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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