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JEE Main & Advanced JEE Main Paper (Held On 22 April 2013)

  • question_answer
    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 Main  Online Paper (Held On 22 April 2013)

    A)  \[5.5\,\,\Omega \]                              

    B)  \[3.5\,\,\Omega \]

    C)  \[4.5\,\,\Omega \]                              

    D)  \[2.5\,\,\Omega \]

    Correct Answer: C

    Solution :

    \[\frac{100}{R+r}=\frac{90}{R}\] \[\Rightarrow \]               \[\frac{R+r}{R}=\frac{10}{9}\] \[\Rightarrow \]               \[1+\frac{0.5}{R}=\frac{10}{9}\] \[\Rightarrow \]               \[\frac{0.5}{R}=\frac{1}{9}\] \[\therefore \]\[R=4.5\,\Omega \]


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