Question

In the figure ε₁ = 4.47 V, ε₂ = 1.05 V, R₁ = 5.46 Ω, R₂ = 2.85 Ω, R₃ = 3.27 Ω, and both batteries are ideal. What is the rate at which energy is dissipated in (a) R₁, (b) R₂, and (c) R₃? What is the power of (d) battery 1 and (e) battery 2?

Answer 1

Let I₁ be current from E₁ and I₂ be current from E₂.

Applying KVL on left loop

E₁-I₁R₁-(I₁+I₂)R₃=0

8.73I₁+3.27I₂=4.47----------------1

Applying KVL on right loop

E₂-I₂R₂-(I₁+I₂)R₃=0

3.27I₁+6.12I₂=1.05----------2

Solving 1 and 2 we get

I₁=0.5598 A

I₂=-0.12754 A

a)

Power dissipated in R1 is

P₁=I₁²R₁=0.5598²*5.46=1.711 Watts

b)

Power dissipated in R2 is

P₂=I₂²R₂=(-0.12754)²*2.85 =0.04636 Watts

c)

Power dissipated in R3 is

P₃=(0.5598-0.12754)²*3.27=0.611 Watts

d)

P_E1=0.5598*4.47=2.502 Watts

e)

P_E2=0.12754*1.05=-0.134 Watts