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In the above question it is given that: Emf of the battery, E= 15V Internal resistance of the battery, r = 0.5 Ω Consider the maximum current drawn from the battery to be I. Therefore, using Ohm’s law, E = Ir => I = E/r => I = 15/0.5 => I = 30 A Clearly, the maximum current drawn from the given battery is 30A
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