Two cells ε1 and ε2 connected in opposition to each other as shown in figure. The cell ε1 is of emf 9 V and internal resistance 3 Ω the cell ε2 is of emf 7 V and internal resistance 7 Ω. The potential difference between the points A and B is

Question

Two cells ε1 and ε2 connected in opposition to each other as shown in figure. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/8d338dc8bf5b690ecd4bd880e302c9da-.jpeg" /> The cell ε1 is of emf 9 V and internal resistance 3 Ω the cell ε2 is of emf 7 V and internal resistance 7 Ω. The potential difference between the points A and B is (A) 8.4 V (B) 5.6 V (C) 7.8 V (D) 6.6 V

Tardigrade Admin · Accepted Answer

I=(Δε/r1+r2)=(9-7/3+7) =(2/10)=0.2 A Potential difference across cell ε1 is =9-0.2 × 3 =9-0.6=8.4 V Potential difference across ε2: VAB=ε2+0.2 r2=7+0.2×7=7 =7+1.4=8.4 V

Q. Two cells $ε_{1}$ and $ε_{2}$ connected in opposition to each other as shown in figure.

The cell $ε_{1}$ is of emf $9 V$ and internal resistance $3 Ω$ the cell $ε_{2}$ is of emf $7 V$ and internal resistance $7 Ω$ . The potential difference between the points $A$ and $B$ is

$8.4 V$

$5.6 V$

$7.8 V$

$6.6 V$

$I = \frac{Δ ε}{r _{1} + r _{2}} = \frac{9 - 7}{3 + 7}$
$= \frac{2}{10} = 0.2 A$
Potential difference across cell $ε_{1}$ is
$= 9 - 0.2 \times 3$
$= 9 - 0.6 = 8.4 V$
Potential difference across $ε_{2}$ :
$V_{A B} = ε_{2} + 0.2 r_{2} = 7 + 0.2 \times 7 = 7$
$= 7 + 1.4 = 8.4 V$

Q. Two cells ε1​ and ε2​ connected in opposition to each other as shown in figure. The cell ε1​ is of emf 9V and internal resistance 3Ω the cell ε2​ is of emf 7V and internal resistance 7Ω. The potential difference between the points A and B is

8.4V

5.6V

7.8V

6.6V