If the internal resistance of the cell is 3 ohm, find the current supplied by the cell in the circuit shown in figure.

Explanation

`r = 3Omega`
`E = 1.5 V`
`R_1 = 6 Omega`
`R_2 = R_3 =12Omega`
All the resistors are in parallel and let their equivalent - resistance be `R_p` .
So `1/R_P =1/R_1 +1/R_2+1/R_3`
`implies1/R_P =1/6 +1/12 +1/12`
`implies 1/R_P =(2+1+1)/12 =4/12=1/3`
`implies R_P =3Omega`
Now `R_P` and r are in series
`:.` Effective resistance of circuit `=R_("eff")`
`=R_(p)+ r = 3+3`
`= 6 Omega`
So `I=E/(R_("eff"))`
`I = (1.5)/6=0.25A`