Question

In the figure ε_{1} = 4.47 V, ε_{2} = 1.05 V,
*R*_{1} = 5.46 Ω, *R*_{2} = 2.85 Ω,
*R*_{3} = 3.27 Ω, and both batteries are ideal. What
is the rate at which energy is dissipated in **(a)**
*R*_{1}, **(b)**
*R*_{2}, and **(c)**
*R*_{3}? What is the power of **(d)**
battery 1 and **(e)** battery 2?

Answer #1

Let I_{1} be current from E_{1} and
I_{2} be current from E_{2}.

Applying KVL on left loop

E_{1}-I_{1}R_{1}-(I_{1}+I_{2})R_{3}=0

8.73I_{1}+3.27I_{2}=4.47----------------1

Applying KVL on right loop

E_{2}-I_{2}R_{2}-(I_{1}+I_{2})R_{3}=0

3.27I_{1}+6.12I_{2}=1.05----------2

Solving 1 and 2 we get

I_{1}=0.5598 A

I_{2}=-0.12754 A

a)

Power dissipated in R1 is

P_{1}=I_{1}^{2}R_{1}=0.5598^{2}*5.46=1.711
Watts

b)

Power dissipated in R2 is

P_{2}=I_{2}^{2}R_{2}=(-0.12754)^{2}*2.85
=0.04636 Watts

c)

Power dissipated in R3 is

P_{3}=(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

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