Maximum Power Transfer

Power delivered to an external load by a real cell is (P = I^2 R = \frac{\varepsilon^2 R}{(R + r)^2}), and differentiating with respect to (R) and setting the derivative to zero shows the maximum occurs when the external resistance equals the internal resistance, (R = r). This is the maximum power transfer theorem. With (r = 3;\Omega), the optimal load is (R = 3;\Omega). The maximum power is then (P_{max} = \frac{\varepsilon^2}{4r} = \frac{6^2}{4 \times 3} = \frac{36}{12} = 3) W. The option 6 (\Omega), 3 W gives the right power but the wrong matching resistance. The option 3 (\Omega), 6 W uses (\varepsilon^2/2r) instead of (\varepsilon^2/4r). The option 1.5 (\Omega), 6 W errs in both the matching condition and the power. This is the standard JEE application of impedance matching in a DC source. A plausibility check confirms it: at (R = r) the current is (I = 6/6 = 1) A, so (P = I^2 R = 1^2 \times 3 = 3) W, exactly matching the maximum-power formula. It is instructive that at this matched condition the efficiency of transfer is only fifty percent, since equal power is simultaneously wasted inside the source, so maximum power delivered and maximum efficiency are distinct goals, and practical power systems usually favour the latter by keeping the load resistance much larger than the source resistance.