A $dc$ source of $emf \,E_1 = 100\,V$ and internal resistance $r = 0.5\,\Omega ,$ 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 ? ................ $\Omega$
JEE MAIN 2013, Medium
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$\frac{100}{R+r}=\frac{90}{R}$
$\Rightarrow \frac{R+r}{R}=\frac{10}{9}$
$\Rightarrow \quad 1+\frac{0.5}{\mathrm{R}}=\frac{10}{9}$
$\Rightarrow \quad \frac{0.5}{\mathrm{R}}=\frac{1}{9}$
$\therefore R=4.5 \,\Omega$
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