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A dc source of emf $\mathrm{E}_1=100 \mathrm{~V}$ and internal resistance $r=0.5 \Omega$, a storage battery of emf $\mathrm{E}_2=90 \mathrm{~V}$ and an external resistance $\mathrm{R}$ are connected as shown in figure. For what value of R no current will pass through the battery?
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Verified Answer
The correct answer is:
$4.5 \Omega$
$4.5 \Omega$
$\frac{100}{R+r}=\frac{90}{R}$
$$
\begin{aligned}
& \Rightarrow \quad \frac{\mathrm{R}+\mathrm{r}}{\mathrm{R}}=\frac{10}{9} \\
& \Rightarrow \quad 1+\frac{0.5}{\mathrm{R}}=\frac{10}{9} \\
& \Rightarrow \quad \frac{0.5}{\mathrm{R}}=\frac{1}{9} \\
& \therefore \mathrm{R}=4.5 \Omega
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad \frac{\mathrm{R}+\mathrm{r}}{\mathrm{R}}=\frac{10}{9} \\
& \Rightarrow \quad 1+\frac{0.5}{\mathrm{R}}=\frac{10}{9} \\
& \Rightarrow \quad \frac{0.5}{\mathrm{R}}=\frac{1}{9} \\
& \therefore \mathrm{R}=4.5 \Omega
\end{aligned}
$$
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