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Homework Help: What a phasor is

  1. Jul 4, 2009 #1
    1. The problem statement, all variables and given/known data
    I am not sure what a phasor is. So when I looked it up on google, it simply said it was a sinusoidal function where [tex]A, \theta, \omega[/tex] are time-invariant. I am not sure what would be good for, or any other relevant information. I also would like to know what [tex]v_x, V_s Cos(\omega t)[/tex] is (as shown on the diagram I've attached), and how one were to obtain the equation down below. I don't have much background in the electrical area, but understand some basic physics.

    My notes says: "We are summing currents here, and the -90° accounts for the
    phase shift between voltage and current in the capacitor".
    Could someone tell me what it means to have a phase shift between voltage and current in a capacitor?

    2. Relevant equations
    According to my notes, When we "Apply KCL at the node containing [tex]v_x[/tex] we get:
    [tex]\frac{v_xCos(\omega t + \theta) - V_sCos(\omega t)}{R}[/tex] + [tex]\frac{v_xCos(\omega t + \theta - 90°)}{\frac{1}{\omega C}}[/tex] [tex] = 0 [/tex]

    Could someone explain this long equation to me.

    Thank you very much,


    Attached Files:

  2. jcsd
  3. Jul 5, 2009 #2
    Re: Phasors

    So I found out that [tex]v_x[/tex] is the voltage at that node [why would they say node, it's simply a point in the circuit], which must have the same voltage as the source- which is [tex]V_Scos(\omega t)[/tex].

    1. Could someone explain to me why [tex]V_S[/tex] is multiplied with [tex]cos(\omega t)[/tex]? What does it mean for a voltage to have a sinusoidal form?

    2. Why will [tex]v_x[/tex] have the same frequency but different phase shift and magnitude if the input is sinusoidal?

    3. Since the circuit is not in parallel, doesn't that mean at all points of the current is the same [by kirchoff's law]?

    4. When we talk about a node, does that mean any given point in a circuit?

    5. I am not sure by what is meant by "The current through a capacitor is equal to the derivative of the voltage across the capacitor (j*w*V_s*cos(w*t), but the j becomes a 90 deg. phase shift) times the capacitance." What does j stand for? I thought phase shifts were signified within the cosine function, as with the w?


  4. Jul 6, 2009 #3


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    Re: Phasors

    I'll answer what I can. Are you taking a class in this, or doing it by self-study?

    I think of a phasor as a complex number, represented by a vector in the complex plane. Some general properties of phasors are:
    • The phasor (again, think vector) rotates about the origin at angular frequency ω
    • The real part of the phasor represents a current or voltage in a circuit.

    Vs cosωt is the voltage of the generator, the left-most element in your circuit. This is commonly referred to as a source voltage, hence the subscript "s".
    Note that the source voltage changes in time due to the cosine term, as is typical of AC voltages, with an amplitude of Vs.

    A node is simply some point or location in a circuit. Kirchoff's Current Law (KCL) says that the sum of all currents entering any node is zero. The equation above is saying:

    "At node 'Vx', the current from the resistor, plus the current from the capacitor, equals zero",​
    where the two terms on the left-hand-side represent the resistor and capacitor currents, respectively.

    Actually, Vx will be different than Vs.

    It just means that the voltage changes in time, as given by the cosωt term. This is typical of AC voltages, such as in the wall outlets that power household electronic devices.

    Without going into too much detail: solving the equations for the circuit involves terms that are proportional to either the source voltage itself, or to the derivative of the source voltage.

    Since the derivative of a cosine will be a sine function at the same frequency, the frequency of every voltage and current in the circuit will be at the same frequency.

    Yes, the current is the same for each of the three circuit elements.


    j is the square root of -1, and is the same as i that is used in math courses. Electrical engineers use j instead of i, perhaps because i is used for current.

    Since q=CV for a capacitor, and the current i=dq/dt, we have
    i = dq/dt = C dV/dt
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