The previous lesson ended with Ohm's law, enough for a single resistor or a tidy chain in series or parallel. Real circuits are rarely that tidy, and once one splits into several loops that share components, Ohm's law alone stalls. This lesson adds the two principles that crack any circuit, and the methods built on them. It ends with two components that do what a resistor cannot: instead of burning energy as heat, a capacitor and an inductor store it and give it back, and storing takes time, which is how the clock first enters a circuit.
Two laws that hold any circuit together
Two laws, written by Gustav Kirchhoff in 1845, are enough between them to analyse any circuit. Neither is exotic; both are bookkeeping, one for charge and one for energy, each insisting nothing goes missing.
Kirchhoff's Current Law (KCL) is the bookkeeping for charge. At any junction, the current flowing in equals the current flowing out. Nothing piles up and nothing vanishes, so every coulomb that arrives must leave by some route. Picture water at a pipe junction: every litre a second in must come out a branch, because it has nowhere else to go.
Kirchhoff's Voltage Law (KVL) is the bookkeeping for energy. Go round any closed loop, adding the rises and subtracting the drops, and the total is zero. Voltage is energy per coulomb, and that energy is not the same at every point: it is highest right after the battery's positive terminal and lower after each resistor, as the resistors turn the difference into heat. The battery lifts every coulomb back up by exactly as much as the resistors dropped it. The sum is therefore zero, not because nothing happened, but because the source covered every drop exactly.
Each resistor has a drop across it: the potential is higher where current enters than where it leaves, and that difference drives the current through. In the diagram, V₁ and V₂ are those two drops, each across its own resistor; the source gives a rise of ℰ volts, the resistors share it as drops, and rise minus drops is zero. The sign is simple: walk the loop in any direction; crossing a component from − to + is a rise, from + to − is a drop. Two sources opposing each other each count toward the total with opposite signs, and only the net drives current.
KCL and KVL look simple but they are powerful. Every method that follows, from mesh analysis to Thévenin equivalents, is just one or both of them applied.
Splitting a voltage, splitting a current
Two configurations appear so often that they deserve names of their own.
A voltage divider solves a common problem: you have a supply and need a lower fraction of it. Put two resistors in series across the supply and tap the output at the junction between them. With only one resistor the far end connects straight to ground, giving 0 V always; the second resistor gives you a point between the full drop and zero, and where that point sits depends on the ratio.
There is one path, so the same current I flows through both:
The output sits directly across R₂, so it equals the drop there:
The larger resistor always takes the larger share, in direct proportion. Check the extremes: R₂ much larger means Vout nears Vin; R₁ much larger means Vout sinks toward zero.
This assumes almost nothing is drawn from the output. Connect a load and it sits in parallel with R₂, shrinking the effective bottom resistance and pulling Vout below the simple fraction. The Thévenin method later is how you handle that properly.
A current divider is the parallel version: a total current arrives at a junction and you want to know how it splits. Both branches span the same two nodes, so both see the same voltage V, and by Ohm's law I₁ = V/R₁ and I₂ = V/R₂. The total is:
Solving for V and substituting back into I₁ = V/R₁:
R₂ in the numerator for I₁ looks backwards but it is right: the easier a branch is, the more current it takes, so what governs I₁ is how easy the other branch is. Equal resistors split equally; a near-zero R₂ steals almost all the current and leaves almost nothing for R₁. That is why a short circuit across a load is dangerous: it drops the parallel resistance to near zero and redirects nearly all the current through the fault.
Solving a circuit through its loops
Ohm's law and the divider rules handle simple circuits but run out of road fast. Once a circuit has more than one loop sharing branches, you hit a chicken-and-egg problem: to use KVL on a loop you need the current in every resistor, but those currents are what you are trying to find. There is nowhere to start.
Mesh analysis breaks the deadlock with a trick. Instead of the real branch currents, imagine one current circulating around each loop, like a car on a racetrack; name them I₁, I₂, and so on, all running clockwise. They are not the real wire currents but a convenient set of unknowns. The payoff is at the branches: a branch in one loop carries that loop's current, while a branch shared by two loops has both running through it in opposite directions, so its real current is the difference.
Now write KVL once around each loop, each drop being current times resistance. A resistor in one loop drops that loop's current times its resistance; a shared resistor drops its resistance times the difference of the two loop currents, since that is what really flows through it. Each loop gives one equation, and with as many equations as unknowns you solve the lot.
The circuit below has two meshes. The 18 V source and R₁ sit in mesh 1. R₂ is the shared branch. R₃ and R₄ sit in mesh 2.
KVL for mesh 1, clockwise. The source rises 18 V. Each drop is current times resistance: R₁ (3 Ω) carries only I₁, dropping 3I₁; R₂ (6 Ω) carries the difference I₁ − I₂, dropping 6(I₁ − I₂):
KVL for mesh 2, clockwise, no source. R₂ (6 Ω) now carries I₂ − I₁, dropping 6(I₂ − I₁); R₃ (2 Ω) and R₄ (4 Ω) carry only I₂:
From the second equation, I₁ = 2I₂. Substituting into the first:
The actual current through R₂ is I₁ − I₂ = 1.5 A in the direction of I₁.
Each loop costs one equation, so mesh analysis is quickest when a circuit has few loops. The two here gave two equations, light work by hand.
Solving a circuit through its junctions
Nodal analysis attacks the same deadlock from the other end: instead of loop currents, work with the voltage at each junction. Pick one junction as ground (0 V); every other voltage is measured from it. Then at each remaining junction apply KCL: all the currents leaving must sum to zero.
The circuit below has one free node, A. The 12 V source feeds into A through R₁ (2 Ω); from A, R₂ (6 Ω) and R₃ (3 Ω) drain to ground. Call the unknown voltage at A simply Vₐ.
Write KCL at A as the sum of every current leaving, using (this node − far node)/R for each branch:
The first term comes out negative when Vₐ < 12 V, meaning current actually enters A from the source. The sign handles the direction automatically; no need to guess before you start.
Multiplying through by 6:
Branch currents: I₁ = (12 − 6)/2 = 3 A into the node, I₂ = 6/6 = 1 A out, I₃ = 6/3 = 2 A out. KCL: 3 = 1 + 2. ✓
Just as each loop costs one equation in mesh, each non-ground junction costs one here. To choose between them: count loops, count free junctions, pick the smaller. This circuit had one free junction against mesh's two loops, so nodal wins.
Two kinds of source
Every source in a circuit is one of two types.
A voltage source fixes the voltage across its terminals and lets the current be whatever the circuit demands. A battery is the common example: connect any load and the voltage stays constant while the current adjusts. An ideal voltage source has zero internal resistance; a real one (a battery or bench supply) has a small internal resistance, which is why its voltage sags slightly under heavy load.
A current source is the mirror: it fixes the current through the circuit and lets the voltage settle where it needs to. Connect any load and the current stays constant while the voltage adjusts. Transistors in their active region and solar cells behave this way. An ideal current source has infinite internal resistance.
The distinction matters most when killing sources to find a Thévenin equivalent. A voltage source set to zero volts is a plain wire, so you replace it with one. A current source set to zero amps passes nothing, so you replace it with a break: an open gap in the wire. Same logic, opposite result.
Any network as one source and one resistor
Here is the problem. You have a circuit and want to try different loads on it: a motor, a sensor, a lamp. Each swap means re-solving the whole circuit.
Thévenin's theorem says you never have to. Any circuit of sources and resistors, however complex, looks from the outside like exactly one battery in series with one resistor. Find those two values once and you can predict what any load sees without touching the original circuit again.
The two values are Vth and Rth, and they describe the circuit as seen from a specific pair of points: the terminals a and b, where the load connects. You choose them. Pick any two points where a load might attach. Importantly, a different choice of terminals on the same circuit gives different Vth and Rth; they are not properties of the circuit alone, but of the circuit as seen from those two points.
In the example, the load connects across R₂. So a is the top of R₂ and b is the bottom. The source and R₁ sit behind those terminals as the network.
Vth is the open-circuit voltage: what you measure across a and b with nothing bridging them. The network is still alive inside; the source drives current through its own loops, resistors drop voltage. Vth is what the network offers at those terminals with no load drawing current. In the example the source drives current through R₁ and R₂ in series even with no load, and since a and b sit across R₂, Vth is R₂'s share: 16 V.
Rth is the resistance the network presents at a and b. To find it, kill all sources first; a live source would drive current into a meter and spoil the reading. Killing means setting what it forces to zero: a voltage source becomes a wire; a current source becomes an open gap. With sources dead, look in from a and b and calculate what the remaining resistors form.
One thing to check: replacing the source with a wire does not short a to b. The wire goes where the source was, inside the network, with resistors still between it and the terminals. In the example, killing the 24 V source leaves R₁ running from terminal a, through the wire, back to terminal b, in parallel with R₂. So Rth is 4 Ω, not zero.
The whole network then collapses to one battery and one resistor, V_th in series with R_th, shown on the left below. Norton's theorem gives the same box the other way up, a current source in parallel with R_th, shown on the right.
With a load R_L connected across terminals a and b, the Thévenin circuit is a voltage divider: R_th plays the role of R₁ and R_L the role of R₂.
The maximum power the circuit can deliver follows from this. A very small R_L causes most of V_th to drop across R_th, leaving little for the load. A very large R_L reduces the current to near zero, so power at the load also approaches zero. The maximum sits between, at R_L = R_th: matching the load to the Thévenin resistance delivers the most power to the load. That is the most power reaching the load, not the most the source puts out, since a smaller R_L draws more current but wastes more of it inside R_th.
Norton's form describes the same network differently: a current source I_N in parallel with R_th, rather than a battery in series. R_th is identical in both.
The two are equivalent because they agree at both extremes. Open the terminals and all of I_N flows through R_th, the only path, putting I_N · R_th across a and b, which equals Vth. Short the terminals and all of I_N takes the short, so the short-circuit current is I_N itself. Agreeing at both ends, they agree for every load between, and convert directly:
To find I_N without first finding Vth: short terminals a and b and calculate the current through the wire. Some circuits give I_N more easily than Vth, so it is worth knowing both routes.
Taking the sources one at a time
When a circuit has more than one source, superposition lets you handle them one at a time. Kill all but one (voltage source becomes a wire, current source becomes an open gap) and find what that single source contributes. Repeat for each source, then add all contributions, respecting direction. The sum is the real answer.
It works because a resistive circuit is linear: double a source and its effect exactly doubles. Ohm's law and Kirchhoff's laws contain no squared terms or products, so sources never interfere; each contributes independently.
One limit: superposition only works for voltage and current, not power. Power is I²R, which is nonlinear, so you cannot find power in each sub-circuit and add them.
Example. A series loop contains a 12 V source, a 6 V source opposing it, and a 2 Ω resistor.
Kill the 6 V source (short it). Only 12 V drives the loop:
Kill the 12 V source (short it). Only 6 V drives the loop, in the opposing direction:
The actual current is I = I' − I'' = 6 − 3 = 3 A. Direct check: net EMF = 12 − 6 = 6 V across 2 Ω = 3 A. ✓
Superposition is especially useful when sources run at different frequencies, which no single combined calculation can handle.
The next part covers the two components that store energy rather than burning it, capacitors and inductors, and the behaviour that storage produces.
A worked example, Thévenin from start to finish
A circuit has a 24 V source, R₁ = 6 Ω in series with the source, and R₂ = 12 Ω connected from the junction of R₁ to the negative terminal. A load R_L = 4 Ω is connected across R₂. Find the Thévenin equivalent seen by the load and the voltage across it.
Step 1. Find V_th. Remove R_L. With the terminals open, no current leaves them, but the source still drives current through R₁ and R₂ in series. The drop across R₂ is the open-circuit voltage V_th.
Step 2. Find R_th. Switch the 24 V source off, replacing it with a wire. Trace from the load terminals: R₂ sits directly across them, and R₁ runs from one terminal back through the shorted source to the other, a second path between the same two points. So R₁ and R₂ are in parallel, which is what the load sees. Not a short, since both resistors are still there.
Step 3. Connect the load. The Thévenin equivalent is a 16 V source in series with 4 Ω. With R_L = 4 Ω:
The Thévenin model makes it trivial to find V_L for any load without re-solving the original circuit each time.
Step 4. Convert to Norton form. The Norton equivalent uses the same R_th = 4 Ω, now in parallel with a current source.
To check, reconnect R_L = 4 Ω to the Norton form. It parallels R_th = 4 Ω for 2 Ω, and 4 A × 2 Ω = 8 V, matching the Thévenin result. Both forms describe the same network; pick whichever is handier.
Where this comes from
- Kirchhoff's laws, mesh and nodal analysis, Thévenin and Norton theorems, superposition: Alexander and Sadiku, Fundamentals of Electric Circuits, ch. 2 to 5.
- Supporting worked examples: Boylestad, Introductory Circuit Analysis, ch. 8 to 10.
Check yourself
1. A node has three wires. I₁ = 6 A flows in, I₂ = 2 A flows in. What does I₃ carry, and in which direction?
By KCL: 6 + 2 = 8 A in, so I₃ = 8 A out.
2. A 9 V source, a 3 Ω resistor, and a 6 Ω resistor are in series. What is the voltage drop across each, and does KVL hold?
I = 9/(3+6) = 1 A. V₃ = 3 V, V₆ = 6 V. Sum = 9 V = supply. KVL: 9 − 3 − 6 = 0 ✓
3. A voltage divider has R₁ = 18 kΩ and R₂ = 2 kΩ, fed from 10 V. What is Vout?
Vout = 10 × 2/(18 + 2) = 1 V.
4. Two resistors in parallel: R₁ = 10 Ω and R₂ = 40 Ω. Total current is 5 A. How much flows through each?
I₁ = 5 × 40/(10+40) = 4 A. I₂ = 5 × 10/(10+40) = 1 A. The smaller resistor takes the larger share.
5. A node A connects to a 9 V source through R₁ = 3 Ω, and to ground through R₂ = 9 Ω. What is the voltage at A?
KCL at A: (Vₐ−9)/3 + Vₐ/9 = 0. Multiply by 9: 3Vₐ − 27 + Vₐ = 0 → 4Vₐ = 27 → Vₐ = 6.75 V.
6. A Thévenin equivalent has V_th = 6 V and R_th = 2 Ω. What load draws maximum power, and what is that power?
Maximum power when R_L = R_th = 2 Ω. I = 6/(2+2) = 1.5 A. P = 1.5² × 2 = 4.5 W.
7. A series loop has a 10 V source and a 4 V source opposing it, with a 3 Ω resistor. Use superposition to find the current.
Kill 4 V source: I' = 10/3 A. Kill 10 V source: I'' = 4/3 A, opposing. Total: I = 10/3 − 4/3 = 2 A. Check: (10−4)/3 = 2 A. ✓