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(B) The terminal potential difference $V$ of a cell is given by the relation $V = E - Ir$, where $E$ is the $e.m.f.$ and $r$ is the internal resistanc

Vedclass · Accepted Answer

$2\,V$, $0.4\,\Omega$

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(B) The terminal potential difference $V$ of a cell is given by the relation $V = E - Ir$, where $E$ is the $e.m.f.$ and $r$ is the internal resistance.Comparing this with the equation of a straight line $y = mx + c$, we get $V = -rI + E$.From the graph, when current $I = 0$, the potential difference $V = 2\,V$. This intercept on the $V$-axis represents the $e.m.f.$ of the cell, so $E = 2\,V$.The slope of the $V-I$ graph is equal to $-r$. The magnitude of the slope is $\frac{	ext{change in } V}{	ext{change in } I} = \frac{2 - 0}{5 - 0} = \frac{2}{5} = 0.4\,\Omega$.Therefore, the internal resistance $r = 0.4\,\Omega$.Thus, the $e.m.f.$ is $2\,V$ and the internal resistance is $0.4\,\Omega$.

For a cell, the graph between the potential difference $(V)$ across the terminals of the cell and the current $(I)$ drawn from the cell is shown in the figure. The $e.m.f.$ and the internal resistance of the cell are

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