A battery of emf E and internal resistance r is connected across a resistance R. Resistance R can be adjusted to any value greater than or equal to zero. A graph is plotted between the current passing through the resistance (I) and potential difference(V) across it. Select the correct alternatives

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Emf of the battery is 10V

V-I graph can never be a straight lines as shown in figure

Internal resistance of the battery is 5 $Ω$

Maximum current that can be taken from the battery is 2A

answer is A, B, C.

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Detailed Solution

Potential difference across resistance is equal to the potential difference across the terminals of the battery. So V=E-Ir. This is an equation of a straight line. Comparing this with the given graph, we can see that E=10V and r=5 $Ω$ . Also
$I_{max} = \frac{E}{r} = \frac{10}{5} = 2 A$ When external resistance R=0.

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