Verified Solution • 2025–26 Session moderate

QJEEM19Ph3c10: JEE Main 2019 Physics - Class 12 JEEM19Ph3c10 Solution: A cell of internal resistance r drives current through an…

A cell of internal resistance r drives current through an external resistance R. The power delivered by the cell to the external resistance will be maximum when :-

Multiple Choice Options

  • A R = 1000 r
  • B R = r
  • C R = 2r
  • D R = 0.001 r

Step-by-Step Verified Solution

Current i = Er+R{E \over {r + R}}

Power generated in R

P = i2R

P=E2R(r+R)2P = {{{E^2}R} \over {{{\left( {r + R} \right)}^2}}}

For maximum power dPdR=0{{dP} \over {dR}} = 0

E2[(r+R)2×1−R×2(r+R)(r+R)4]=0{E^2}\left[ {{{{{\left( {r + R} \right)}^2} \times 1 - R \times 2(r + R)} \over {{{\left( {r + R} \right)}^4}}}} \right] = 0

⇒\Rightarrow r = R or R = r

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