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  • Two batteries of emf ε_1 and ε_2 (ε_2 > ε_1) and internal resistances r_1 and r_2 respectively are connected in parallel as shown in Figure

Two batteries of emf \varepsilon _{1} and \varepsilon _{2}\left ( \varepsilon_ {2}>\varepsilon _{1} \right )and internal resistances r1 and r2 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 1.
C. The \varepsilon _{eq} is given by \varepsilon _{eq}=1+2always.
D. \varepsilon _{eq} is independent of internal resistances  r1 and r2.

Answers (1)

The cells of emf ε_{1} and ε_{2} with internal resistance r_{1} and r_{2} respectively.

In parallel combination,

the relation between equivalent emf and equivalent resistance is,

\frac{\epsilon_{eq}}{r_{eq}} = \frac{\epsilon_{1}}{r_{1} } + \frac{\epsilon_{2}}{r_{2} }

The resistance in parallel combination is,

r_{eq} = \frac{r_{1}.r_{2} }{r_{1} + r_{2}}

From the above equation,

\epsilon_{eq} = \frac{\epsilon_{1}r_{2}+\epsilon_{2}r_{1} }{r_{1} + r_{2}}

In question,

{\epsilon_{1} > {\epsilon_{2}

Then The equivalent emf εeq of two cells is between ε1 and ε_{2}.

The correct option is A. 

Posted by

infoexpert24

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