Question

In the figure ε1 = 4.47 V, ε2 = 1.05 V, R1 = 5.46 Ω, R2...

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?

Homework Answers

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)

PE1=0.5598*4.47=2.502 Watts

e)

PE2=0.12754*1.05=-0.134 Watts

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