Question: Q. 3. 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 :
(a) The equivalent emf $\varepsilon_{e q}$ of the two cells is between $\varepsilon_{1}$ and $\varepsilon_{2}$, i.e. $\varepsilon_{1}<\varepsilon_{e q}<\varepsilon_{2}$.
(b) The equivalent emf $\varepsilon_{e q}$ is smaller than $\varepsilon_{1}$.
(c) The $\varepsilon_{e q}$ is given by $\varepsilon_{e q}=\varepsilon_{1}+\varepsilon_{2}$ always.
(d) $\varepsilon_{e q}$ is independent of internal resistances $r_{1}$ and $r_{2}$.
[NCERT Exemplar]
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Solution:
Ans. Correct option : (a)
Explanation: As we know that the equivalent emf in parallel combination of cells is :
$$ \varepsilon_{e q}=\frac{\left(\varepsilon_{1} r_{2}+\varepsilon_{2} r_{1}\right)}{\left(r_{1}+r_{2}\right)} $$
so, it is clear that part ’ $c$ ’ and ’ $d$ ’ are incorrect by formula. According to this formula only option (a), is correct.
