The current in a simple series cirucit is 5A. When an additional resistance of `2Omega` is introduced, the current is reduced to 4A. Calculate the resistance o fthe original circuit. Assume that the applied potential difference is the same in both the cses.

Explanation

If V is the applied potential difference and R is the original resistance, then V//R = 5 and V//(R +2) = 4. Dividing we get
`(R +2)/(R ) = (5)/(4) or R = 8 Omega`