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 \[{{E}_{diff}}=\frac{\frac{{{E}_{1}}}{{{r}_{1}}}+\frac{{{E}_{2}}}{{{r}_{2}}}+\frac{{{E}_{3}}}{{{r}_{3}}}}{1/r}=\frac{\frac{1}{1}+\frac{2}{2}+\frac{3}{1}}{5/2}\] \[=\frac{\frac{2+2+6}{2}}{5/2}=\frac{10/2}{5/2}=\frac{5}{5}\times 2=2v\]You need to login to perform this action.
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