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Question

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Units below are correct. b, c, e and g are correct too. Please don't answer unless you're 100% sure your answer is correct. I only have one attempt. Thanks

In the figure the ideal batteries have emfs epsilont = 21.4 V, epsilon2 = 9.48 V, and epsilon3 = 4.80 V, and the resistances are each 2.40 Ohm. What are the (a) size and (b) direction (left or right) of current i1? (c) Does battery 1 supply or absorb energy, and (d) what is its power? (e) Does battery 2 supply or absorb energy, and (f) what is its power? (g) Does battery 3 supply or absorb energy, and (h) what is its power?

Explanation / Answer

At the Bottom of the figure Two resistors are in parallel

R' =2.4*2.4/(2.4+2.4)

R'=1.2 ohms

this is in series with another resistance R=2.4 ohms

R" =1.2+2.4 =3.6 ohms

Applying KVL

E3+E1-E2+I1*R"=0

4.8+21.4-9.48-I1*3.6=0

I1=4.644 A

so

a)

I1= 4.644 A

b)

Right

c)

Since current I1 flowing from battery E1 is positive ,so Battery E1 is supplying Energy

d)

P1=E1*I1 =21.4*4.644

P1=99.39 Watts

e)

The equivalent resistance which is parallel to E2 is

R2' =2.4*2.4/(2.4+2.4) =1.2 ohms

I' =E2/R2' =9.48/1.2 =7.9 A

so current from Batter E2 is

I2 =I"-I1 =7.9-4.644

I2=3.256 A

so current from battery 2 is positive ,it is supplying energy

f)

P2=E2*I2=9.48*3.256

P2=30.87 Watts

g)

the equivalent resistance which is parallel to E3 is

1/R3 =1/2.4 +1/2.4 +1/(2.4+2.4)

R3=0.96 ohms

I" =E3/R3=4.8/0.06

I"=5 A

so current from battery E3 is

I3 =5+4.644

I3=9.644 A

so Current from Batter 3 is positive ,it is supplying energy

P3=E3*I3=4.8*9.644

P3=46.29 Watts

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