Question
Illustrate-combination of cells e.g., three cells, in parallel, explaining the combination briefly. Obtain an expression for current ‘i’ in the combination.
✓
Answer
Combination of cells in parallel: Let there be n cells, each of e.m.f. E and internal resistance r, be connected in parallel as shown in the diagram.
Total e.m.f. across the point A and B is E
Total internal resistance
$\frac{1}{ r _1}=\frac{1}{ r }+\frac{1}{ r }+\frac{1}{ r } \ldots .=\frac{ n }{ r }$ (if there are $n$ cells)
or $r_1=\frac{r}{n}$
But total resistance of circuit $=R+\frac{r}{n}=\frac{n R+r}{n}$
$\therefore$ Current $i =\frac{\text { Total e.m.f }}{\text { Total resistance }}$
$=\frac{E}{R+\frac{r}{n}}$
$=\frac{ nE }{ nR +r}$
It is advantageous to join cells in parallel when the internal resistance of each cell is quite high as compared to the external resistance R.
Then, Approximately, $i =\frac{ nE }{ r }$ (neglecting $R$ ).
Total e.m.f. across the point A and B is E
Total internal resistance
$\frac{1}{ r _1}=\frac{1}{ r }+\frac{1}{ r }+\frac{1}{ r } \ldots .=\frac{ n }{ r }$ (if there are $n$ cells)
or $r_1=\frac{r}{n}$
But total resistance of circuit $=R+\frac{r}{n}=\frac{n R+r}{n}$
$\therefore$ Current $i =\frac{\text { Total e.m.f }}{\text { Total resistance }}$
$=\frac{E}{R+\frac{r}{n}}$
$=\frac{ nE }{ nR +r}$
It is advantageous to join cells in parallel when the internal resistance of each cell is quite high as compared to the external resistance R.
Then, Approximately, $i =\frac{ nE }{ r }$ (neglecting $R$ ).
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