question_answer

Two identical non-ideal batteries are connected in parallel.

Consider the following statements.

(i) The equivalent e.m.f. is smaller than either of the two e.m.f.

(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 :

Let two cells of emfs \[{{\varepsilon }_{1}}\] and \[{{\varepsilon }_{2}}\] and of internal resistance \[{{r}_{1}}\] and \[{{r}_{2}}\] respectively are connected in parallel.

The equivalent e.m.f. is given by

\[{{\varepsilon }_{eq}}=\frac{{{\varepsilon }_{1}}{{r}_{2}}+{{\varepsilon }_{2}}{{r}_{1}}}{{{r}_{1}}+{{r}_{2}}}\]                               (i)

The equivalent internal resistance is given by

\[\frac{1}{{{r}_{eq}}}=\frac{1}{{{r}_{1}}}+\frac{1}{{{r}_{2}}}\] or \[{{r}_{eq}}=\frac{{{r}_{1}}{{r}_{2}}}{{{r}_{1}}+{{r}_{2}}}\]         (ii)

Let us consider, two cells connected in parallel of same emf and same internal resistance r.

From equation (i), we get \[{{\varepsilon }_{eq}}=\frac{\varepsilon r+\varepsilon r}{r+r}=\varepsilon \]

From equation (ii), we get

\[{{r}_{eq}}=\frac{{{r}^{2}}}{r+r}=\frac{r}{2}\]