AC Circuits and Two-Port Networks

AC power triangle (reactive power)

The power triangle relates the three powers of a sinusoidal load. The complex power is \bar S = P + jQ, where P = V_{rms}I_{rms}\cos\phi is the active (real) power in W and Q = V_{rms}I_{rms}\sin\phi is the reactive power in VAr. The apparent power S = V_{rms}I_{rms} in VA satisfies S^2 = P^2 + Q^2, so the three sides form a right triangle. The power factor is \cos\phi = P/S. For an inductive (lagging) load \phi is positive and the reactive power is drawn by the inductor, while for a capacitive (leading) load the reactive power is of the opposite sign. Example: a load with S = 500 VA and P = 300 W has |Q| = \sqrt{500^2 - 300^2} = 400 VAr and \cos\phi = 0.6 lagging.

When the power factor of a network is needed, first combine the impedances (series impedances add, parallel branches are combined through admittances), then take \cos of the angle of the total impedance. The sign of the net reactance decides whether the power factor is leading or lagging. Example: Z = 3 + j4\ \Omega gives \cos\phi = 0.6 lagging, while Z = 8 - j6\ \Omega gives \cos\phi = 0.8 leading.

The active power supplied by a source equals the real part of the complex power, P = \text{Re}(\bar S), because ideal reactive elements store and return energy without dissipating it. Only resistances consume active power, so a source feeding a reactive network supplies P equal to the total power dissipated in the resistors.

Series RLC resonance and Q factor

A series RLC circuit has its resonant frequency where the inductive and capacitive reactances cancel, \omega_0 = 1/\sqrt{LC}. At this frequency the impedance is purely resistive and equal to R, so the current is maximum. The quality factor is Q = \omega_0 L/R = \frac{1}{R}\sqrt{L/C} = \frac{1}{\omega_0 CR}, and the bandwidth is B = \omega_0/Q. Example: for L = 10 mH, C = 1 \muF and R = 10\ \Omega, \omega_0 = 10^4 rad/s and Q = \omega_0 L/R = 10.

For a band-pass filter taken across the resistor, the same Q defines the sharpness of the peak. Pole and residue analysis is a common tool for such second-order networks. The residue of a rational function at a simple pole p is \lim_{z\to p}(z-p)f(z). Example: for f(z) = 1/[z(z+4)] the residue at z = 0 is 1/4.

When L and R carry measurement uncertainties, the uncertainty in Q is found from the sensitivity of Q. Since Q = \frac{1}{R}\sqrt{L/C} and C is fixed by \omega_0 and L, Q varies as \sqrt{L}/R. The worst-case relative error is then \delta Q/Q \approx \tfrac{1}{2}(\delta L/L) + \delta R/R. Example: L = 50 mH with \pm 2\% and R = 20\ \Omega with \pm 2.5\% give \delta Q/Q = 1\% + 2.5\% = 3.5\%.

RL circuit frequency response

In a series RL circuit the impedance is Z = R + j\omega L, so |Z| = \sqrt{R^2 + \omega^2L^2} and the impedance angle is \phi = \tan^{-1}(\omega L/R). The current lags the source voltage by \phi. The cutoff (corner) frequency is \omega_c = R/L. Across the resistor the output is a low-pass response, and across the inductor it is a high-pass response.

A common inverse problem gives the lag angle and asks for the frequency. Example: if the current lags by 30^\circ, then \omega L/R = \tan 30^\circ, which gives \omega = R\tan 30^\circ/L. For 45^\circ lag, \omega L = R exactly, so f = R/(2\pi L).

Unknown inductance is often measured with a balanced AC bridge. At balance the arm reactance X is found from the bridge condition, and the inductance follows from L = X/\omega, where \omega = 2\pi f is the supply angular frequency.

Non-sinusoidal waveform RMS and power

A periodic waveform that contains a DC term and several harmonics has an rms value found from its components. For a current i = I_0 + \sum_n I_{mn}\cos(n\omega t + \theta_n), the rms value is I_{rms} = \sqrt{I_0^2 + \sum_n I_{mn}^2/2}. Each harmonic contributes half the square of its peak value, and the DC term contributes its square directly. Example: i = 3 + 4\cos(t) A has I_{rms} = \sqrt{9 + 8} = \sqrt{17} A.

Average power in a circuit with non-sinusoidal voltage and current is the sum of the power of each frequency component. Only components at the same frequency produce average power; a voltage harmonic and a current harmonic at different frequencies give zero average power. A DC voltage with no DC current contributes nothing. For a matching pair, P_n = \tfrac{1}{2}V_{mn}I_{mn}\cos(\theta_{vn} - \theta_{in}). Example: a 100 V peak voltage at \omega and a 2 A peak current at \omega with a 60^\circ phase difference give P = \tfrac{1}{2}(100)(2)\cos 60^\circ = 50 W.

Source output impedance loading

A practical source has an internal (output) impedance, and any load connected across its terminals forms a voltage divider with it. If the open-circuit voltage is V_{oc} and the source resistance is R_s, the loaded output is V_{out} = V_{oc}\,Z_L/(R_s + Z_L). Connecting a capacitor Z_C lowers the output magnitude, and the ratio is |V_{out}|/V_{oc} = |Z_C|/\sqrt{R_s^2 + |Z_C|^2}.

To find an unknown reactance from a measured voltage drop, set r = |V_{out}|/V_{oc} and solve |Z_C| = R_s\,r/\sqrt{1 - r^2}. Example: with R_s = 100\ \Omega and r = 0.6, |Z_C| = 100\times 0.6/0.8 = 75\ \Omega. Once |Z_C| is known, the capacitance follows from |Z_C| = 1/(\omega C).

Resonance condition for purely real input admittance

The input admittance Y_{in} = G + jB is purely real when its susceptance B vanishes. For two parallel branches, R + j\omega L and R - jX with X = 1/(\omega C), setting B = 0 gives \omega L X^2 - (R^2 + \omega^2L^2)X + \omega L R^2 = 0. Its roots are X = \omega L and X = R^2/(\omega L).

The first root is the familiar case where the two branches are complex conjugates, so Y_{in} = 2R/(R^2 + \omega^2L^2) is real. Example: for R = 2\ \Omega and \omega L = 3\ \Omega, the capacitive reactance X = 3\ \Omega makes Y_{in} = 4/13 S, which is real. The capacitance is then C = 1/(\omega^2 L).