1. Use superposition to calculate the magnitudes of all currents in the circuit shown below (show all work):

Question

h ttps://us-static.z-dn.net/files/dfb/358f6d54eff63b6f60f1d929137beacf.png
there's no space between the h and ttps

I did:
V1= ((r1||r2)/(r1+r2||r3))*Vs1
v1 = (((1kΩ*680Ω)/(1kΩ+680Ω))/(3.3kΩ +((1kΩ*680Ω)/(1kΩ+680Ω)))*10v
v1= 2.3v

v2 = ((r1||r2)/(r1+r2||r3))*Vs2
v2 = (((1kΩ*680Ω)/(1kΩ+680Ω))/(3.3kΩ +((1kΩ*680Ω)/(1kΩ+680Ω)))*5v
v2= 1.161v

v = v1 +v2
v= 2.3v + 1.161v
v=3.461v

I1= v1/r1
I1 = 3.461V / 3.3KΩ
I1 = 1.05A

I2 = 3.461V / 1KΩ
I2 = 0.003461A

I3 = 3.461V / 680Ω
I3 = 0.00509A

Is my work correct?

Answer 1

It is easy to check.
Put those currents in and make sure the voltages check out.
However
i1 goes into that top middle point
i2 and i3 come out of that point
oh!
I1 = 3.461V / 3.3KΩ
I1 = 1.05A YOU FORGOT K = 1000
I think your i1 = 0.0105
still does not quite work
Therefore i2+i3 must equal i1 unless electrons are evaporating.

Answer 2

what do you mean by "Therefore i2+i3 must equal i1 unless electrons are evaporating."

Answer 3

the question said to find i1, i2 and i3 and that it.

Answer 4

I get i2 = 0.001418

Answer 5

in the junction at the top middle, the total current in must equal the total current out.
i1 = i2 + i3
like period

Answer 6

That is the easy way to check your arithmetic.

Answer 7

what I did (not your way)
i3 = i1 - i2
10 = 3300 i1 + 680 i3
5 = 680 i3 - 1000 i1
three equations, three unknowns

Answer 8

10 = 330 i1 +680(i1-i2)
5 = 680 (i1-i2) -1000 i2
so
10 = 1010 i1 -680 i2
5 = 680 i1 -1680 i2 multiply this line by 1.4853 which is 1010/680
then
10 = 1010 i1 - 680 i2
7.426 = 1010 i1 - 2495 i2
---------------------------------------subtract
2.574 = 1815 i2 etc etc etc

Answer 9

can you simplify what you are saying because i am so lost and confused.