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Two batteries of emf $\varepsilon_1$ and $\varepsilon_2\left(\varepsilon_2>\varepsilon_1\right)$ and internal resistances $r_1$ and $r_2$ respectively are connected in parallel as shown in figure.
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Two equivalent emf $\varepsilon_{\text {eq }}$ of the two cells is between $\varepsilon_1$ and $\varepsilon_2$, i.e., $\varepsilon_1 < \varepsilon_{\mathrm{eq}} < \varepsilon_2$
Two equivalent emf $\varepsilon_{\text {eq }}$ of the two cells is between $\varepsilon_1$ and $\varepsilon_2$, i.e., $\varepsilon_1 < \varepsilon_{\mathrm{eq}} < \varepsilon_2$
As we know the equivalent $\operatorname{emf}\left(\varepsilon_{\mathrm{eq}}\right)$ in the parallel combination
$$
\varepsilon_{\text {eq }}=\frac{\varepsilon_2 r_1+\varepsilon_1 r_2}{r_1+r_2}
$$
So according to formula the equivalent emf $\varepsilon_{\mathrm{eq}}$ of the two cells in parallel combination is between $\varepsilon_1$ and $\varepsilon_2$. Thus $\left(\varepsilon_1\right.$ $ < \varepsilon_{\text {eq }} < \varepsilon_2$ ).
$$
\varepsilon_{\text {eq }}=\frac{\varepsilon_2 r_1+\varepsilon_1 r_2}{r_1+r_2}
$$
So according to formula the equivalent emf $\varepsilon_{\mathrm{eq}}$ of the two cells in parallel combination is between $\varepsilon_1$ and $\varepsilon_2$. Thus $\left(\varepsilon_1\right.$ $ < \varepsilon_{\text {eq }} < \varepsilon_2$ ).
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