Current Electricity: Tricky Questions Solved (11)

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Question

A battery of EMF 12 V12\text{ V} and internal resistance 1 Ω1\,\Omega is connected to an external resistance RR. (a) Find RR for which the power delivered to the external circuit is maximum. (b) Find this maximum power. (c) Compute the efficiency of the battery at maximum power transfer.

Solution — Step by Step

Current in the circuit: I=εR+rI = \dfrac{\varepsilon}{R + r}. Power delivered to external RR:

P=I2R=ε2R(R+r)2P = I^2 R = \frac{\varepsilon^2 R}{(R + r)^2}

Differentiate PP with respect to RR and set to zero:

dPdR=ε2⋅(R+r)2−R⋅2(R+r)(R+r)4=0\frac{dP}{dR} = \varepsilon^2 \cdot \frac{(R+r)^2 - R\cdot 2(R+r)}{(R+r)^4} = 0

Numerator: (R+r)−2R=0  ⟹  R=r(R+r) - 2R = 0 \implies R = r. So R=1 ΩR = 1\,\Omega.

Plug R=r=1 ΩR = r = 1\,\Omega back:

Pmax⁡=122×1(1+1)2=1444=36 WP_{\max} = \frac{12^2 \times 1}{(1+1)^2} = \frac{144}{4} = 36\text{ W}

Total power from battery =εI=12×6=72 W= \varepsilon I = 12 \times 6 = 72\text{ W} (where I=12/2=6 AI = 12/2 = 6\text{ A}). Efficiency:

η=PexternalPtotal=3672=50%\eta = \frac{P_{\text{external}}}{P_{\text{total}}} = \frac{36}{72} = 50\%

R=1 ΩR = 1\,\Omega, Pmax⁡=36 WP_{\max} = 36\text{ W}, η=50%\eta = 50\%.

Why This Works

The maximum power transfer theorem says external load power peaks when load resistance equals source resistance. The catch — efficiency is only 50% at this point. Half the energy is wasted heating the battery.

That is why power systems do NOT operate at maximum power transfer; they aim for high efficiency (load ≫\gg internal resistance), accepting less peak power per unit voltage.

Alternative Method

AM-GM inequality avoids calculus. Write P=ε2/(R+r)2/R=ε2/(R+r2/R+2r)P = \varepsilon^2 / (R + r)^2 / R = \varepsilon^2 / (R + r^2/R + 2r). Minimise R+r2/RR + r^2/R, which by AM-GM is minimum when R=r2/RR = r^2/R, i.e. R=rR = r. Same answer, faster for objective papers.

Common Mistake

Confusing “maximum power” with “maximum efficiency.” For maximum efficiency, R→∞R \to \infty (negligible loss in rr), but then current and power both go to zero. JEE Main 2023 had a four-option question where one trap option used the "R≫rR \gg r" condition — students who memorised “high RR = good” picked the wrong choice.

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