Phasor Analysis of AC Circuits

• Assumptions for phasor analysis:
  • The system is Linear Time-Invariant (LTI) with a sinusoidal input
  • The system has no switching behavior and has reached steady-state at $t=0$
• Therefore, the complete response is composed completely of the steady-state response

AC Circuits

• Current and voltage vary sinusoidally with time

Phasor Analysis of AC Circuits

• Steps:
  1. Convert time-domain sources to phasors
  2. Use Ohm’s Law, KVL, KCL in phasor domain
  3. Solve for phasor voltages/currents
  4. Convert phasors back to time domain
• Shortcut: The source current is $i(t) = \frac{V_{m}}{|Z_{\text{eq}}|}\cos(\omega t+\phi+\theta)$ where
  • $\theta = \arg(Y_{\text{eq}})$ for parallel
  • $\theta = -\arg(Y_{\text{eq}})$ for series

Current Through RL Series Circuit

Given:
$v(t) = V_m \cos(\omega t + \phi)$

Impedance:
$Z = R + j\omega L$

Phasor voltage:
$\tilde{V} = V_m \angle \phi$

Phasor current:
$\tilde{I} = \frac{\tilde{V}}{Z} = \frac{V_m \angle \phi}{R + j\omega L}$

Convert to polar form:
$|\tilde{I}| = \frac{V_m}{\sqrt{R^2 + (\omega L)^2}}$
$\angle \tilde{I} = \phi - \tan^{-1} \left( \frac{\omega L}{R} \right)$

Time-domain current:

1In [1]: R, V, L, ω, φ = symbols('R V L ω φ')
2In [2]: i = symbols('i')
3In [9]: eq = Eq(i, V/(sqrt(R**2 + (ω*L)**2))*cos(ω*t+φ-atan(ω*L/R)))
4In [15]: eq.subs({V:100, ω:100, φ:rad(30), R:10, L:0.1})
5Out[15]:
6                        ⎛                            π⎞
7i = 7.07106781186548⋅cos⎜100⋅t - 0.785398163397448 + ─⎟
8                        ⎝                            6⎠

Current Through RC Series Circuit

Given:
$v(t) = V_m \cos(\omega t + \phi)$

Impedance:
$Z = R - j\frac{1}{\omega C}$

Phasor voltage:
$\tilde{V} = V_m \angle \phi$

Phasor current:
$\tilde{I} = \frac{\tilde{V}}{Z} = \frac{V_m \angle \phi}{R - j\frac{1}{\omega C}}$

Convert to polar form:
$|\tilde{I}| = \frac{V_m}{\sqrt{R^2 + \left( \frac{1}{\omega C} \right)^2}}$
$\angle \tilde{I} = \phi + \tan^{-1} \left( \frac{1}{\omega RC} \right)$

Time-domain current:

1In [1]: R, V, C, ω, φ = symbols('R V C ω φ')
2In [2]: i = symbols('i')
3In [9]: eq = Eq(i, V/(sqrt(R**2 + (1/(ω*C))**2)) * cos(ω*t + φ + atan(1/(ω*R*C))))
4In [11]: eq.subs({V:100, ω:100, φ:rad(30), R:10, C:100e-6})
5Out[11]:
6                         ⎛        π                   ⎞
7i = 0.995037190209989⋅cos⎜100⋅t + ─ + 1.47112767430373⎟
8                         ⎝        6                   ⎠

Current Through RLC Series Circuit

Given:
$v(t) = V_m \cos(\omega t + \phi)$

Impedance:
$Z = R + j\omega L - j\frac{1}{\omega C} = R + j \left( \omega L - \frac{1}{\omega C} \right)$

Phasor voltage:
$\tilde{V} = V_m \angle \phi$

Phasor current:
$\tilde{I} = \frac{V_m \angle \phi}{R + j \left( \omega L - \frac{1}{\omega C} \right)}$

Magnitude and phase:
$|\tilde{I}| = \frac{V_m}{\sqrt{R^2 + \left( \omega L - \frac{1}{\omega C} \right)^2}}$
$\angle \tilde{I} = \phi - \tan^{-1} \left( \frac{\omega L - \frac{1}{\omega C}}{R} \right)$

Time-domain current:

1In [1]: R, V, L, C, ω, φ = symbols('R V L C ω φ')
2In [2]: i = symbols('i')
3In [3]: eq = Eq(i, V/(sqrt(R**2 + (ω*L - 1/(ω*C))**2)) * cos(ω*t + φ - atan((ω*L - 1/(ω*C))/R)))
4In [6]: eq.subs({V:100, ω:100, φ:rad(30), R:10, L:0.1, C:100e-6})
5Out[6]:
6                        ⎛        π                 ⎞
7i = 1.10431526074847⋅cos⎜100⋅t + ─ + 1.460139105621⎟
8                        ⎝        6                 ⎠

Current Through RL Parallel Circuit

Given:
$v(t) = V_m \cos(\omega t + \phi)$

Impedances:

• Resistor: $Z_R = R$
• Inductor: $Z_L = j\omega L$

Admittances:

• $Y_R = \dfrac{1}{R}$
• $Y_L = \dfrac{1}{j\omega L} = -j \dfrac{1}{\omega L}$

Total admittance:

Phasor voltage:
$\tilde{V} = V_m \angle \phi$

Phasor current:

Magnitude and phase:

Time-domain current:

Current Through RC Parallel Circuit

Given:
$v(t) = V_m \cos(\omega t + \phi)$

Impedances:

• Resistor: $Z_R = R$
• Capacitor: $Z_C = -j\dfrac{1}{\omega C}$

Admittances:

• $Y_R = \dfrac{1}{R}$
• $Y_C = j\omega C$

Total admittance:

Phasor voltage:
$\tilde{V} = V_m \angle \phi$

Phasor current:

Magnitude and phase:

Time-domain current:

Current Through RLC Parallel Circuit

Given:
$v(t) = V_m \cos(\omega t + \phi)$

Impedances:

• $Z_R = R$
• $Z_L = j\omega L$
• $Z_C = -j\dfrac{1}{\omega C}$

Admittances:

• $Y_R = \dfrac{1}{R}$
• $Y_L = -j\dfrac{1}{\omega L}$
• $Y_C = j\omega C$

Total admittance:

Phasor voltage:
$\tilde{V} = V_m \angle \phi$

Phasor current:

Magnitude and phase:

Time-domain current:

Impedance

• Equivalent of resistance for AC: $Z = \frac{V}{I}$
• Complex value: magnitude and phase

Resistor: $Z = R$
Inductor: $Z = j \omega L$
Capacitor: $Z = -j \frac{1}{\omega C}$
Units: Ohms ($\Omega$)

Admittance

• Reciprocal of impedance: $Y = \frac{1}{Z}$
• Used for parallel circuits: $I = YV$
• Units: Siemens (S)

Resonant Frequency

In series RLC: resonance when net reactance is zero

Solve for $\omega_0$:

In Hz:

At resonance:

• $Z = R$
• Current is maximized
• Inductor and capacitor voltages cancel

Combo Circuit Analysis

Example 03.1.

AC Bridge Circuit

• Bridge circuits are used to determine unknown impedances, with four impedances arranged in a bridge configuration
• The bridge is balanced when the ratio of impedances in one branch equals the ratio in the other branch
  • For the circuit below:

Unknown Impedance Derivation

Source Current Derivation

Norton/Thevenin Equivalent Examples

Example 03.2.

References

• Course handout 3: Analysis of AC Circuits Using Phasor Approach