Question

Two cells `e_(1) and e_(2)` connected in opposition to each other as shown in figure. The cell of emf 9 V and internal resistance `3Omega` the cell is of emf 7V and internal resistance `7Omega`. The potential difference between the points A and B is

A. 8.4V
B. 5.6V
C. 7.8V
D. 6.6V

Answer 1

Correct Answer - A
`I=(Deltaepsi)/(r_(1)+r_(2))=(9-7)/(3+7)=(2)/(10)=0.2A`
Potential difference across cell `epsi_(1)` is `=9-0.2xx3=9-0.6=8.4V`
Potential difference across `epsi_(2)`,
`V_(AB")=epsi_(2)+0.2 r_(2)=7+0.2xx7=7+1,4=8.4V`