Current Electricity

JEE Main level, single correct, medium

198 of 380

Question 198 of 380

A dc source of emf E1=100 V\mathrm{E}_1=100 \mathrm{~V} and internal resistance r=0.5Ωr=0.5 \Omega, a storage battery of emf E2=90 V\mathrm{E}_2=90 \mathrm{~V} and an external resistance R\mathrm{R} are connected as shown in figure. For what value of R no current will pass through the battery?

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Answer: (C) 4.5Ω4.5 \Omega

100R+r=90R\frac{100}{R+r}=\frac{90}{R}
⇒R+rR=109⇒1+0.5R=109⇒0.5R=19∴R=4.5Ω\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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