A dc source of emf $\mathrm{E}_1=100 \mathrm{~V}$ and internal resistance $r=0.5 \Omega$, a storage battery…

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?
  1. $5.5 \Omega$
  2. $3.5 \Omega$
  3. $4.5 \Omega$
  4. $2.5 \Omega$

Solution

$\frac{100}{R+r}=\frac{90}{R}$ $ \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} $

Asked in: JEE Main 2013 (22 Apr Online)

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