- #1
Joshb60796
- 62
- 0
Do I have the right idea? I'm not getting what is in the solutions manual but it's been wrong on past chapter and I'm beyond frustration.
You have two components with a common current that has a given defined direction. The expected potential drop is in the direction of the current flow, and that is what is indicated by the "+ -" on the capacitor. So in this case the labeling which defines how to interpret the potential across the capacitor is in accord with the assumed current direction.Joshb60796 said:Because the polarity of the capacitor in reference to the current flow. If the current flow Ic was in the direction towards the negative terminal of the capacitor then I wouldn't have added the minus sign to the current. Is this a wrong assumption on my part?
Well then, the book solution is not correct for the given circuit. The total impedance is Z = 5 - 0.385j Ohms, so its angle is -4.4°. No way can you get -27.5° from there.Joshb60796 said:Ok I see that now. I think I was confusing ideas from transients. My main problem is even when I didnt have the minus sign in there, I wasn't getting the book solution so I don't know if my process is flawed or not. The book solution is Ic=88.7sin(20t-27.5°)mA and Vc=2.31sin(20t+62.5°)Volts with the angles in degrees.
Yup. Looks good to me!Joshb60796 said:Ok, I have been able to get that. Then I find the current in the loop using ohm's law I = V/R which gives me 0.997sin(20t+4.4°) Amps. Then I use the current through the capacitor which I just found and the Z of the capacitor which is 0.385∠-90° to find the voltage across the capacitor, using ohms law V=I*R which gives me 0.384sin(20t-85.6°) Volts.
Don, the problem statement does not mention anything about the physical nature of the capacitor or its construction, only that it has a certain value of capacitance in millifarads. The + and - indicated on the diagram and labeled Vc defines how to interpret the capacitor voltage polarity (i.e., it defines Vc).donpacino said:just something to think about josh. there are many different types of capacitors. the one used in your homework problem is an electrolytic capacitor. They are polarized and must be biased correctly (one side having higher voltage than the other. The is the reason for the plus and the minus. that being said when doing homework problems such as these, unless stated otherwise, assume all capacitors are ideal.
http://en.wikipedia.org/wiki/Types_of_capacitor
Phasors are complex numbers used in circuit analysis to represent the amplitude and phase of a sinusoidal voltage or current. They allow us to simplify complex calculations and visualize the behavior of circuits in the frequency domain.
To convert from the time-domain to phasor representation, we use the Fourier transform. This converts a time-varying signal into a complex phasor with magnitude and phase. To convert back from phasor to time-domain, we use the inverse Fourier transform.
Impedance is the opposition to the flow of current in a circuit, while admittance is the ease with which current can flow. In phasor analysis, impedance is represented by a complex number with magnitude and phase, while admittance is represented by a complex number with magnitude and negative phase.
No, phasors can only be used for sinusoidal signals. For non-sinusoidal signals, we use techniques such as Fourier series or Fourier transform to analyze the signal in the frequency domain.
In parallel circuits, the phasors for each branch are added to find the total current, while the phasors for each branch in series are added to find the total voltage. This is done using the rules for complex numbers, such as the angle and magnitude addition rules.