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# Stuck here, help me understand: 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:

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

Views

$i=\frac{E}{R+r}$

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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