Question

A circuit consists of three batteries of emf $E_1=1V,E_2=2V$ and $E_3=3V$ 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 $Q$
$E_{\text{diff }}=\frac{\frac{E_1}{r_1}+\frac{E_2}{r_2}+\frac{E_3}{r_3}}{1/r}=\frac{\frac{1}{1}+\frac{2}{2}+\frac{3}{1}}{5/2}$
$=\frac{\frac{2+2+6}{2}}{5/2}=\frac{10/2}{5/2}=\frac{5}{5}\times 2=2V$