A cell of emf $E$ is connected across a resistance $R$ . The potential difference across the terminals of the cell is $V$ . What is the internal resistance of the cell?

E - \frac{V}{R}

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\frac{E - V}{V}

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\frac{E - V}{R} \times V

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\frac{E - V}{V} \times R

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Solution

The correct option is D $\frac{E - V}{V} \times R$
We know
$E = I (R + r) . . . . . . (1)$
$V = I R . . . . . . . (2)$
Dividing 1 by 2 we get,
$\frac{E}{V} = \frac{I (R + r)}{I R}$
$= \frac{R + r}{R}$
$= 1 + \frac{r}{R}$
$\frac{E}{V} - 1 = \frac{r}{R}$
$r = R (\frac{E - V}{V})$

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A cell of emf E is connected across a resistance R. The potential difference across the terminals of the cell is V. What is the internal resistance of the cell?

The correct option is D E−VV×RWe knowE=I(R+r)......(1)V=IR.......(2)Dividing 1 by 2 we get,EV=I(R+r)IR =R+rR =1+rREV−1=rRr=R(E−VV)

A cell of emf $E$ is connected across a resistance $R$ . The potential difference across the terminals of the cell is $V$ . What is the internal resistance of the cell?

The correct option is D $\frac{E - V}{V} \times R$
We know
$E = I (R + r) . . . . . . (1)$
$V = I R . . . . . . . (2)$
Dividing 1 by 2 we get,
$\frac{E}{V} = \frac{I (R + r)}{I R}$
$= \frac{R + r}{R}$
$= 1 + \frac{r}{R}$
$\frac{E}{V} - 1 = \frac{r}{R}$
$r = R (\frac{E - V}{V})$