• # question_answer Two cells of emfs, 2E and E, and internal resistances, 2r and r respectively, are connected in parallel to each other and with an external resistance R. Obtain the expression for the equivalent emf and equivalent internal resistance of this combination.

Suppose, the cell combination is connected with resistance R (say) as shown in figure below. Circuit diagram Applying Kirchhoff's first law at junction B, we get                         $I={{I}_{2}}+{{I}_{2}}$                    ?(i) Applying KVL rule in the loop ABCDA, we get                         $2E-IR-{{I}_{1}}(2r)=0$ $\Rightarrow$               $IR=2E-2{{I}_{1}}r$                ?(ii) Again applying KVL rule in the loop EFCDE, we get $E-IR-{{I}_{2}}r=0$ or         $IR=E-{{I}_{2}}r$                               ?(iii) or         $IR=E-(I-{{I}_{1}})r$               [from Eq. (i)] or         $IR=E-Ir+{{I}_{1}}r$ or         $IR+Ir=E+{{I}_{1}}r$ or         $I(R+r)=E+{{I}_{1}}r$                        ?(iv) Multiplying Eq. (iv) by 2 and adding with Eq. (ii) we get             $2I(R+r)+IR=(2E-2{{I}_{1}}r)+(2E+2{{I}_{1}}r)$ or         $3IR+2Ir=4E$ or         $I(3R+2)=4E$ or         $I=\frac{4E}{3\left( R+\frac{2r}{3} \right)}$ or $I=\frac{\left( \frac{4E}{3} \right)}{R+\frac{2r}{3}}$        ?(v) If equivalent emf and equivalent internal resistance of battery be ${{E}_{eq}}$ and ${{r}_{eq}}$ respectively, then current                         $I=\frac{{{E}_{eq}}}{R+{{r}_{eq}}}$ On comparing Eqs. (v) and (vi), we get Equivalent emf,  ${{E}_{eq}}=\frac{4E}{3}$ Internal resistance, ${{r}_{eq}}=\frac{2r}{3}$ Alternative Method In parallel combination of cells of emf ${{E}_{1}}$ and internal resistance ${{r}_{1}}$ and emf ${{E}_{2}}$ and internal resistance ${{r}_{2}},$ the equivalent emf is given as             $\frac{{{E}_{eq}}}{{{r}_{eq}}}=\frac{{{E}_{1}}}{{{r}_{1}}}+\frac{{{E}_{2}}}{{{r}_{2}}}$ and equivalent resistance is given as $\frac{1}{{{r}_{eq}}}=\frac{1}{{{r}_{1}}}+\frac{1}{{{r}_{2}}}$                               ?(ii) $\Rightarrow$   $\frac{1}{{{r}_{eq}}}=\frac{1}{r}+\frac{1}{2r}=\frac{3}{2r}$ $\Rightarrow$   ${{r}_{eq}}=\frac{2r}{3}$                   [using Eq. (i)] $\Rightarrow$   $\frac{{{E}_{eq}}}{2r}=\frac{2E}{2r}+\frac{E}{r}=\frac{2E}{r}$ $\Rightarrow$   ${{E}_{eq}}=\frac{4E}{3}$
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