Q190 / 200 · WBJEE 2014 · Physics
A circuit consists of three batteries of emf $E_{1}=1 \mathrm{V}, E_{2}=2 \mathrm{V}$ and $E_{3}=3 \mathrm{V}$ and internal resistances $1 \Omega, 2 \Omega$ and $1 \Omega$ respectively which are connected in parallel as shown in the figure. The potential difference between points $P$ and $Q$ is 
- A$1.0 \mathrm{V}$
- Correct answerB$2.0 \mathrm{V}$
- C$2.2 \mathrm{V}$
- D$3.0 \mathrm{V}$
Solution
$1 \Omega .2 \Omega$ and $1 \Omega$ are in parallel
So, the required internal resistance
$$
\begin{array}{l}
\frac{1}{r}=\frac{1}{r_{1}}+\frac{1}{r-{2}}+\frac{1}{r-{3}} \\
\frac{1}{r}=\frac{1}{1}+\frac{1}{2}+\frac{1}{1} \\
\frac{1}{r}=\frac{2+1+2}{2} \Rightarrow r=\frac{2}{5} \Omega
\end{array}
$$
The potential difference between points $P$ and $Q$
$\begin{aligned} E_{\text {dif }} &=\frac{\frac{E_{1}}{r_{1}}+\frac{E_{2}}{5_{2}}+\frac{E_{3}}{5_{3}}}{1 / r}=\frac{1}{1}+\frac{2}{2}+\frac{3}{1} \\ &=\frac{2+2+6}{5 / 2} \\=& \frac{2}{5 / 2}=\frac{10 / 2}{5 / 2}=\frac{5}{5} \times 2=2 \mathrm{V} \end{aligned}$
So, the required internal resistance
$$
\begin{array}{l}
\frac{1}{r}=\frac{1}{r_{1}}+\frac{1}{r-{2}}+\frac{1}{r-{3}} \\
\frac{1}{r}=\frac{1}{1}+\frac{1}{2}+\frac{1}{1} \\
\frac{1}{r}=\frac{2+1+2}{2} \Rightarrow r=\frac{2}{5} \Omega
\end{array}
$$
The potential difference between points $P$ and $Q$
$\begin{aligned} E_{\text {dif }} &=\frac{\frac{E_{1}}{r_{1}}+\frac{E_{2}}{5_{2}}+\frac{E_{3}}{5_{3}}}{1 / r}=\frac{1}{1}+\frac{2}{2}+\frac{3}{1} \\ &=\frac{2+2+6}{5 / 2} \\=& \frac{2}{5 / 2}=\frac{10 / 2}{5 / 2}=\frac{5}{5} \times 2=2 \mathrm{V} \end{aligned}$Hence, the correct answer is option: 'B'.