DC Circuit Analysis: From Ohm's Law to Thevenin

Lesson 1 · DC Circuit Fundamentals & the Analysis Workflow

Level: Beginner

• Recall voltage, current, resistance and power with correct units and symbols.
• Distinguish ideal voltage and current sources and apply the passive sign convention.
• Follow a repeatable workflow for analysing any DC circuit and verify it with a power balance.

Before any theorem, you need the vocabulary. Voltage (V, volts) is energy per unit charge — the "push" available between two points. Current (I, amperes) is the rate charge actually flows, and by convention it is the direction positive charge would move (opposite to electron flow). Resistance (R, ohms) opposes that flow, and power (P, watts) is the rate at which electrical energy is transferred: P = V·I. A circuit is just a closed path for current, made of sources that deliver energy (an ideal voltage source holds a fixed voltage no matter the current; an ideal current source holds a fixed current no matter the voltage) and elements that absorb or store it, such as resistors.

The one convention that keeps every calculation consistent is the passive sign convention: if you define current entering the positive terminal of an element, then P = V·I is the power the element absorbs (a resistor always absorbs). If current enters the negative terminal — as with a source delivering energy — P = V·I comes out positive for power delivered. Hold onto this: it is how you'll later prove a whole circuit balances.

Analysis then becomes a routine. You almost always solve a DC circuit in the same five moves, and you check the answer by confirming that the total power delivered by the sources equals the total power absorbed by the resistors — if it doesn't, a sign or a value is wrong somewhere.

1. Label — name the nodes, choose a reference (ground), and mark a current direction and a voltage polarity on each element.
2. Reduce — combine series and parallel resistors where you can (Lesson 4).
3. Write equations — apply KCL at nodes and KVL around loops (Lessons 2–3), or use Thevenin/Norton/superposition (Lessons 5–7).
4. Solve — solve the resulting algebra for the unknown currents and voltages.
5. Verify — check the power balance: Σ P(delivered) = Σ P(absorbed).

Lesson 1 quiz

Lesson 2 · Kirchhoff's Current Law (KCL)

Level: Beginner → Intermediate

• Identify nodes/junctions and assign current directions with a sign convention.
• Write and solve KCL equations, including when a computed current is negative.
• Apply KCL to multi-branch nodes as a foundation for nodal analysis.

A node is any point where two or more elements connect, and electrically all the points directly joined by a wire share the same node. Kirchhoff's Current Law is simply the statement that charge cannot pile up at a node: whatever current flows in must flow out. Written as a sum, the algebraic sum of all currents at a node is zero. The practical form is "sum of currents entering = sum of currents leaving." Both are the same equation; pick one and stick to it.

Directions are a choice you make before solving. Draw an arrow on each branch; if the number you solve for is negative, the real current simply runs the other way. That is not an error — it's the sign convention doing its job. This is why KCL is the engine behind nodal analysis: you write one KCL equation per unknown node voltage, expressing each branch current in terms of the node voltages, and solve the system.

Start simple and grow the complexity. A two-branch node is trivial; a node with five branches is the same rule with more terms. The only discipline required is to keep "entering" and "leaving" straight, which the sign convention handles for you once you commit to it.

Lesson 2 quiz

Lesson 3 · Kirchhoff's Voltage Law (KVL)

Level: Beginner → Intermediate

• Define a loop, a traversal direction, and a consistent voltage sign convention.
• Write KVL equations for single and multi-loop (mesh) circuits.
• Solve a two-mesh circuit for its mesh currents and branch voltages.

Kirchhoff's Voltage Law is the conservation of energy applied to a loop: if you walk once around any closed path, the net change in electric potential must be zero. So the sum of all voltage rises equals the sum of all voltage drops (or the algebraic sum is zero). A source that you traverse from − to + is a rise; a resistor you traverse in the direction of its current is a drop of I·R.

Like KCL, you choose a direction before solving. Pick a loop current direction (say clockwise); as you traverse each element, assign a sign based on whether you meet a rise or a drop. If a solved current comes out negative, it simply flows the other way. For multi-loop circuits, assign one mesh current to each independent loop and write one KVL equation per loop — where a resistor is shared by two meshes, its current is the difference of the two mesh currents.

This "mesh" approach is the workhorse for hand analysis of planar circuits. It turns a tangle of wires into a small system of linear equations, which is exactly the kind of thing that scales nicely — and it sets up the nodal/mesh duality you'll meet again in Thevenin and Norton.

Lesson 3 quiz

Lesson 4 · Series & Parallel Circuit Analysis

Level: Beginner → Intermediate

• Combine series and parallel resistors into a single equivalent resistance.
• Apply voltage and current division to find individual values without full nodal analysis.
• Reduce mixed series-parallel networks step by step and verify with KVL/KCL.

Many networks reduce to a single equivalent resistor, and that single number unlocks everything. In series the same current flows through each element, so resistances simply add: R_eq = R1 + R2 + … . In parallel the same voltage appears across each element and the currents split, so the conductances add: 1/R_eq = 1/R1 + 1/R2 + … . For two parallel resistors this collapses to the handy R_eq = (R1·R2)/(R1+R2) — always smaller than the smaller resistor, which is a good sanity check.

Once you've reduced a network, two shortcuts do most of the work. The voltage divider gives the voltage across one series resistor: V_x = V_total · (R_x / R_total). The current divider gives the current in one of two parallel branches: I_x = I_total · (R_other / (R_x + R_other)) — note the "other" resistor on top, because current prefers the easier (lower-resistance) path. These avoid writing full KCL/KVL systems for the most common topologies.

Real problems mix both: a parallel pair in series with another resistor, and so on. The method is always "reduce from the inside out, then work back out": collapse the most deeply nested parallel or series group, repeat until one equivalent remains, find the source current, then propagate voltages and currents back through the stages using the dividers. KVL and KCL are your verification at every step.

Lesson 4 quiz

Lesson 5 · Thevenin's Theorem

Level: Intermediate

• Reduce any linear two-terminal network to a single voltage source in series with a resistor.
• Find the Thevenin voltage (open-circuit) and Thevenin resistance (sources deactivated).
• Use the equivalent to find the voltage and current delivered to any load.

Thevenin's theorem is the most powerful simplification in DC analysis: any linear network of sources and resistors, viewed from two terminals, is equivalent to a single ideal voltage source Vth in series with a single resistor Rth. You don't need to solve the whole internal network every time you change the load — you characterise the network once, then every load is a trivial two-element circuit.

Vth is the open-circuit voltage at the terminals (the voltage you'd measure with nothing connected). Rth is the equivalent resistance seen at the terminals with all independent sources deactivated: voltage sources become short circuits (0 V) and current sources become open circuits (0 A). Dependent sources, if present, stay in place. Because the network is linear, these two numbers — one open-circuit, one resistance — fully describe its external behaviour.

Once you have Vth and Rth, connecting a load R_L gives I_L = Vth/(Rth + R_L) and V_L = Vth·R_L/(Rth + R_L) — just a voltage divider. The same idea underlies maximum-power transfer (best when R_L = Rth) and is the bridge to Norton, which describes the same network as a current source in parallel with a resistance.

Lesson 5 quiz

Lesson 6 · Norton's Theorem

Level: Intermediate

• Reduce a linear two-terminal network to a current source in parallel with a resistance.
• Find the Norton (short-circuit) current and Norton resistance.
• Convert between Thevenin and Norton forms and analyze a load in either.

Norton's theorem is the dual of Thevenin's: the same two-terminal network is equally well described by an ideal current source I_N in parallel with a resistance R_N. Where Thevenin thinks "a voltage that sags under load," Norton thinks "a current that splits between the internal resistance and the load." Either form gives identical results at the terminals — they are two faces of the same equivalent.

I_N is the short-circuit current: the current that would flow if you connected the two terminals with a wire. R_N is found exactly as Rth — deactivate the independent sources and read the resistance at the terminals — so R_N = Rth. The conversion between the two is direct and always valid for a linear network: I_N = Vth / R_N and Vth = I_N · R_N, with the resistance unchanged.

With a load R_L in parallel with R_N, the load current follows from the current divider: I_L = I_N · R_N/(R_N + R_L), and V_L = I_L · R_L. Which form is "easier" depends on the problem: Norton is natural when the network looks current-driven or when you're finding a short-circuit current; Thevenin is natural when you're finding an open-circuit voltage. Knowing both, and how to flip between them, makes load analysis almost automatic.

Lesson 6 quiz

Lesson 7 · Superposition Theorem

Level: Intermediate

• Analyze a multi-source circuit by letting each independent source act alone.
• Deactivate sources correctly (voltage → short, current → open).
• Recognize that superposition applies to voltages and currents, not to power.

The analysis of a multi-source circuit can be effectively conducted by allowing each independent source to operate in isolation. In this context, it is essential to deactivate the sources appropriately; specifically, a voltage source should be transformed into a short circuit, while a current source should be treated as an open circuit. It is also crucial to understand that the principle of superposition is applicable to voltages and currents, but not to power.

Superposition takes advantage of linearity, positing that in a circuit containing multiple independent sources, the voltage or current at any given point can be expressed as the algebraic sum of the contributions from each source, while all other independent sources are deactivated. Dependent sources, however, should remain in the circuit. Thus, one must solve a simplified problem for each individual source, subsequently integrating the components while adhering to their respective polarities and directions.

The term "algebraic" functions significantly in this context, as the contribution from one source may counterbalance that from another; therefore, it is imperative to combine signed values rather than mere magnitudes. Furthermore, this theorem is constrained by specific limitations; it holds validity exclusively for linear quantities. Voltage and current demonstrate linear relationships with respect to the sources, in contrast to power, which is characterized by a quadratic relationship (as illustrated by P = I²R or V²/R). Consequently, one cannot derive the total power dissipated in a resistor by simply summing the powers contributed by each source in isolation; rather, it is essential to first ascertain the total current or voltage and subsequently calculate the power.

The superposition principle is particularly beneficial when a network presents challenges for reduction, yet allows for the analytical convenience of isolating single sources. Furthermore, it fosters the development of intuition required for nodal and mesh analysis, where, in essence, the effects of multiple sources are combined incrementally within the analysis.

Lesson 7 quiz

Lesson 8 · Integrated Analysis & Assessment

Level: Intermediate — challenge

• Choose the most efficient method for a given problem.
• Chain reduction, Thevenin/Norton and superposition on a multi-source network.
• Pass a comprehensive assessment covering all seven previous lessons.

The skill that separates "can run a formula" from "can analyzee a circuit" is choosing the right tool. A few heuristics cover almost everything: if you need the voltage or current at two terminals of a complicated network (or want to try many loads), use Thevenin/Norton. If several independent sources act and the circuit resists reduction, use superposition. If the network is mostly series and parallel, just reduce and use the dividers. If you want a node voltage with a clear reference, write KCL (nodal); for a loop or mesh, write KVL. Most real problems combine two or three of these.

Comprehensive assessment

Five questions across the module. Multiple-choice questions are scored; the calculation questions are self-checked with Show answer.

Score: 0 / 3