Question

Two batteries A and B each of emf 2 volt are connected in series to external resistance $R=1\Omega$. Internal resistance of A and B are $1.9\Omega$ and $0.9\Omega$ respectively. What is the potential difference between the terminals of battery A ?

• A. 2 V
• B. 3.8 V
• C. 0
• D. None of these

Answer 1

The correct option is C 0

$i=\frac{E_1+E_2}{R+r_1+r_2}=\frac{2+2}{1+1.9+0.9}=\frac{4}{3.8}$. Hence $V_A=E_A-i{r}_A=2-\frac{4}{3.8}1.9=0$