If the internal resistance of the cell is 1.5 ohm, find the current supplied by the cell in the circuit shown in figure.

Explanation

E = 1.0 V
`R = 5.2Omega`
`R_2 = R_3 = 10Omega`
`r = 1.5 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`
Now `R_p and r` are in series
`:.` Effective resistance of the circuit `=R_("eff") = r+R_p`
`R_("eff") = 1.5+ 2.5 = 4Omega`
`I=E/(R_("eff"))=1/4`
`I = 0.25 A`