Emf of the battery, E = 12 V
Internal resistance of the battery, r = 0.4 Ω
Maximum current drawn from the battery = I
According to Ohm’s law,
E = Ir
I = E / r
= 12 / 0.4
= 30 A
The maximum current drawn from the given battery is 30 A.
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Emf of the battery, E = 12 V Internal resistance of the battery, r =0.4 Ω Maximum current drawn from the battery = I According to Ohmâs law, E = Ir,I = E / r = 12 / 0.4 = 30 A
E=12V R=0.4Ω I=? According to ohm's law E=IR => I=E/R => 12/0.4 => 30A The max current that can be drawn from the battery is 30A
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