Question

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.

Answer 1

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