Question

Two cells having an internal resistance of `0.2 Omega` and `0.4 Omega` are connected in parallel, the voltage across the battery is 1.5 V. If the emf of one cell is 1.2 V, then the emf of second cell is
A. 2.1 V
B. 2.7 V
C. 3 V
D. 4.2 V

Answer 1

Correct Answer - A
`r_(1)=0.2 Omega, r_(2)=0.4 Omega`
V=lR
`R=(r_(1)r_(2))/(r_(1)+r_(2)), l=l_(1)+l_(2) implies V=(r_(1)r_(2))/(r_(1)+r_(2))(l_(1)+l_(2))`
`l_(1)=E_(1)//r_(1) " and " l_(2)=E_(2)//r_(2)`
`V=((E_(1))/(r_(1))+(E_(2))/(r_(2)))(r_(1)r_(2))/(r_(1)+r_(2))=((E_(2)r_(1)+E_(1)r_(2)))/(r_(1)r_(2))((r_(1)r_(2))/(r_(1)+r_(2)))`
` V=(E_(2)r_(1)+E_(1)r_(2))/(r_(1)+r_(2)) implies 1.5=(E_(2)xx0.21.2xx0.4)/(0.6)`
`0.9=0.2 E_(2)+0.48 implies (0.42)/(0.2)=E_(2) implies E_(2)=2.1 V`