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.

Answer:

            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)] \[\Rightarrow \]   \[\frac{{{E}_{eq}}}{2r}=\frac{2E}{2r}+\frac{E}{r}=\frac{2E}{r}\] \[\Rightarrow \]   \[{{E}_{eq}}=\frac{4E}{3}\]