Phasor Voltage and Capacitive Impedance

Phasor basics and the setup

• We are analyzing a phasor representation of an AC circuit element, focusing on how voltage and current relate when a capacitor is involved.

• Key phrases from the transcript:

  • "+ j times sine negative 90 degrees" hints at using the imaginary unit j to encode a 90° phase shift.

  • "Cosine of negative 90 degrees is zero" recalls trigonometric values: $\cos(-90^{\circ}) = 0.$

  • The phrase "our phasor voltage" signals we are expressing voltages as phasors in the complex plane.

• Phasor diagram concept:

  • There are two axes: the real (horizontal) axis and the imaginary axis represented by j (the vertical axis).

  • The angle corresponds to the phase of the phasor relative to a reference.

  • When the real part is zero and the imaginary part is nonzero (on the j axis), the angle is ±90°; the transcript notes that the inverse tangent (\tan(\theta) = \text{Imag}/\text{Real}) becomes undefined when Real = 0.

• Practical interpretation of the undefined tangent:

  • If a phasor lies on the imaginary axis (Real = 0), the angle is ±90°, which can be remembered as: you go 90° from the real axis to reach the imaginary axis.

  • The memory aid provided: "j is just the vertical axis, and you have to go 90 degrees to get there."

• Phase rotation by multiplication with j:

  • Multiplying a phasor by (j) rotates it by +90° in the complex plane.

  • Hence, multiplying the current phasor by (j) yields a 90° phase shift.

• Introduction of the voltage with a phase reference:

  • The speaker introduces an "initial phase voltage v" as part of setting up the phasor relationship.

• Summary of the observed phase behavior:

  • By introducing the factor (j) to the current, the corresponding voltage phasor acquires a +90° phase shift relative to the current phasor.

  • The movement in the phasor diagram is described as going in the positive phase direction, which correlates to adding phase to the current.

• Transition to a compact expression for the relationship:

  • The voltage across a capacitor can be written as

  • $V = \left( \frac{1}{j \omega C} \right) I = -j \frac{1}{\omega C} I.$

  • This shows the impedance of a capacitor is $Z_C = \frac{1}{j \omega C} = -\frac{j}{\omega C}.$

• Phase lead/lag interpretation for a capacitor:

  • In this context, the current leads the voltage by 90° (i.e., the voltage phasor lags the current phasor by 90°), which is consistent with the negative imaginary factor in the impedance.

  • The transcript notes: "the current here is ahead in terms of… because we're adding the phase to it" and "it needs the voltage by 90 degrees"—pointing to the voltage lagging behind the current by 90°.

• Directional intuition in the phasor plane:

  • The discussion emphasizes visualizing the positive phase direction as the CCW rotation (toward larger angles) in the complex plane.

  • The phrase "we're going in this direction in the positive phase direction" reinforces that interpretation.

• Key takeaway about the capacitor phasor relationship:

  • For a capacitor, the impedance is

  • $Z_C = \frac{1}{j\omega C} = -\frac{j}{\omega C}.$

  • The voltage phasor is a -90° rotation of the current phasor scaled by (\frac{1}{\omega C}) in magnitude.

• Step-by-step derivation outline (based on the transcript): 1) Start with sine/cosine identities at -90°: (\sin(-90^{\circ}) = -1), (\cos(-90^{\circ}) = 0). 2) Form the phasor for the voltage using the capacitor relation, noting the imaginary unit rotates phase by 90° when multiplying by (j).

  • This yields (V = \left( \frac{1}{j \omega C} \right) I = -j \frac{1}{\omega C} I).
    3) Interpret the phase: the current leads the voltage by 90°, which matches the negative imaginary impedance for the capacitor.
    4) Use the phasor diagram intuition to locate the voltage relative to the current on the plane and understand the 90° lag.

• Worked example visualization (conceptual, not numeric):

  • If the current phasor is at angle (\theta_I) with magnitude (|I|), then

  • The capacitor's impedance provides a magnitude scaling of (\frac{1}{\omega C}) and a phase shift of -90°, so the voltage phasor is at angle (\thetaV = \thetaI - 90^{\circ}) with magnitude (\frac{|I|}{\omega C}).

• Practical and exam-oriented note:

  • Always remember: for a capacitor, I leads V by 90°, and Z_C = (1/(j\omega C)) = -(j/(\omega C)).

  • The algebraic form V = -j (1/(\omega C)) I compactly captures both the magnitude scaling and the phase shift.

• Final tip to remember the relationship shared in the transcript:

  • The key relationship to memorize is $V = \left( \frac{1}{j \omega C} \right) I = -j \frac{1}{\omega C} I,$ and the associated phase behavior: current leads voltage by 90° for a capacitor.