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I am having a problem applying the principle of linear superposition to the circuit below. My main problem arises in not know what to do with the capacitor and the inductor in parallel as well as the differing angular frequencies, how does differing angular frequencies change the summation of the two sources? Does it even change the summation?

I understand that I have to open circuit the current source and short circuit the voltage source and so I assumed in order to calculate the voltage source contribution I would get

$$30\sin10t *\frac{((1/0.4)+(1/1))^{-1}}{((1/0.4)+(1/1))^{-1})+4}$$

Also would I be correct in saying that due to the current division theorem, the voltage across the capacitor is also the same voltage across the inductor?

Do the same for the current source and add the two?

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  • \$\begingroup\$ You must use the reactances: jωL and 1/jωC for the inductor and capacitor. The ω values will be 10 rad/sec and 5 rad/sec for the voltage and current sources, respectively. Treat each source separately, and add the results - as you correctly say. And yes, the L and C have the same voltage across them. \$\endgroup\$
    – Chu
    Commented Jun 22, 2015 at 0:17
  • \$\begingroup\$ Hi thanks for the response, how would I go about using the reactances? Would I put the capacitor in parallel with the inductor in series with the resistor? \$\endgroup\$
    – Stan
    Commented Jun 22, 2015 at 0:29
  • \$\begingroup\$ Treat the reactances as if they were resistances - all the circuit analysis rules are exactly the same, except you are now dealing with complex numbers. When you've obtained the two equations under superposition, convert these back to functions of time before combining them to find \$V_o (t)\$ \$\endgroup\$
    – Chu
    Commented Jun 22, 2015 at 6:31
  • \$\begingroup\$ Just out of curiosity, what book is this from? I'm trying to figure out where this trend of using arrows for voltages came from. \$\endgroup\$
    – Adam Haun
    Commented Jun 22, 2015 at 18:56

2 Answers 2

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not know what to do with the capacitor and the inductor in parallel

Either write the differential equations with two storage elements, or solve it as a phasor circuit with two impedances in parallel and convert back to time domain.

how does differing angular frequencies change the summation of the two sources?

It doesn't. A + B is still A + B. Since \$\sin(\omega_1 t)\$ and \$\sin(\omega_2 t)\$ are orthogonal functions when \$\omega_1\ne\omega_2\$, there's no way to simplify \$\sin(\omega_1 t) + \sin(\omega_2 t)\$; it's already the simplest form you'll be able to write.

due to the current division theorem, the voltage across the capacitor is also the same voltage across the inductor?

It has nothing to do with the current division theorem. It's simply the consequence of being connected in parallel and Kirchoff's voltage law. No matter what path you take from node X to node Y, you must get the same potential difference between X and Y.

Do the same for the current source and add the two?

For a basic rundown of how to solve circuits by superposition, see here.

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how does differing angular frequencies change the summation of the two sources? Does it even change the summation?

You clearly didn't understand the Principle of Linear Superposition, because if you use it you cancel 1 source to find the response of the other source. So than you don`t have any problem causing by the different angular frequency.

I would recommend you read against theory of the Principle of Linear Superposition.

Regards, MathieuL.

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