Ncert Exercises & Solutions

Two batteries of emf $\varepsilon_1$ and $\varepsilon_2$ $(\varepsilon_2 > \varepsilon_1)$ respectively are connected in parallel as shown in the figure.

1.	The equivalent emf $\varepsilon_{eq}$ of the two cells is between $\varepsilon_1$ and $\varepsilon_2$ i.e, $\varepsilon_1<\varepsilon_{e q}<\varepsilon_2$
2.	The equivalent emf $\varepsilon_{eq}$ is smaller than $\varepsilon_1$
3.	The $\varepsilon_{eq}$ is given by $\varepsilon_{e q}=\varepsilon_1+\varepsilon_2$ always
4.	$\varepsilon_{eq}$ is independent of internal resistances $r_1$ and $r_2$

Hint: $\varepsilon_{{eq}}=\frac{\varepsilon_2 {r}_1+\varepsilon_2 {r}_2}{{r}_1+{r}_2}$

Step: Find the relationship between $\varepsilon_1,\varepsilon_{eq}$ and $\varepsilon_2.$
The equivalent emf $\varepsilon_{eq}$ of batteries in a parallel combination is given by;
$\Rightarrow \varepsilon_{{eq}}=\frac{\varepsilon_2 {r}_1+\varepsilon_2 {r}_2}{{r}_1+{r}_2}$
Since $r_1$ and $r_2$ is positive, the value of $\varepsilon_{eq}$ lies between $\varepsilon_1$ and $\varepsilon_2.$
It is also stated that $(\varepsilon_2 > \varepsilon_1)$ then, $\varepsilon_1<\varepsilon_{e q}<\varepsilon_2.$
Hence, option (1) is the correct answer.

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