1. Use superposition to calculate the magnitudes of all currents in the circuit shown below (show all work):
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?
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.
what do you mean by "Therefore i2+i3 must equal i1 unless electrons are evaporating."
the question said to find i1, i2 and i3 and that it.
I get i2 = 0.001418
in the junction at the top middle, the total current in must equal the total current out.
i1 = i2 + i3
like period
That is the easy way to check your arithmetic.
what I did (not your way)
i3 = i1 - i2
10 = 3300 i1 + 680 i3
5 = 680 i3 - 1000 i1
three equations, three unknowns
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
can you simplify what you are saying because i am so lost and confused.
To calculate the magnitudes of all currents in the circuit, the correct approach would be to use the superposition principle.
First, disconnect the voltage source Vs1 and calculate the current I1 through the 3.3kΩ resistor due to Vs2.
1. Calculate the voltage at the top node (v1') due to the current source Is:
v1' = Is * (r1 || r2)
v1' = 5V * (1kΩ || 680Ω)
v1' = 5V * (1kΩ * 680Ω) / (1kΩ + 680Ω)
v1' = 2.585V
2. Calculate the voltage at the top node (v1'') due to the voltage source Vs2:
v1'' = Vs2 * (r1 || r2) / (r1 + r2 || r3)
v1'' = 5V * (1kΩ * 680Ω) / (1kΩ + 680Ω || 3.3kΩ)
v1'' = 5V * (1kΩ * 680Ω) / (1kΩ + 680Ω || (3.3kΩ * (1kΩ * 680Ω) / (1kΩ + 680Ω)))
v1'' = 2.585V
3. Calculate the voltage at the top node (v1) due to both Vs1 and Vs2:
v1 = v1' + v1'' = 2.585V + 2.585V = 5.17V
4. Calculate the current I1 through the 3.3kΩ resistor:
I1 = v1 / r1 = 5.17V / 3.3kΩ = 1.566mA
Next, disconnect the voltage source Vs2 and calculate the current I2 through the 1kΩ resistor due to Vs1.
1. Calculate the voltage at the top node (v2') due to the current source Is:
v2' = Is * (r1 || r2)
v2' = 10V * (1kΩ || 680Ω)
v2' = 10V * (1kΩ * 680Ω) / (1kΩ + 680Ω)
v2' = 5.236V
2. Calculate the voltage at the top node (v2'') due to the voltage source Vs1:
v2'' = Vs1 * (r1 || r2) / (r1 + r2 || r3)
v2'' = 10V * (1kΩ * 680Ω) / (1kΩ + 680Ω || 3.3kΩ)
v2'' = 10V * (1kΩ * 680Ω) / (1kΩ + 680Ω || (3.3kΩ * (1kΩ * 680Ω) / (1kΩ + 680Ω)))
v2'' = 5.236V
3. Calculate the voltage at the top node (v2) due to both Vs1 and Vs2:
v2 = v2' + v2'' = 5.236V + 5.236V = 10.472V
4. Calculate the current I2 through the 1kΩ resistor:
I2 = v2 / r2 = 10.472V / 1kΩ = 10.472mA
Finally, calculate the current I3 through the 680Ω resistor using the total voltage V = v1 + v2:
V = v1 + v2 = 5.17V + 10.472V = 15.642V
I3 = V / r3 = 15.642V / 680Ω = 22.98mA
Correcting your calculations:
v1 = 2.585V (not 2.3V)
I1 = 1.566mA (not 1.05A)
I2 = 10.472mA (not 0.003461A)
I3 = 22.98mA (not 0.00509A)
Therefore, based on the correct calculations, the current values you obtain would be:
I1 = 1.566mA
I2 = 10.472mA
I3 = 22.98mA