Two cells of emf 3 V and 2 V and internal resistances `1.5 Omega` and `1 Omega` respectively are connected in parallel across `3 Omega` resistor such that they tend to send current through resistor in the same direction. Calculate potential difference across `3 Omega` resistor.

Explanation

`E_1=3V " " E_2=2V , " " r_1=1.5Omega " " r_2=1Omega " " R=3Omega`
Equivalent internal resistance of the combination,
`r_(eq)=(r_1r_2)/(r_1+r_2)=(1.5xx1)/(1.5+1)=0.6Omega`
Equivalent emf of the combination,
`E_(eqv)/r_(eq)=E_1/r_1+E_2/r_2 , E_(eqv)/0.6=3/1.5+2/1, E_(eq)`=2.4 V
Current through the `3Omega` resistor
`I=E_(eq)/(R+r_(eq))=2.4/(3+0.6)`
`I=2/3A`
Potential differences across `3Omega` resistor
`V=IR=(2//3)(3)=2V`