Current Electricity Ques 45
- Find the $\operatorname{emf}(V)$ and internal resistance $(r)$ of a single battery which is equivalent to a parallel combination of two batteries of emfs $V _1$ and $V _2$ and internal resistances $r _1$ and $r _2$ respectively, with polarities as shown in figure
(1997C, 5M)
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Answer:
Correct Answer: 45.$( V=\frac{V _1 r _2-V _2 r _1}{r _1+r _2} ,r=\frac{r _1 r _2}{r _1+r _2} )$
Solution:
Formula:
- (a) Equivalent emf $(\boldsymbol{V})$ of the battery
PD across the terminals of the battery is equal to its emf when current drawn from the battery is zero. In the given circuit,
Current in the internal circuit,
$ i=\frac{\text { Net emf }}{\text { Total resistance }}=\frac{V _1+V _2}{r _1+r _2} $
Therefore, potential difference between $A$ and $B$ would be
$ \begin{aligned} & V _A-V _B=V _1-i r _1 \\ \therefore \quad & V _A-V _B=V _1-(\frac{V _1+V _2}{r _1+r _2} )r _1=\frac{V _1 r _2-V _2 r _1}{r _1+r _2} \end{aligned} $
So, the equivalent emf of the battery is
$ V=\frac{V _1 r _2-V _2 r _1}{r _1+r _2} $
Note that if $V _1 r _2=V _2 r _1: V=0$.
If $V _1 r _2>V _2 r _1: V _A-V _B=$ Positive i.e. $A$ side of the equivalent battery will become the positive terminal and vice-versa.
(b) Internal resistance ( $r$ ) of the battery
$r _1$ and $r _2$ are in parallel. Therefore, the internal resistance $r$ will be given by
$ 1 / r=1 / r _1+1 / r _2 \text { or } r=\frac{r _1 r _2}{r _1+r _2} $
BETA
