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

Click For Summary
SUMMARY

The discussion focuses on using superposition to find the voltage V(t) in an AC circuit involving a voltage source and reactive components. The impedance values for the inductor and capacitor are given as Z(L) = 4j and Z(C) = 5j, leading to a total impedance of 6.87 <60°. The user initially miscalculated the voltage division due to incorrect signs for the reactive components' impedances. The correct approach involves recognizing that inductive impedances are positive while capacitive impedances are negative, as derived from the formulas Z_L = jωL and Z_C = 1/(jωC).

PREREQUISITES
  • Understanding of phasor forms in AC circuit analysis
  • Knowledge of voltage and current division techniques
  • Familiarity with complex impedance calculations
  • Basic principles of reactive components: inductors and capacitors
NEXT STEPS
  • Study the concept of complex impedance in AC circuits
  • Learn about the application of superposition in circuit analysis
  • Research the rules for determining the signs of reactive component impedances
  • Explore voltage division and current division in more complex AC circuits
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in AC circuit analysis and troubleshooting.

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
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!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
19
Views
6K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K