Question

A circuit consists of three batteries of emf $E_{1}=1 \,V , E_{2}=2 \,V$ and $E_{3}=3 \,V$ 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 V
• B. 2.0 V
• C. 2.2 V
• D. 3.0 V

Answer 1

Answer: (B)

$1\, \Omega, 2 \,\Omega$ and $1 \,\Omega$ are in parallel

So, the required internal resistance
$\frac{1}{r}=\frac{1}{r_{1}}+\frac{1}{r_{2}}+\frac{1}{r_{3}}$
$\frac{1}{r}=\frac{1}{1}+\frac{1}{2}+\frac{1}{1}$
$\frac{1}{r}=\frac{2+1+2}{2}$
$ \Rightarrow r=\frac{2}{5} \,\Omega$
The potential difference between points $P$ and $Q$
$E_{\text {diff }} =\frac{\frac{E_{1}}{r_{1}}+\frac{E_{2}}{r_{2}}+\frac{E_{3}}{r_{3}}}{1 / r}=\frac{\frac{1}{1}+\frac{2}{2}+\frac{3}{1}}{5 / 2}$
$= \frac{\frac{2+2+6}{2}}{5 / 2}=\frac{10 / 2}{5 / 2}=\frac{5}{5} \times 2=2\, V$