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

Explanation

`E = 0.75V `
`R_1 = 4Omega`
`R_2 = R_3 = 8 Omega`
`r =1 Omega, I=?`
`R_1, R_2 and R_3` are in parallel and let their net resistance be `R_p`
So `1/R_P = 1/R_1 +1/R_2 +1/R_3`
`1/R_P=1/4 +1/8 +1/8`
`implies R_P =2Omega`
Now r and `R_p` are in series Effect resistance of the circuit = `R_("eff") = r+R_P`
`R_("eff")=1+2=3Omega`
`I = E/(R_("eff")`
`= I = (0.75)/3 = 0.25A`