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Asked in GUJCET 2026 · Maximum power transfer
Given: E = 6.0 V, r = 0.2 Ω.
Idea: the power drawn from the battery is the rate at which its emf does work, P = EI. It is greatest when the current is greatest.
The current is greatest when the terminals are joined by a wire of negligible resistance: Iₘₐₓ = E/r = (6.0)/(0.2) = 30 A.
Pₘₐₓ = EIₘₐₓ = (E²)/r = 6.0×30 = 180 W, all of it dissipated inside the battery.
Note: the largest power an external resistor can take is (E²)/(4r) = 45 W, at R = r. That is a different quantity, and the paper does not offer it.
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