The battery shown in the figure has an e.m.f. | vedclass

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(D) $1$. For the terminal voltage $V$: The relation is $V = E - Ir$. This is a linear equation of the form $y = mx + c$,where the slope is $-r$ and th

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All of the above are correct

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(D) $1$. For the terminal voltage $V$: The relation is $V = E - Ir$. This is a linear equation of the form $y = mx + c$,where the slope is $-r$ and the intercept is $E$. Thus,the graph of $V$ versus $I$ is a straight line with a negative slope.$2$. For the external power $P$: The power dissipated in the external circuit is $P = VI = (E - Ir)I = EI - I^2r$. This is a downward-opening parabola with respect to $I$,passing through the origin $(0,0)$ and $(E/r, 0)$.$3$. For the fraction $\eta$: The thermal power generated in the cell is $P_{thermal} = I^2r$. The total electrical power generated in the cell is $P_{total} = EI$. The fraction $\eta$ is given by $\eta = \frac{P_{thermal}}{P_{total}} = \frac{I^2r}{EI} = \frac{r}{E}I$. This is a linear equation of the form $y = mx$,representing a straight line passing through the origin.Since all three relationships are correctly described by their respective graphs,the correct option is $(d)$.

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