question_answer

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

Correct Answer: B

Solution :

 \[1\,\Omega ,\,2\,\Omega \]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 \[{{E}_{diff}}=\frac{\frac{{{E}_{1}}}{{{r}_{1}}}+\frac{{{E}_{2}}}{{{r}_{2}}}+\frac{{{E}_{3}}}{{{r}_{3}}}}{1/r}=\frac{\frac{1}{1}+\frac{2}{2}+\frac{3}{1}}{5/2}\] \[=\frac{\frac{2+2+6}{2}}{5/2}=\frac{10/2}{5/2}=\frac{5}{5}\times 2=2v\]