If a battery of emf E and internal resistance r is connected across a load of resistance R. Shot that the rate at which energy is dissipated in R is maximum when R = r and this maximur power is P = E^2//4r.
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Answer:
Total resistance of the circuit R
eq
=R+r
Thus current in the circuit I=
R
eq
E
=
R+r
E
Power dissipated P=I
2
R=
(R+r)
2
E
2
R
For maxima,
dR
dP
=0
∴ E
2
[
(R+r)
4
(R+r)
2
×1−R(2)(R+r)×1
]=0
OR
(R+r)
3
E
2
×(r−R)=0
∴ r−R=0 ⟹R=r
Maximum power dissipated P
max
=
(r+r)
2
E
2
r
=
4r
E
2
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