question_answer

Two identical non-ideal batteries are connected in parallel. Consider the following statements.

(i) The equivalent emf is smaller than either of the two emfs.

(ii) The equivalent internal resistance is smaller than either of the two internal resistances.

A) Both (i) and (ii) are correct

B) (i) is correct but (ii) is wrong

C) (ii) is correct but (i) is wrong

D) Both (i) and (ii) are wrong

Correct Answer: C

Solution :

(c) (ii) is correct but (i) is wrong Let two cells of emf’s \[{{\varepsilon }_{1}}\]and \[{{\varepsilon }_{2}}\] and of internal resistances \[{{r}_{1}}\] and \[{{r}_{2}}\] respectively are connected in parallel. The equivalent emf is given by             \[{{\varepsilon }_{eq}}=\frac{{{\varepsilon }_{1}}{{r}_{2}}+{{\varepsilon }_{2}}{{r}_{1}}}{{{r}_{1}}+{{r}_{2}}}\]                               …(1) The equivalent internal resistance is given by             \[\frac{1}{{{r}_{eq}}}=\frac{1}{{{r}_{1}}}+\frac{1}{{{r}_{2}}}\] Or         \[{{r}_{eq}}=\frac{{{r}_{1}}{{r}_{2}}}{{{r}_{1}}+{{r}_{2}}}\]                                 …(2) Let us consider, two cells connected in parallel of same emf \[\varepsilon \] and same internal resistance r. From eq. (1), we get             \[{{\varepsilon }_{eq}}=\frac{\varepsilon r+\varepsilon r}{r+r}=\varepsilon \] From eq. (2), we get             \[{{r}_{eq}}=\frac{{{r}^{2}}}{r+r}=\frac{r}{2}\]