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Asked in GUJCET 2020 · Maximum power transfer
Given: E = 12 V, r = 0.4 Ω.
Idea: the power drawn from the battery is the rate at which its emf does work, P = EI. It is largest when the current is largest.
The largest current flows when the terminals are joined by a wire of negligible resistance (R = 0): Iₘₐₓ = E/r = (12)/(0.4) = 30 A.
Pₘₐₓ = EIₘₐₓ = (E²)/r = 12×30 = 360 W, all of it turned into heat inside the battery.
Note: the most an external resistor can receive is (E²)/(4r) = 90 W, at R = r. That is a different quantity, and it is not offered.
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