in the ratio `(4)/(3)` and `(2)/(3)`, then the ratio of the currents passing through the wires will be

Answer

4

Explanation

Given that,
`(l_(1))/(l_(2))=(4)/(3)` and `(r_(1))/(r_(2))=(2)/(3)`
Here, `l_(1)` and `l_(2)` are the length of the wires while `r_(1)` and `r_(2)` are the radii of the wires.
Now, we know that
`V=IRimpliesIR=` constant
`impliesI_(1)R_(1)=I_(2)R_(2)`
or `(I_(1))/(I_(2))=(R_(2))/(R_(1))" "......(i)`
But we know that the resistance of the wire is
`R=(rhol)/(A)`
Hence, from Eq (i)
`(I_(1))/(I_(2))=(rhol_(2)//A_(2))/(rhol_(1)//A_(1))`
Here, `R_(1)=(rhol_(1))/(A_(1)),R_(2)=(rhol_(2))/(A_(2))` and `rho_(1)=rho_(2)=rho`
Because both wires are of same material.
`:.(I_(1))/(I_(2))=(l_(2)A_(1))/(l_(1)A_(2))`
`implies(I_(1))/(I_(2))=(l_(2)pir_(1)^(2))/(l_(1)pir_(2)^(2))`
Here, `(l_(2))/(l_(1))=(3)/(4)` and `(r_(1))/(r_(2))=(2)/(3)`
`implies(I_(1))/(I_(2))=(3)/(4)xx((2)/(3))^(2)`
Or `(I_(1))/(I_(2))=(1)/(3)=(m)/(n)`
`impliesm+n=4`