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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Internal resistance of the battery is $5 Ω$ .

Emf of the battery is 10 V.

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

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

answer is A.

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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 = 10 V and r = 5 $Ω$ . Also

$I_{max} = \frac{E}{r} = \frac{10}{5} = 2 A$

When external resistance R = 0.

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