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JEE Main 2016 · Physics · Work, Energy and Power · easy

A battery of internal resistance r delivers the same power to an external resistance of 2 Ω as to one of 8 Ω. The maximum power the battery can…

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A battery of internal resistance rr delivers the same power to an external resistance of 2 Ω2\ \Omega as to one of 8 Ω8\ \Omega. The maximum power the battery can deliver to any external resistor is 25 W25\ \mathrm{W}. The power it delivers to the 2 Ω2\ \Omega resistor is
  1. A. 25 W25\ \mathrm{W}
  2. B. 20 W20\ \mathrm{W}
  3. C. 22.2 W22.2\ \mathrm{W}
  4. D. 18 W18\ \mathrm{W}
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Answer: (C) 22.2 W22.2\ \mathrm{W}

Power in RR is P=E2R(R+r)2P=\dfrac{E^2R}{(R+r)^2}. Equal power for R1R_1 and R2R_2 requires r2=R1R2r^2=R_1R_2, so r=2×8=4 Ωr=\sqrt{2\times8}=4\ \Omega. The maximum power occurs when R=rR=r and equals E24r=25\dfrac{E^2}{4r}=25, giving E2=400E^2=400 and E=20 VE=20\ \mathrm{V}. Then for the 2 Ω2\ \Omega resistor the current is 202+4=103 A\dfrac{20}{2+4}=\dfrac{10}{3}\ \mathrm{A} and P=(103)2×2=2009≈22.2 WP=\left(\dfrac{10}{3}\right)^2\times2=\dfrac{200}{9}\approx22.2\ \mathrm{W}. The same value is obtained for 8 Ω8\ \Omega: (2012)2×8=22.2 W\left(\dfrac{20}{12}\right)^2\times8=22.2\ \mathrm{W}, confirming the choice r=4 Ωr=4\ \Omega.
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