Question

Two cells $A\text{and}B$ each of $2\text{V}$ are connected in series to an external resistance $R=1\Omega$ as shown in figure. If the internal resistance of $A$ is $r_A=1.9\Omega$ and that of $B$ is $r_B=0.9\Omega$, then the potential difference between the terminals of $A$ is

• A. $1.75\text{V}$
• B. $1.5\text{V}$
• C. $1.25\text{V}$
• D. zero

Answer 1

The correct option is D zero


$I=\frac{\text{Total voltage}}{\text{Total resistance}}$

$\therefore I=\frac{V_A+V_B}{(R+r_A+r_B)}$

Substituting the data given in the question gives,

$I=\frac{2+2}{1+1.9+0.9}=\frac{4}{3.8}\text{A}$

Potential difference of $A$ is , $V_A=E_A-I{r}_A$

$\Rightarrow V_A=2-(\frac{4}{3.8}\times 1.9)=2-2=0\text{V}$