Circuit Analysis Tutoring

Circuit analysis is the first real electrical engineering course, and it sets the pace for the rest of the degree. The rules are few: Ohm's law, Kirchhoff's current and voltage laws, and the element equations for resistors, capacitors, and inductors. The hard part is choosing a method. Nodal or mesh? Superposition or a Thevenin equivalent? Time domain or phasors? Guessing wrong usually costs points at the setup stage, not in the arithmetic. Sessions focus on reading a circuit, picking the shortest method, and checking the answer before moving on.

Circuits I usually covers resistive networks, op amps, and first-order and second-order transients. Circuits II moves on to AC steady state, complex power, frequency response, and Laplace-domain analysis, with two-port networks near the end of many syllabi. Sessions follow your textbook, whether that is Nilsson and Riedel, Alexander and Sadiku, or Hayt, along with your professor's problem sets. We also go through lab results and simulation runs when your course includes them.

A typical circuit problem, worked

A 24 V source is in series with a 6 Ω resistor. At the node after that resistor, a 12 Ω resistor goes to ground and a 4 Ω resistor goes to output terminal a. Terminal b is ground. Find the Thevenin equivalent at a and b, then the current and power in a 24 Ω load.

Open-circuit voltage first. With nothing connected at a and b, no current flows in the 4 Ω resistor, so $V_{th}$ is the voltage across the 12 Ω resistor: $\dfrac{24 \times 12}{6 + 12} = 16\ \text{V}$.

Now turn the source off by replacing it with a short. Looking in from a, the 6 Ω and 12 Ω resistors are in parallel: $\dfrac{6 \times 12}{18} = 4\ \Omega$. Add the 4 Ω in series and $R_{th} = 8\ \Omega$.

Connect the 24 Ω load. $I = \dfrac{16}{8 + 24} = 0.5\ \text{A}$, so the load voltage is 12 V and the load power is $0.5^{2} \times 24 = 6\ \text{W}$.

Check against the original circuit. The node after the 6 Ω resistor sits at 14 V, and 14 V across $4 + 24 = 28\ \Omega$ gives the same 0.5 A. Maximum power would reach the load at $R_L = R_{th} = 8\ \Omega$, where $P = \dfrac{V_{th}^{2}}{4 R_{th}} = 8\ \text{W}$.

Where students usually get stuck

• Sign errors in nodal and mesh equations, usually from an unlabeled current direction or voltage polarity
• Turning off dependent sources when finding $R_{th}$. They stay in the circuit, so use a test source or the open-circuit and short-circuit method instead.
• Initial conditions in transients. Capacitor voltage and inductor current cannot jump, so they carry across the switching instant.
• Mixing peak and RMS values in phasor and power problems
• Using the op amp virtual short when there is no negative feedback path