Three cells of emf 3V, 4V, and 6V are connected in parallel, If their internal resistances are `1 Omega`, `2 Omega` and `1 Omega` find the `epsilon_(eff), r_(eff)`, and the current in the external load R = 1.6 `Omega` .

Explanation

The cells are conneted in parallel. The equivalent call emf is
`e_(eff) = (Sigmaepsilon_i//r_i)/(Sigma1//r_1) =((3)/(1)+(-(4)/(2))+(6)/(1))/((1)/(2)+(1)/(2)+(1)/(1)) = 2.8V`
The internal resistance of the cell is
`(1)/(r_("eff"))=sum(1)/(r_(i))=(1)/(1)+(1)/(2)+(1)/(1)=(5)/(2)`
or `r_(eff) = (2)/(5)Omega`

The current in the circuit is
`i = (epsilon)/(R +r) = (2.8)/((8)/(5)+(2)/(5))= 1.4A`