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

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(C) For two cells connected in parallel with $emfs$ $E_1, E_2$ and internal resistances $r_1, r_2$,the equivalent $emf$ $(E_{eq})$ is given by the for

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$\frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$ and $\frac{r_1 r_2}{r_1 + r_2}$

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(C) For two cells connected in parallel with $emfs$ $E_1, E_2$ and internal resistances $r_1, r_2$,the equivalent $emf$ $(E_{eq})$ is given by the formula:$E_{eq} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{\frac{E_1 r_2 + E_2 r_1}{r_1 r_2}}{\frac{r_1 + r_2}{r_1 r_2}} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$The equivalent internal resistance $(r_{eq})$ for parallel connection is given by:$\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} = \frac{r_1 + r_2}{r_1 r_2}$$r_{eq} = \frac{r_1 r_2}{r_1 + r_2}$Thus,the equivalent $emf$ is $\frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$ and the equivalent internal resistance is $\frac{r_1 r_2}{r_1 + r_2}$.

Two cells of $emfs$ $E_1$ and $E_2$ and internal resistances $r_1$ and $r_2$ are connected in parallel. The $emf$ and internal resistance of the equivalent source is

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