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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:

 

  • Option 1)

    R=0.001 r

  • Option 2)

    R=1000 r

  • Option 3)

    R=2r

  • Option 4)

    R=r

 

Answers (1)

best_answer

i=\frac{E}{R+r} 

   

P across R=i^{2}R=\frac{E^{2}}{\left ( R+r \right )^{2}}R

For maximum= \frac{dp}{dR}=0

\frac{E^{2}\left ( R+r \right )^{2}-2\left ( R+r \right )E^{2}R}{\left ( R+r \right )^{4}}=0

E^{2}\left (\left ( R+r \right )^{2}-2\left ( R+r \right )R \right )=0

R=r

 


Option 1)

R=0.001 r

Option 2)

R=1000 r

Option 3)

R=2r

Option 4)

R=r

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