A battery of em.f. 6 volt and internal resistance 0.4 ohm is connected to a parallel combination of two resistors of resistances 9 ohm and 6 ohm. Calculate the terminal voltage of the battery.

Explanation

`E = 6 V`
`r = 0.4 omega`
`R_1 = 9 Omega`
`R_2 = 6 Omega`
`V=?`
`:. R_1 and R_2` are in parallel and let their resistance be R.
So `1/R=1/R_1+1/R_2=1/9+1/6`
`implies 1/R=(2+3)/18=5/18` `implies R =18/5 =3.6Omega`
Now `r=(E/V-1)R`
`implies 0.4 =(6/V-1)3.6`
`implies 6/V-1 = (0.4)/(3.6)=1/9`
`implies 6/V =1/9 +1 = (1+9)/9`
`implies 6/V = 10/9`
`implies V = (6 xx9)/10 = 5.4 ` volt