Solve AC Nodal Voltage V(t) Homework Statement

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SUMMARY

The discussion focuses on solving an AC nodal voltage problem using phasor analysis and superposition. The voltage supplies in question operate at the same angular frequency (ω), allowing for the application of superposition despite their phase differences. Key equations include the voltage function v(t) = Vmaxcos(ωt + φ) and the impedance formulas ZL = jωL and ZC = 1/jωC. The participants emphasize the importance of converting circuit elements to phasor form and suggest using Kirchhoff's Voltage Law (KVL) or superposition for analysis.

PREREQUISITES
  • Understanding of phasor representation in AC circuits
  • Familiarity with Kirchhoff's Voltage Law (KVL)
  • Knowledge of complex impedance for inductors and capacitors
  • Ability to apply superposition theorem in circuit analysis
NEXT STEPS
  • Study phasor conversion techniques for AC circuit analysis
  • Learn about the superposition theorem in the context of AC circuits
  • Explore Kirchhoff's Voltage Law (KVL) applications in phasor circuits
  • Investigate complex impedance calculations for various circuit components
USEFUL FOR

Electrical engineering students, circuit analysts, and anyone involved in AC circuit design and analysis will benefit from this discussion.

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Homework Statement


I've attached the question.
Nodal Voltage question.png

So I'm not sure which method should be used to solve this. I was thinking superposition, but the voltage supplies are are both at the same ω value.

Homework Equations


v(t) = Vmaxcos(ωt + \phi)
ZL = jωL
ZC = 1/jωL

The Attempt at a Solution


convert everything to phasor form
Vs1 = 20\angle0°
Vs2 = 20cos(1000t - 90°) = 20\angle-90°
10mH = j10 = 10\angle90°
0.1mF = -j10 = 10\angle-90°

Then I'm not really sure what to do next, do I use superposition or mesh analysis or etc. and then how does that work with phasors?
So would appreciate a push in the right direction.
 
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Since you have marked all node voltages, you could use: current through the inductor + current through the capacitor = current in the resistor
and with only one unknown, solve this to find Vx
 
In terms of phasors, you can let Vs2 = 20, then Vs1 = 20ejπ/2.

You can then solve using kvl or you can use superposition to solve for one of the sources at a time, then add the results.

Superposition is allowable if the sources are independent, which here they are.

Had the frequencies been different you would have had to use superposition, solving two separate problems individually, with different complex impedances for each problem. Superpositoon would have to be done in the time domain.
 

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