Two resistors are connected in series with 5V battery of negligible internal resistance. A current of 2 A flows through each resistor. If they are connected in parallel with the same battery a current of `(25)/(3)A` flows through combination. Calculate the value of each resistance.

Explanation

`I_g=(IS)/(G+S)`….(I)
`I_gG=(I-I_g)S`
or `S=(I_gG)/(I-I_g) " " I_gG=IS-I_gS`
`therefore I_g=(IS)/(G+S)`
`I_gG+I_gS=IS , " " (G+S)I_g=IS`
from equation (1) `I_g prop I`
Since S and G are constants, the scale can be graduated to give the main current directly.