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Physics

Combination of Resistances

n identical c...

Question

$n$ identical cells are joined in series with two cells A and B with reversed polarities. Emf of each cell is $E$ and the internal resistance is $r$ . Potential difference across cell A and B is :( $n > 4$ )

\frac{2 E}{n}

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2 E (1 - \frac{1}{n})

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\frac{4 E}{n}

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2 E (1 - \frac{2}{n})

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Solution

The correct option is D $2 E (1 - \frac{2}{n})$
Net EMF = $(n - 2) E - 2 E$
= $(n - 4) E$
$I = \frac{(n - 4) E}{n r}$
Potential difference across A and B = $E + I_{r}$
$= E + \frac{(n - 4) E}{n r} \times r$
$= E [1 + (\frac{n - 4}{n})]$
$= E [\frac{n + n - 4}{n}]$
$= 2 E [\frac{n - 2}{n}]$
$= 2 E [1 - \frac{2}{n}]$

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n identical cells are joined in series with two cells A and B with reversed polarities. Emf of each cell is E and the internal resistance is r. Potential difference across cell A and B is :(n>4)

The correct option is D 2E(1−2n)Net EMF = (n−2)E−2E =(n−4)E I=(n−4)EnrPotential difference across A and B = E+Ir=E+(n−4)Enr×r=E[1+(n−4n)]=E[n+n−4n]=2E[n−2n]=2E[1−2n]

$n$ identical cells are joined in series with two cells A and B with reversed polarities. Emf of each cell is $E$ and the internal resistance is $r$ . Potential difference across cell A and B is :( $n > 4$ )