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Question

Physics Question on Current electricity

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

A

R = 1000 r

B

R = 0.001 r

C

R = 2r

D

R = r

Answer

R = r

Explanation

Solution

Current i=Er+Ri = \frac{E}{r + R}
Power generated in R
P=i2RP = i^2 R
P=E2R(r+R)2P = \frac{E^2R}{(r + R)^2}
for maximum power dPdR=0\frac{dP}{dR} = 0
E2[(r+R)2×1R×2(r+R)(r+R)4]=0E^{2} \left[\frac{\left(r+R\right)^{2} \times1-R \times2\left(r+R\right)}{\left(r+R\right)^{4}}\right] = 0
r=R\Rightarrow r =R