Question 182 of 380

The combination of two identical cells, whether connected in series or parallel combination provides the same current through an external resistance of $2\Omega$. The value of internal resistance of each cell is

A$2\Omega$B$4\Omega$C$6\Omega$D$8\Omega$

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Answer: (A) $2\Omega$

When the cells are connected in series ${i}_{s}=\dfrac{2\epsilon }{2r+R}=\dfrac{2\epsilon }{2r+2}$

When they are connected in parallel, the emf is ${\epsilon }_{p}=\dfrac{\dfrac{\epsilon }{r}+\dfrac{\epsilon }{r}}{\dfrac{1}{r}+\dfrac{1}{r}}=\epsilon$ and the effective internal resistance will be ${r}_{p}=\dfrac{r\times r}{r+r}=\dfrac{r}{2}$. Therefore, the current will be,
${i}_{p}=\dfrac{\epsilon }{R+\dfrac{r}{2}}=\dfrac{\epsilon }{2+\dfrac{r}{2}}=\dfrac{2\epsilon }{4+r}$
Since, ${i}_{s}={i}_{p} \Rightarrow \dfrac{2\epsilon }{2r+2}=\dfrac{2\epsilon }{4+r}\Rightarrow r=2\Omega$