Answers includedSolutionCurrent flowing in the circuit Let I

Solution

Current flowing in the circuit

Let I be the current flowing in the circuit ,from kirchoff\'s loop rule

E₂-I(R₁+R₂) -E₁=0

I=E₂-E₁/(R₁+R₂) =(8-4.5)/(12+R₂)

I=3.5/(12+R₂)

SInce

P₁=I²R₁

=>0.75 = I²*12

I=0.25 A

a)

I=3.5/(12+R₂)

0.25 (12+R₂)=3.5

3+0.25R₂=3.5

R₂ =0.5/0.25 =2 ohms

b)

Energy dissipated in R₂ is

P₂=I²R₂ =0.25²*2

P₂=0.125 W =0.13 W (approx)

c)

Here Battery 2 supplies energy since E₂>E₁

P₁=E₁I=8*0.25=2 W

d)

Battery 1 consumes energy

P₂=E₂I =4.5*0.25

P₂=1.125 W =1.1 W