# Superposition with AC voltage source and DC current source

• eehelp150
In summary, the superposition principle is used to determine the current i(t) by converting inductors and capacitors to impedance, simplifying the circuit, and calculating the total impedance. The final answer is -1.5 - 0.5j amps. In the time domain, the current is represented as I(t) = sqrt(2)/2 * cos(10t - 45°) - 2.
eehelp150

## Homework Statement

Using the superposition principle, determine the current i(t.

2. Homework Equations

Zc = 1/jwC
Zl = jwL
V = I*R
I = V/Z

## The Attempt at a Solution

First, I converted inductor/capacitor to impedance:
L = 1.5H -> jwL = j * 10 * 1.5 = 15j
C = 10mf = 1/(jwC) = 1/(j*10*10*10^-3) = -0.1j

Then, I transformed the 3A DC current source and 10 ohm resistor into a DC voltage source:

I then shorted the 30V DC source to find i(t) with only Vs(t) = 10cos(10t)V
After shorting the 30V DC source, C2 and R3 are in parallel. Their impedance is:

(0.1j * 10)/(10-0.1j) = -j/(10-0.1j)
After rationalizing: (0.1 - j)/100.01

Now that, L2 and R4 are in series, total impedance:
15j + 5 + (0.1 - j)/100.01 = (500.15-1499.15j)/100.01 = 15.8 < -71.55°.
I = V/Z = (10<0°)/(15.88<-71.55° ) = 10/15.88<71.55°A

Now I short the AC source:
The DC source had a frequency of 0 rad/s, so the impedance of C3 is infinity (1/0) and L3 is 0 (j*0*L).
The inductor is shorted and the capacitor is opened. There is a voltage source of 30V and a combined resistance of 15ohms. V=IR -> I = 30/15 = 2A<0°
[(500.15/100.01 + 2) + (1499.15j/100.01)]
Is 16.54<64.96° A the correct answer?

Last edited:

eehelp150
gneill said:
Is it -10j instead of -0.1j?

eehelp150 said:
Is it -10j instead of -0.1j?
It is indeed!

eehelp150
gneill said:
It is indeed!
My new current for part 1 is 10/14.14 <-45°
Is the final answer (2.5 + 0.5j) A?

That looks good!

eehelp150
gneill said:
That looks good!
Oops, hang on. I think your signs may be incorrect. The diagram defines the current direction as flowing out of the AC source. So the DC contribution to that must be negative.

eehelp150
eehelp150 said:
My new current for part 1 is 10/14.14 <-45°
Is the final answer (2.5 + 0.5j) A?
Also, if the AC current angle is -45°, its imaginary part should be negative...

eehelp150
gneill said:
Also, if the AC current angle is -45°, its imaginary part should be negative...
It was negative on my paper, I don't know why I typed positive.
(-1.5 - 0.5j)A, correct?

Yes, that looks good.

eehelp150
gneill said:
Yes, that looks good.
Thank you so much for your help!

gneill said:
Yes, that looks good.
the final answer is -1.5 - 0.5j amps
If w = 10, would the answer in time domain be:
$$\sqrt{2.5} * cos(10t + 18.435°)A$$
?

The DC and AC components must be kept separate. There will be a constant DC current with an AC component superimposed on it, or if you like, the AC signal will be offset by a DC bias.

gneill said:
The DC and AC components must be kept separate. There will be a constant DC current with an AC component superimposed on it, or if you like, the AC signal will be offset by a DC bias.
How would I write it out? I need the answer in the time domain.

eehelp150 said:
How would I write it out? I need the answer in the time domain.
A DC current is just a fixed value in the time domain; a number. So your result should have the form:

##I(t) = I_{dc} + I_{ac}##

where ##I_{ac}## is your cosine expression (the part that varies with time).

gneill said:
A DC current is just a fixed value in the time domain; a number. So your result should have the form:

##I(t) = I_{dc} + I_{ac}##

where ##I_{ac}## is your cosine expression (the part that varies with time).
So:

I(t) = sqrt(2)/2 * cos(10t - 45°) - 2

Yes. That looks better!

## 1. What is superposition in the context of AC voltage source and DC current source?

Superposition in this context refers to the principle that states when multiple sources of voltage and current are present in a circuit, the total voltage or current is the sum of the individual voltages or currents. This means that the effects of each source can be analyzed separately and then combined to determine the total response of the circuit.

## 2. How does superposition work in practical applications?

In practical applications, superposition allows engineers to simplify complex circuits by analyzing the effects of each source individually. This makes it easier to troubleshoot and design circuits, as well as calculate the total voltage, current, and power in a circuit.

## 3. What are the limitations of superposition in circuits?

Superposition only applies to linear circuits, meaning the components in the circuit obey Ohm's law and have a constant resistance value. Additionally, it cannot be used to analyze circuits with dependent sources, such as transistors or operational amplifiers.

## 4. Can superposition be applied to both AC and DC circuits?

Yes, superposition can be applied to both AC and DC circuits. However, in AC circuits, the individual voltages and currents must be expressed in terms of phasors, which represent the magnitude and phase of the AC signal.

## 5. How can superposition be used to solve circuit problems?

To use superposition to solve circuit problems, the circuit is first broken down into smaller sub-circuits, each with only one source active and the other sources turned off. The response of each sub-circuit is then calculated using simple circuit analysis techniques, such as Ohm's law or Kirchhoff's laws. Finally, the individual responses are combined to determine the total response of the circuit.

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