AC circuit, voltage expressed by time

Click For Summary

Discussion Overview

The discussion revolves around an AC circuit problem involving a voltage source and current expressed as time-dependent functions. Participants explore how to find the impedance of the circuit in both polar and rectangular forms, while also considering whether the circuit is inductive or capacitive. The scope includes mathematical reasoning and technical explanations related to phasors and sinusoidal functions.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about the time-dependent expressions for voltage and current in AC circuits and questions how to start solving for impedance.
  • Another participant suggests that converting both voltage and current to the same trigonometric function could simplify the problem.
  • A participant proposes converting the voltage and current to phasor form to facilitate calculations, providing specific transformations for both expressions.
  • There is a correction regarding angle calculations in the phasor transformations, with a participant questioning the numerical values derived from the expressions.
  • One participant calculates the impedance using the phasor forms and concludes that the circuit is capacitive based on the phase relationship between voltage and current.

Areas of Agreement / Disagreement

Participants generally agree on the approach of converting to phasors, but there are discrepancies in the calculations and interpretations of the angles involved. The discussion remains unresolved regarding the correctness of the angle calculations and the final classification of the circuit.

Contextual Notes

There are unresolved issues related to the accuracy of angle conversions and the implications of the phase relationships in determining circuit characteristics. Participants have not reached a consensus on the correct values or classifications.

Mutaja
Messages
239
Reaction score
0

Homework Statement



A voltage source e(t) = 50 sin(2000t + 30°) V supplies the current i(t) = 0.5 cos(2000t) A in a circuit. Find the impedance in the circuit, in both polar and rectangular form. Is the circuit inductive or capacitive?

Homework Equations





The Attempt at a Solution



I have a problem here, right of the bat I'm not used to seeing voltage and current being given as an expression including time. Well, in AC circuits, that is. I've looked through my book, but I can't find any examples on this.

There isn't much work to show here as I'm confused as to where I should start.

I RMS involved? I picture a sinus wave with current and voltage waves, that's where that idea came from.

Any input, at all, would be greatly appreciated.
 
Physics news on Phys.org
Consider how much simpler life would be if both the voltage and current were expressed in terms of the same trig function...
 
gneill said:
Consider how much simpler life would be if both the voltage and current were expressed in terms of the same trig function...

So I can convert the cosine to sine, or vice versa. I'm just confused by the whole expression for some reason.

The voltage and current is expressed by time. Given that I understand that, which frankly I don't, how can I use that to find the impedance?

What I understand about the expressions is that you can read the values of the current or voltage at a given time. I assume I can use this to find the RMS value, or average value.
 
The idea is to convert everything to phasors. Then the usual phasor manipulations will be at your disposal.
 
gneill said:
The idea is to convert everything to phasors. Then the usual phasor manipulations will be at your disposal.

So e(t) = 50 sin(2000t + 30°) -> e(t) = 50 cos(2000t + (30° - \frac{∏}{2}))
In phasor form this is 50 ej(1/2 - ∏/2 which is 50V ∠1.07°


i(t) = 0.5 cos(2000t) in phasor form is 0.5 ej0 which translates to 0.5A∠90°

Am I onto something here? I'm fully aware there will be mistakes here even if I've gone though my numbers twice. I'm not used to these calculations.
 
That's the right idea. Check your angle calculations. pi/2 is 90°, so 30° - 90° is not likely to be 1.07°. Also, e0 does not yield an angle of 90°. What's e0 numerically?
 
gneill said:
That's the right idea. Check your angle calculations. pi/2 is 90°, so 30° - 90° is not likely to be 1.07°. Also, e0 does not yield an angle of 90°. What's e0 numerically?

e(t) = 50 sin(2000t + 30°) -> e(t) = 50 cos(2000t + (30° - ∏/2))
In phasor form this is 50 ej(1/2 - ∏/2) which is 50V ∠-60°


i(t) = 0.5 cos(2000t) in phasor form is 0.5 ej0 which translates to 0.5A∠0°

Z = \frac{V}{I} = \frac{50V∠-60°}{0.5A∠0°} = 100Ω∠-60°

The circuit is capacitive since the voltage lags the current by 60°.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 8 ·
Replies
8
Views
8K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
15
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K