Finding i(t) in a Circuit with Phasors: Introductory Exercise

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SUMMARY

The discussion focuses on finding the current i(t) in a circuit using phasors, given the voltage source v_s = 2 cos(w x + π/2), and the reactances X_C = -1 Ω and X_L = 1 Ω. The user applied voltage dividers and derived the current expression as I = (Z_C v_s (Z_L + R)) / ((Z_L + R)(Z_C + Z_L + R)(Z_C(Z_L + R) / (Z_C + Z_L + R) + R)). The calculated result was i(t) = (2/√5) cos(wt + 0.46), but the expected phase was -2.68, indicating a misunderstanding of the phase shift associated with the voltage source.

PREREQUISITES
  • Understanding of phasor analysis in AC circuits
  • Familiarity with complex impedance (Z_C and Z_L)
  • Knowledge of voltage divider rules in electrical circuits
  • Basic trigonometric functions and their application in circuit analysis
NEXT STEPS
  • Study phasor representation of AC voltages and currents
  • Learn about complex impedance and its role in circuit analysis
  • Explore the concept of phase shifts in AC circuits
  • Review voltage divider circuits and their applications in phasor analysis
USEFUL FOR

Electrical engineering students, circuit designers, and anyone interested in mastering AC circuit analysis using phasors.

okh
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Homework Statement


In this circuit, find [itex]i(t)[/itex], knowing that [itex]v_s= 2 \cos \left(w x+\frac{\pi }{2}\right)[/itex], and that, at the source's frequency, [itex]X_C= -1 Ω[/itex] and [itex]X_L = 1 Ω[/itex].
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Homework Equations


Basic phasors and dividers equations.
[itex]Z_C = jX_C[/itex]
[itex]Z_L = jX_L[/itex]

The Attempt at a Solution


I used dividers. The source divides between R and the parallel of C and the series of L and R.
[itex] I=\frac{Z_C v_s \left(Z_L+R\right)}{\left(Z_L+R\right) \left(Z_C+Z_L+R\right) \left(\frac{Z_C \left(Z_L+R\right)}{Z_C+Z_L+R}+R\right)}[/itex]
Solving with [itex]R=1, Z_c=-j, Z_l=j, v_s=2j[/itex] I get [itex]i(t)=\frac{2}{\sqrt{5}}*cos(wt+0.46)[/itex], while the correct phase should be [itex]-2.68[/itex]. Basically I get the symmetrical cosine wave with respect to the x axis.
 
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I agree with your result. It seems that they took the given voltage source phase ##\pi/2## to be a negative phase shift for some reason.
 
Thank you. Yeah, that may be the reason.
 

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