For two cells in parallel, the equivalent emf $\varepsilon_{eq}$ and e…

Question

For two cells in parallel, the equivalent emf $\varepsilon_{eq}$ and equivalent internal resistance $r_{eq}$ satisfy which compact relation?

Accepted Answer

C. $\dfrac{\varepsilon_{eq}}{r_{eq}} = \dfrac{\varepsilon_1}{r_1} + \dfrac{\varepsilon_2}{r_2}$

1. The parallel combination gives $\varepsilon_{eq} = (\varepsilon_1 r_2 + \varepsilon_2 r_1)/(r_1 + r_2)$.
2. Dividing through by $r_{eq} = r_1 r_2/(r_1 + r_2)$ simplifies the ratio.
3. The result is $\varepsilon_{eq}/r_{eq} = \varepsilon_1/r_1 + \varepsilon_2/r_2$.
4. So C; B is the series emf rule, and A and D do not follow from the derivation.
_Source: NCERT Class 12 Physics Ch 3 "Current Electricity", p.16_

Answer

$\varepsilon_{eq} r_{eq} = \varepsilon_1 r_1 + \varepsilon_2 r_2$

Answer

$\varepsilon_{eq} = \varepsilon_1 + \varepsilon_2$

Answer

$\dfrac{r_{eq}}{\varepsilon_{eq}} = \dfrac{r_1}{\varepsilon_1} + \dfrac{r_2}{\varepsilon_2}$