Question

Two batteries of emf $\varepsilon_{1}$ and $\varepsilon_{2}\left(\varepsilon_{2}>\varepsilon_{1}\right)$i) and internal resistances $r_1$ and $r_2$ respectively are connected in parallel as shown in figure

• A. The equivalent emf $\varepsilon_{eq}$ of the two cells is between $\varepsilon_{1}$ and $\varepsilon_{2}$, i.e. $\varepsilon_{1} < \varepsilon_{eq} < \varepsilon_{2}$.
• B. The equivalent emf $ \varepsilon_{eq} $ is smaller than $ \varepsilon_{1}$
• C. The $ \varepsilon_{eq} $ is given by $ \varepsilon_{eq} = \varepsilon_{1} +\varepsilon_{2}$ always
• D. $ \varepsilon_{eq} $ is independent of internal resistances $r_1$ and $r_2$

Answer 1

Answer: (A)

A. The equivalent emf $\varepsilon_{ eq }$ of the two cells is between $\varepsilon_{1}$ and $\varepsilon_{2}$,
i.e., $\varepsilon_{1}<\varepsilon_{ eq }<\varepsilon_{2}$
The cells of emf $\varepsilon_{1}$ and $\varepsilon_{2}$ with internal resistance $r_{1}$ and $r_{2}$ respectively.
In parallel combination,
the relation between equivalent emf and equivalent resistance is,
$\frac{\varepsilon_{ eq }}{ r _{ eq }}=\frac{\varepsilon_{1}}{ r _{1}}+\frac{\varepsilon_{2}}{ r _{2}}$
The resistance in parallel combination is,
$r _{ eq }=\frac{ r _{1} \cdot r _{2}}{ r _{1}+ r _{2}}$
From the above equation,
$\varepsilon_{ eq }=\frac{\varepsilon_{1} r _{2}+\varepsilon_{2} r _{1}}{ r _{1}+ r _{2}}$
In question,
$\varepsilon_{2}>\varepsilon_{1}$
Then The equivalent emf $\varepsilon_{ eq }$ of two cell is between $\varepsilon_{1}$ and $\varepsilon_{2}$.