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Question

Physics Question on Resistance

The maximum power dissipated in an external resistance R, when connected to a cell of emf E and internal resistance r, will be

A

E2r\frac {E^2}{r}

B

E22r\frac {E^2}{2r}

C

E23r\frac {E^2}{3r}

D

E24r\frac {E^2}{4r}

Answer

E24r\frac {E^2}{4r}

Explanation

Solution

Current         i=Er+R \, \, \, \, \, \, \, \, i = \frac {E}{r + R}
Power         p=i2R \, \, \, \, \, \, \, \, p = i^2 R
⇒    p=E2R(r+R)2\Rightarrow \, \, \, \, p = \frac {E^2R}{(r + R)^2}
Power will be maximum when r = R
pmax=E24rp_{max} = \frac {E^2}{4r}