Maximum Power Transfer

Question

A battery of emf 12 volts and internal resistance 3 ohm drives a variable load resistor, so for what load value is the power delivered to the load maximum?

RankGuru · Accepted Answer

The power delivered to an external load by a source with internal resistance $r$ is maximised when the load resistance equals the internal resistance, a result that follows from analysing $P = I^2 R = \left(\frac{\varepsilon}{R+r}ight)^2 R$ and building on the source concepts in NCERT Class 12, Chapter 3 (Current Electricity). Differentiating this expression with respect to $R$ and setting it to zero yields the maximum power transfer condition $R = r$. Since $r = 3$ ohm, the optimal load is $R = 3$ ohm. The option 6 ohm is wrong because although larger loads give higher terminal voltage, the current falls, reducing delivered power beyond the matched point. The option 0 ohm gives maximum current but zero power in the load, as all power dissipates internally. The option 12 ohm, matching the emf number, has no physical basis for power matching. A plausibility check confirms the answer: at $R = r = 3$ ohm the current is $\frac{12}{6} = 2$ A and the load power is $2^2 	imes 3 = 12$ W, and testing nearby values like 2 ohm or 4 ohm gives slightly less than 12 W, confirming the maximum.

Maximum Power Transfer

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