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

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 2K views
iharuyuki
Messages
15
Reaction score
1

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.
 
Physics news on Phys.org
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.