| A dc source of emf \[{{E}_{1}}=100\,V\] and internal resistance \[\operatorname{r}=0.5\Omega ,\] a storage battery of emf \[{{\operatorname{E}}_{2}}=90\operatorname{V}\] and an external resistance R are connected as shown in figure. For what value of R no current will pass through the battery? [JEE ONLINE 22-04-2013] |
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A) \[5.5\,\,\Omega \]
B) \[3.5\,\,\Omega \]
C) \[4.5\,\,\Omega \]
D) \[2.5\,\,\Omega \]
Correct Answer: C
Solution :
| [c] |
| \[\frac{100}{R+r}=\frac{90}{R}\] |
| \[\Rightarrow \] \[\frac{R+r}{R}=\frac{10}{9}\] |
| \[\Rightarrow \] \[1+\frac{0.5}{R}=\frac{10}{9}\] |
| \[\Rightarrow \] \[\frac{0.5}{R}=\frac{1}{9}\] |
| \[\therefore \] \[R=4.5\,\Omega \] |
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