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A dc source of emf $E_{1} = 100$ V and internal resistance r = 0.5 $Ω$ , a storage battery of emf $E_{2} = 90$ V and an external resistance R are connected as shown in figure. For what value of R no current will pass through the battery ?

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$4 .5 Ω$

$3 .5 Ω$

$5.5 Ω$

$2 .5 Ω$

answer is C.

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Detailed Solution

To have no current in battery,

$\begin{matrix} \frac{ε}{V} = \frac{R + γ}{R} \\ \frac{10}{90} = \frac{R + 0.5}{R} \\ \frac{10}{9} = \frac{R + 0.5}{R} \\ 10 R = 9 R + 4.5 \\ R = 4.5 Ω . \end{matrix}$

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