COMEDK Exam 2022 Solved Paper

Given, emf of cell, E=2V

Load resistance, RL=1.5Ω
Current flowing the circuit is
I=

E

R+r

Power transferered to the load, R,
P=I2R=(

E

R+r

)2.R=

E2R

(R+r)2

Power P will br maximum, if

dP

dR

=0
⇒

d

dR

E2R

(R+r)2

=0
⇒

E2(R+r)2−E2.R.2(R+r)

(R+r)3

=0
⇒ E2(R+r)[R+r−2R]=0
⇒ E2(R+r)(r−R)=0
⇒ r−R=0
⇒ r=R
∴ Maximum power, P=

E2R

(R+r)2

=

E2R

(R+R)2

=

E2

4R

⇒ P=

22

4

×1.5=

1

3

=0.33W