Superposition: Find V(t) in complex (AC) circuit

AI Thread Summary
The discussion focuses on solving for the voltage V(t) in a complex AC circuit using superposition and impedance calculations. The user initially calculates total impedance incorrectly, leading to an incorrect voltage division result. Feedback highlights the importance of correctly identifying the signs of reactive component impedances, noting that inductive impedances are positive while capacitive impedances are negative. The user corrects their calculations after adjusting the capacitor's impedance sign, resulting in the correct answer. Understanding the rules for determining impedance signs is crucial for accurate circuit analysis.
iharuyuki
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Homework Statement



upload_2015-4-16_15-42-26.png

Homework Equations


phasor forms
voltage division
current division

The Attempt at a Solution



Using superposition, considering only the varying voltage source.

Z (L) = 4j
Z (C) = 5j

Total impedance:
4 is parallel with 5 = 2.44 + 1.95j
series with 1 + 4j
Total impedance: 3.44 + 5.95j = 6.87 <60°

Voltage source: 10<0°

Voltage division Vo(t) = 10<0° * (1/(6.87 <60°)) = 1.45<-60°
= 1.45cos (2t - 60°)

However this is off from the cos part of the correct answer.

What's wrong with my working?

Thank you very much.
 
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Check the signs of your reactive component impedances.
 
Thank you very much for your response.

With the capacitor's impedance set to negative and the correct answer pops up.

What is the rule to determine whether an impedance is positive or negative?
 
iharuyuki said:
Thank you very much for your response.

With the capacitor's impedance set to negative and the correct answer pops up.

What is the rule to determine whether an impedance is positive or negative?
In general inductive impedances are positive while capacitive impedances are negative.

This comes from the formulas for impedances of inductors and capacitors:

##Z_L = j \omega L##
##Z_C = \frac{1}{j \omega C}##

When the j in the denominator of the capacitive impedance is "moved" to the numerator, its sign changes.
 
I see, thank you!
 
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