For two cells having the same EMF E and inter | vedclass

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(D) When the two cells are connected in series, the total $EMF$ is $2E$ and the total internal resistance is $2r$. The current $i_{1}$ through the ext

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(D) When the two cells are connected in series, the total $EMF$ is $2E$ and the total internal resistance is $2r$. The current $i_{1}$ through the external resistor $R = 6 \ \Omega$ is given by: $i_{1} = \frac{2E}{R + 2r} = \frac{2E}{6 + 2r}$.When the two cells are connected in parallel, the total $EMF$ remains $E$ and the total internal resistance is $\frac{r}{2}$. The current $i_{2}$ through the external resistor $R = 6 \ \Omega$ is given by: $i_{2} = \frac{E}{R + \frac{r}{2}} = \frac{E}{6 + \frac{r}{2}}$.Given that $i_{1} = i_{2}$, we equate the expressions: $\frac{2E}{6 + 2r} = \frac{E}{6 + \frac{r}{2}}$.Dividing both sides by $E$ and simplifying: $\frac{2}{6 + 2r} = \frac{1}{6 + \frac{r}{2}}$.Cross-multiplying: $2(6 + \frac{r}{2}) = 6 + 2r$.$12 + r = 6 + 2r$.Solving for $r$: $r = 12 - 6 = 6 \ \Omega$.

For two cells having the same $EMF$ $E$ and internal resistance $r$, the current passing through an external resistor of $6 \ \Omega$ is the same when both cells are connected either in parallel or in series. The value of internal resistance $r$ is . . . . . . $ \ \Omega$.

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