Two cells of emfs E_1 and E_2 and internal re | vedclass

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(D) The equivalent emf of two cells connected in parallel is given by the formula:$E_{eq} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \

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$E_1 < E_{eq} < E_2$ and $E_{eq}$ is nearer $E_1$

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(D) The equivalent emf of two cells connected in parallel is given by the formula:$E_{eq} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$This can be rewritten as:$E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2} = \frac{E_1 r_2 + E_1 r_1 - E_1 r_1 + E_2 r_1}{r_1 + r_2} = \frac{E_1(r_1 + r_2) + r_1(E_2 - E_1)}{r_1 + r_2} = E_1 + \frac{r_1}{r_1 + r_2}(E_2 - E_1)$Since $E_2 > E_1$ and $r_1, r_2 > 0$,it is clear that $E_{eq} > E_1$.Similarly,$E_{eq} = E_2 - \frac{r_2}{r_1 + r_2}(E_2 - E_1)$. Since $E_2 > E_1$ and $r_1, r_2 > 0$,it is clear that $E_{eq} Thus,$E_1 Given $r_2 > r_1$,the term $\frac{r_1}{r_1 + r_2} Therefore,$E_{eq}$ is closer to $E_1$ than to $E_2$.

Two cells of emfs $E_1$ and $E_2$ and internal resistances $r_1$ and $r_2$ ($E_2 > E_1$ and $r_2 > r_1$) respectively,are connected in parallel as shown in the figure. The equivalent emf of the combination is $E_{eq}$. Then

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