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I was given this problem in a basic engineering circuit analysis class. I really don't know where to start. enter image description here

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    \$\begingroup\$ Hi John! Welcome here. Your question is too broad. You're literally asking us to tell you all that you've learnt in that class. \$\endgroup\$ – Marcus Müller Feb 10 at 21:46
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    \$\begingroup\$ This is obviously a homework problem so we can't provide a complete answer especially since you haven't shown any work. However, I will suggest you use the technique of superposition since you have 3 independent sources. For each source, replace the other sources with the appropriate value (I assume you know this) and calculate the capacitor voltage due to that source. Add up the voltages from each of the three sources and that will give you the answer. \$\endgroup\$ – Barry Feb 10 at 21:49
  • \$\begingroup\$ Thank you. I wasn't really asking for a full answer; I just wanted to know how to get started. \$\endgroup\$ – John Stewart Feb 10 at 21:55
  • \$\begingroup\$ You could also write the mesh equations out and solve them. If you do, note don't try to make two meshes that each run through the current source. The unknown voltage across it makes it tricky. It is easier to just have one loop that runs through it and make a super loop that includes the first loop but bypasses the current source. That makes it easier since the you don't need to solve the voltage drop across the current source, and the current source instantly solves the current for one of the loops. Easier if you use the S-domain instead of time-domain unless you feel like differential eqtns \$\endgroup\$ – DKNguyen Feb 10 at 21:57
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There's no initial conditions nor any turning switches, one must assume this is a steady state problem where starting transients can be assumed to have died off.

You have three different sources which all have different frequency. This can be solved quite easily with superposition.

You have 10V DC voltage source. To calculate its effect to Vo(t) remove the current source and replace the voltage source with a wire. The inductor is a wire for steady state DC.

The current source feeds L, C and R in parallel. You must solve the caused voltage component with phasors and present the result as a sinusoidal function of time.

The voltage source V1 feeds through 6 Ohm resistor a parallel LC circuit. Solve again the caused component of Vo with phasors and present the result as a sinusoidal function of time.

Converting the voltage source V1 + the 6 Ohm resistor to a current source could be considered as clever, because it's possible to reuse the fomulas of the previous step, only with different frequency and voltage.

The final Vo(t) is the sum of the calculated three components.

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Well, we are trying to solve the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Using KCL, we can write:

$$ \begin{cases} \text{I}_2=\text{I}_\text{a}+\text{I}_1\\ \\ \text{I}_\text{a}=\text{I}_2+\text{I}_3\\ \\ \text{I}_1=\text{I}_\text{b}+\text{I}_\text{c}\\ \\ 0=\text{I}_\text{b}+\text{I}_\text{c}+\text{I}_3 \end{cases}\tag1 $$

Using KVL, we can write:

$$ \begin{cases} \text{I}_\text{a}=\frac{\text{V}_\text{a}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1}{\text{R}_2}\\ \\ \text{I}_\text{c}=\frac{\text{V}_\text{c}-\text{V}_1}{\text{R}_3} \end{cases}\tag2 $$

Now, we can solve this system of equations using the (complex) s-domain. Using this method we need to use:

  • $$\text{R}_2=\frac{1}{\text{sC}}\tag3$$
  • $$\text{R}_3=\text{sL}\tag4$$
  • $$\text{V}_\text{a}=\frac{\hat{\text{v}}_\text{a}\text{s}}{\text{s}^2+\omega_\text{a}^2}\tag5$$
  • $$\text{I}_\text{b}=\frac{\hat{\text{i}}_\text{b}\omega_\text{b}}{\text{s}^2+\omega_\text{b}^2}\tag6$$
  • $$\text{V}_\text{c}=\frac{\hat{\text{v}}_\text{c}}{\text{s}}\tag7$$

Using your values we get, for \$\text{V}_1\left(t\right)\$:

enter image description here


I used the following code:

FullSimplify[
 Solve[{I2 == Ia + I1, Ia == I2 + I3, I1 == Ib + Ic, 
   0 == Ib + Ic + I3, Ia == (Va - V1)/R1, I2 == V1/R2, 
   Ic == (Vc - V1)/R3}, {Ia, Ic, I1, I2, I3, V1}]]

Implementing your values:

FullSimplify[
 Solve[{I2 == Ia + I1, Ia == I2 + I3, 
   I1 == (((4*2)/(s^2 + 2^2))) + Ic, 
   0 == (((4*2)/(s^2 + 2^2))) + Ic + I3, 
   Ia == (((12*s)/(s^2 + 3^2)) - V1)/6, I2 == V1/(1/(s*(1/12))), 
   Ic == ((10/s) - V1)/(s*2)}, {Ia, Ic, I1, I2, I3, V1}]]

Solution:

{{Ia -> (2 (-180 - 113 s^2 - 3 s^4 + s^6))/(
   s (4 + s^2) (9 + s^2) (6 + s (2 + s))), 
  Ic -> (360 + s (-252 + s (2 + s) (41 + s (-12 + 5 s))))/(
   s (4 + s^2) (9 + s^2) (6 + s (2 + s))), 
  I1 -> (360 + s (180 + s (226 + s (137 + s (14 + 13 s)))))/(
   s (4 + s^2) (9 + s^2) (6 + s (2 + s))), 
  I2 -> (180 + 
    s^2 (137 + s (8 + s (13 + 2 s))))/((4 + s^2) (9 + s^2) (6 + 
      s (2 + s))), 
  I3 -> -((360 + s (180 + s (226 + s (137 + s (14 + 13 s)))))/(
    s (4 + s^2) (9 + s^2) (6 + s (2 + s)))), 
  V1 -> (12 (180 + s^2 (137 + s (8 + s (13 + 2 s)))))/(
   s (4 + s^2) (9 + s^2) (6 + s (2 + s)))}}
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