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Question

Physics Question on Current electricity

A battery of e. m. f. EE and internal resistance rr is connected to an external resistance RR the condition for maximum power transfer is

A

r<Rr < R

B

r>Rr>R

C

r=1/Rr=1/R

D

r=Rr=R

Answer

r=Rr=R

Explanation

Solution

We know that the current in the circuit
I=ER+rI=\frac{E}{R+r}
and power delivered to the resistance RR is
P=I2R=E2R(R+r)2P=I^{2} R=\frac{E^{2} R}{(R+r)^{2}}
It is maximum when dPdR=0\frac{d P}{d R}=0
dPdR=E2[(r+R)22R(r+R)(r+R)4]=0\frac{d P}{d R}=E^{2}\left[\frac{(r+R)^{2}-2 R(r +R)}{(r +R)^{4}}\right]=0
or (r+R)2=2R(r+R)(r+ R)^{2}=2 R(r+ R) or R=rR=r