Two batteries of emf varepsilon_1 and varepsi | vedclass

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(D) For two batteries connected in parallel,the equivalent emf $\varepsilon_{eq}$ and equivalent internal resistance $r_{eq}$ are given by the formula

Vedclass · Accepted Answer

The equivalent emf $\varepsilon_{eq}$ of the two cells is between $\varepsilon_1$ and $\varepsilon_2$,i.e.,$\varepsilon_1 < \varepsilon_{eq} < \varepsilon_2$.

Vedclass · Answer

(D) For two batteries connected in parallel,the equivalent emf $\varepsilon_{eq}$ and equivalent internal resistance $r_{eq}$ are given by the formulas:$\frac{\varepsilon_{eq}}{r_{eq}} = \frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 r_2}$and$\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} \implies r_{eq} = \frac{r_1 r_2}{r_1 + r_2}$Substituting $r_{eq}$ into the first equation:$\varepsilon_{eq} = \left( \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 r_2} ight) 	imes \left( \frac{r_1 r_2}{r_1 + r_2} ight)$$\varepsilon_{eq} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2}$Since $\varepsilon_2 > \varepsilon_1$,the equivalent emf $\varepsilon_{eq}$ is a weighted average of $\varepsilon_1$ and $\varepsilon_2$,which implies that $\varepsilon_1 < \varepsilon_{eq} < \varepsilon_2$.

Two batteries of emf $\varepsilon_1$ and $\varepsilon_2$ (where $\varepsilon_2 > \varepsilon_1$) and internal resistances $r_1$ and $r_2$ respectively are connected in parallel as shown.

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