Circuit Analysis

Node voltage analysis of resistive networks

Node voltage analysis solves a resistive network by choosing one node as the reference (ground) and treating the voltages of all other nodes as unknowns. The current through a resistor R between nodes a and b is (V_a - V_b)/R, so applying Kirchhoff's current law (KCL) at each non-reference node gives a linear equation in the node voltages. A network with N nodes therefore yields N-1 independent equations, which are solved by ordinary linear algebra.

Several rules make the setup faster and prevent common mistakes.

• Wire (ideal short) — a zero-resistance connection forces its two ends to the same potential, so they must be merged into one node. The current in the wire is unknown in advance and is absorbed into the KCL of the merged node.
• Current source — a current source injects a fixed current into its node, so it appears as a known constant on one side of that node's KCL equation.
• Single-wire sub-network — if a part of the circuit connects to the rest only through one wire, KCL at the junction forces the current in that wire to zero.
• Known branch current — if a resistor carries a known current I, its voltage drop is IR, which fixes the potential of one node directly.
• Supernode — a voltage source placed between two non-reference nodes makes them behave as one. Apply KCL around the enclosing surface and add the constraint that the difference of the two node voltages equals the source voltage.
• Dependent source — a controlled source is treated as an independent source whose value is an unknown multiple of a node voltage or branch current. Its controlling quantity is written in node voltages and added as an extra constraint.
• Choice of reference — never write KCL at the reference node, since its equation is redundant.

Example: a 12 V source feeds a node through a 4\,\Omega resistor, a 6\,\Omega resistor returns the node to ground, and a 3 A current enters the node from outside. KCL gives (12 - V)/4 + 3 = V/6. Multiplying by 12 gives 36 - 3V + 36 = 2V, so V = 14.4 V.

Two-port Y-parameters

A two-port network is described by its terminal voltages and currents, with each port current taken as entering the port. The short-circuit admittance (Y) parameters relate them through I_1 = y_{11}V_1 + y_{12}V_2 and I_2 = y_{21}V_1 + y_{22}V_2. Each parameter is found by shorting one port and exciting the other. Setting V_2 = 0 gives y_{11} = I_1/V_1, the input driving-point admittance, and y_{21} = I_2/V_1, the forward transfer admittance. Setting V_1 = 0 gives y_{12} = I_1/V_2 and y_{22} = I_2/V_2. All Y-parameters are measured in siemens (S).

Key points:

• Method of calculation — write nodal equations at the port nodes, eliminate internal nodes, and read the coefficients of V_1 and V_2 in each port-current equation.
• Reciprocity — a network built only from passive elements satisfies y_{12} = y_{21}. Dependent sources usually break this symmetry, so y_{12} \ne y_{21} signals a non-reciprocal network.
• Controlled sources between ports — a dependent source linking nodes of the two ports changes the transfer terms, and the coefficient is read from the nodal equation at the output port.
• Sign convention — currents are defined entering the ports, so a sign error in one port current changes the sign of the whole row.
• Series and shunt elements — a series admittance y between the ports gives \begin{bmatrix} y & -y \\ -y & y \end{bmatrix}, and a shunt admittance to ground adds only to the diagonal terms.
• Conversion to Z — [Z] = [Y]^{-1}. Once the Y-matrix is known, the impedance matrix follows by inversion, which is needed for Thevenin or maximum-power calculations.

Example: a 4\,\Omega series resistor has y = 0.25 S, so its Y-matrix is \begin{bmatrix} 0.25 & -0.25 \\ -0.25 & 0.25 \end{bmatrix} S.

Phasor impedance of inductor and capacitor

In sinusoidal steady state, a voltage or current is represented by a phasor, and passive elements become complex impedances. For an inductor, Z_L = j\omega L, so the current lags the voltage by 90^\circ. For a capacitor, Z_C = 1/(j\omega C) = -j/(\omega C), so the current leads the voltage by 90^\circ. The admittances are the reciprocals: Y_L = -j/(\omega L) and Y_C = j\omega C. Here \omega = 2\pi f in rad/s.

Useful rules:

• Multiplication by j — multiplying a phasor by j rotates it by +90^\circ, and multiplying by -j rotates it by -90^\circ. An inductive reactance therefore has a positive imaginary part, and a capacitive one has a negative imaginary part.
• Sine and cosine references — convert \sin(\omega t) to \cos(\omega t - 90^\circ) before reading a phasor angle. Otherwise the phase is off by 90^\circ.
• Branch current from admittance — the current is I = YV. Its magnitude is |Y||V| and its phase is the angle of Y plus the angle of V.
• Reactance at a frequency — X_L = \omega L and X_C = 1/(\omega C). In two branches with resistance and reactance, the phase of each branch current relative to the source is \tan^{-1}(X/R), so a condition on the sum of two phase angles becomes a product condition on the element values.

Example: a 2 H inductor at \omega = 3 rad/s has Z = j6\,\Omega. With I = 2\angle 0^\circ A, the voltage is V = IZ = 12\angle 90^\circ V, which leads the current by 90^\circ.