Electrical Circuits

Series RLC resonance

A series RLC circuit has impedance Z = R + jX, where the net reactance is X = \omega L - \frac{1}{\omega C}, so |Z| = \sqrt{R^2 + X^2}. The current is largest when X = 0, the condition called resonance, where the inductive and capacitive reactances cancel and the circuit behaves as a pure resistance. The resonant angular frequency is \omega_0 = \frac{1}{\sqrt{LC}}, and the rms current at resonance equals V/R. When R and C are fixed and only L is varied, the current peaks at the value of L that satisfies X_L = X_C, so L = \frac{X_C}{\omega} = \frac{1}{\omega^2 C}. The reactance X_C can be found from the measured capacitor voltage as X_C = V_C / I at that operating point. For a fixed capacitor, the capacitor voltage in a series RC circuit is a low-pass function, with amplitude |V_C| = \frac{V}{\sqrt{1 + (\omega RC)^2}}. This amplitude is largest at zero frequency and falls monotonically as \omega increases, so the break point \omega = 1/RC is where the amplitude has dropped to 1/\sqrt{2} of its maximum, not where it peaks. The resistor and capacitor voltages in a series RC circuit have equal amplitudes when |Z_R| = |X_C|, that is, R = \frac{1}{\omega C}, which gives \omega = \frac{1}{RC}. When the resistor voltage equals the source voltage in amplitude, |Z| = R must hold, which means the net reactance is zero. Setting \omega L = \frac{1}{\omega C} then gives \omega_0 = \frac{1}{\sqrt{LC}}. At resonance the rms current equals V_{rms}/R, because the impedance is purely resistive and the reactive terms cancel. Keep the units consistent: \omega must be in rad/s, and \omega = 2\pi f when f is given in Hz.

Node voltage analysis

Node voltage analysis writes Kirchhoff's current law (KCL) at each non-reference node, expresses each branch current as a voltage difference divided by the branch impedance, and solves the resulting linear system. Choose the reference (ground) node first. A node tied to ground through a voltage source has a known voltage and needs no equation, while a voltage source between two unknown nodes requires a supernode. A current leaving a node is counted positive in its KCL sum, and V_B - V_N = E means terminal B is at the higher potential. Example: a 10 V source feeds a node through 2 \Omega, and the same node has a 3 \Omega resistor to ground. KCL gives \frac{V - 10}{2} + \frac{V}{3} = 0, so V = 6 V. For a bridge-type network with two unknown nodes, write two KCL equations and solve them simultaneously. The current in the bridging branch is then the difference of the two node voltages divided by the branch resistance, and its sign depends on the reference direction chosen. Many items under this heading test prerequisite mathematics rather than topology. A probability density must integrate to 1 over its domain, and \int_{-\infty}^{\infty}\operatorname{sech}x\,dx = \pi, which fixes the normalising constant. The curl operator acts only on vector fields, so \nabla\times(\nabla\cdot\vec{V}) is undefined because \nabla\cdot\vec{V} is a scalar. By contrast, \nabla\cdot(\nabla\times\vec{V}) = 0 is a valid identity.

Sinusoidal steady-state phasor analysis

In sinusoidal steady state, each voltage or current is written as a phasor, and the elements are replaced by impedances: Z_R = R, Z_L = j\omega L, and Z_C = \frac{1}{j\omega C}. Phasor analysis then follows the rules of DC resistive circuits with complex quantities. A sine is converted to a cosine by \sin t = \cos(t - \frac{\pi}{2}) before its phasor is taken. For a parallel RLC network, the nodal admittance is Y = G + j\left(\omega C - \frac{1}{\omega L}\right). At the frequency where \omega^2 LC = 1 the susceptances cancel, so Y = G and the node voltage phasor equals the source current phasor divided by G. For sinusoidal voltage V_m\sin(\omega t + \theta_v) and current I_m\sin(\omega t + \theta_i), the average power is P = \frac{V_m I_m}{2}\cos\theta, where \theta = \theta_v - \theta_i is the phase difference. The product \frac{V_m I_m}{2} is the apparent power, and \cos\theta is the power factor. Do not confuse the average power with the peak instantaneous power, which is V_m I_m. The magnitude of the input impedance of a circuit is the modulus of the complex sum of its series and parallel parts. A capacitor in parallel with a resistor gives a complex impedance whose real and imaginary parts must be combined before taking the magnitude.

AC bridge balance conditions

An AC bridge is balanced when the detector current is zero, which requires the product of the impedances in opposite arms to be equal in both magnitude and phase: Z_1 Z_3 = Z_2 Z_4. Because this is a complex equation, it gives two real conditions, obtained by equating the real and imaginary parts. The Owen bridge uses a capacitor in one arm and a series resistor–capacitor arm to measure an inductor, and its balance conditions are L_x = R_2 R_3 C_1 and R_x = \frac{R_2 C_1}{C_3}. The balance is independent of frequency, which is a principal advantage of this bridge. Bridges used for inductance measurement include the Maxwell–Wien, Hay and Owen bridges, each having at least one inductive arm. The Schering bridge compares two capacitive arms and is used for capacitance and dissipation factor measurement of capacitors and insulators. Because it contains no inductive arm, it cannot measure inductance.

Power in resistive networks

The power dissipated in a resistor is P = I^2 R = \frac{V^2}{R}, so when the power in one resistor is known, its voltage and current follow directly as V = \sqrt{PR} and I = \sqrt{P/R}. These values can then be propagated through the rest of a resistive network using Ohm's law and KCL. Example: a 4 \Omega resistor dissipates 9 W, so its voltage is \sqrt{36} = 6 V, and its current is 1.5 A. In a network with several sources, the power in a resistor is proportional to the square of the node voltage across it. Each source enters the node voltage with a weight set by the conductances connected to it. Therefore the power in a resistor is proportional to the square of a weighted sum of the source voltages, and the weights come from the ratio of the connected conductances.

Maximum power transfer theorem

The maximum power transfer theorem states that a load absorbs maximum power from a linear source when the load resistance equals the Thevenin resistance of the source network, R_L = R_{th}. The resulting maximum power is P_{max} = \frac{V_{th}^2}{4R_{th}}. To find R_{th}, turn off the independent sources by shorting voltage sources and opening current sources, then combine the resistors in series and parallel. An ideal voltage source has zero internal resistance, so it contributes nothing to R_{th}. When the load is matched, the load current is I_L = \frac{V_{th}}{R_{th} + R_L} = \frac{V_{th}}{2R_{th}}. Note that this is half the short-circuit current, V_{th}/R_{th}, so a value equal to the short-circuit current is not the current at maximum power.