You switch the kettle on and, for an instant, the kitchen lights dip. Nothing is wrong. The kettle draws about 10 amps, those amps flow through the same wires that feed the lights, and every wire has a little resistance, so a few volts quietly go missing on the way. The lights see the leftover.
Feel your phone charger after an hour: warm. Some of the energy it handles never reaches the battery; it turns into heat in the charger's own resistances. If a 20 W charger is 90% efficient, 2 W of heat is the price. Both mysteries have the same answer, and it fits on a postcard. That postcard is circuit analysis.
Who takes it: every electrical and computer engineer, usually in year 2, and many mechanical and aerospace students in a shorter form.
What it's for: predicting the voltage and current anywhere in a circuit, before building it.
Comes after: Physics II: Electricity & Magnetism (34/07) and Differential Equations (34/03).
Leads to: Electronics (34/35), Signals & Systems (34/37) and Power Systems & Electric Machines (34/39). Computer scientists meet it again in Computer Architecture (34/25), where every gate is a circuit.
Honest line: you won't be able to design a power supply after this. You will see why a thin wire carrying a lot of current gets hot, why a sensor reading sags when you connect it to something, and why every radio has a coil and a capacitor in it.
Picture water in a pipe. Voltage is the pressure pushing, current is how much flows past a point each second, and resistance is how narrow and rough the pipe is. A narrow pipe needs more pressure for the same flow. Georg Ohm published in 1827 that for metals the relation is proportional.
The second line is the one a graduate carries: heat in a resistance grows with the square of the current. Double the current and you get four times the heat. A 60 W load at 5 V needs 12 A. Push that through a cable of 0.2 Ω and the cable burns 12² × 0.2 = 28.8 W, nearly half the power you meant to deliver. The same 60 W at 20 V needs 3 A and wastes 3² × 0.2 = 1.8 W, 3%. This is why fast chargers negotiate their voltage upward: a fixed power, a lower current, less heat in the cable.
60 W at 5 V needs 12 A and the cable burns 28.8 W.
Here the voltage at the load is 5 V; the cable heat is I² × R. The last line of the readout shows the same job at 9, 12, 20 and 48 V.
Gustav Kirchhoff published two rules in 1845, and every circuit calculation is built from them. Both are bookkeeping. Current law: charge does not pile up at a junction, so the current in equals the current out. Voltage law: go once around any loop and the voltage rises and falls add up to zero, because a point can only have one voltage.
Two resistors in series share one current, so their resistances add and the voltage splits in proportion. Two in parallel share one voltage, so the currents add, and the combined resistance is always smaller than either. Series is a queue; parallel is extra checkout lanes.
With 5 V across two equal 10 kΩ resistors, the middle is 2.5 V. That is how a sensor, a volume knob or a battery-level gauge turns a resistance into a voltage. Now the catch every beginner meets once: connect something to the middle and it joins in parallel with R2. A 10 kΩ load turns the lower half into 5 kΩ, and the output falls to 1.67 V, a third lower. Measuring or loading a node changes it.
Unloaded: 2.5 V.
Input 5 V. Slide the load toward R2's size and watch the second bar sink. A load ten times larger than R2 costs only a few percent.
Real circuits are not one divider. For the rest, the method is mechanical and it is the one every program uses. Pick a ground, call it 0 V. Give every other junction a voltage, an unknown. Then write Kirchhoff's current law at each: the currents leaving through every resistor add to zero (or to whatever a current source pushes in). Each current is a voltage difference divided by a resistance, so every equation is linear.
That gives one equation per unknown, which is a system you solved in Linear Algebra (34/04). Fifty nodes means fifty equations in a matrix, and the circuit simulator that designs a chip solves exactly that, millions of times. The matrix is symmetric, with each node's total conductance on the diagonal and the minus-conductances between neighbours off it: you can write it by looking at the picture.
In the circuit below, with Vs = 12 V and a 2 mA source, the two equations (in mS and mA) are 2A − 0.5B = 12 and −0.5A + 1.5B = 2. They give A = 6.91 V and B = 3.64 V. The readout checks that the currents leaving each node cancel.
R1 = 1 kΩ, R2 = 2 kΩ, R4 = 1 kΩ. Make R3 tiny and A and B become one node; make it huge and they separate.
Here is the idea that makes big circuits manageable. Take any pile of sources and resistors with two terminals sticking out. From outside, the whole pile behaves like one ideal voltage source Vth in series with one resistance Rth. Léon Thévenin published this in 1883 (Hermann von Helmholtz had it earlier, in 1853), and the Norton version, a current source in parallel, is the same thing turned over. You find Vth by asking what the terminals read with nothing attached, and Rth by asking how much the voltage falls when you draw current.
The wall socket is the clearest example. Its Thévenin voltage is 230 V (or 120 V) and its Thévenin resistance is the wiring back to the transformer, plausibly a few tenths of an ohm. Flicker standards test appliances against a reference loop impedance of about 0.4 Ω plus a small reactive part (IEC 61000-3-3, as listed by test-equipment makers); here we use 0.5 Ω. A 2.4 kW kettle is 22 Ω, so it draws 230 ÷ 22.5 = 10.2 A. The share of the 230 V that reaches the kettle is 22 ÷ 22.5, so the kettle sees about 225 V and everything else on that wire loses the missing 5 V while it runs. That is the dimming lights.
There is a second lesson hiding here. How much power can a source with internal resistance give to a load? Too small a load and the current is large but the load's voltage is tiny; too large and the voltage is high but the current vanishes. The best is the middle:
At that match half the power is wasted inside the source. A radio receiver or a speaker amplifier chasing every last bit of a weak signal accepts this trade. A power station never does: it wants a load far larger than its own resistance so that almost nothing is lost, which is also why the mains is built with the lowest resistance the money allows.
The curve is power as a share of the maximum, against the load size from one hundredth to one hundred times Rth.
Resistors have no memory. A capacitor stores charge, and its current is proportional to how fast its voltage changes (I = C dV/dt). An inductor stores energy in a magnetic field, and its voltage is proportional to how fast its current changes (V = L dI/dt). Put one of them with a resistor and Kirchhoff's law becomes a differential equation, the first one you solved in Differential Equations (34/03). The answer is the exponential everyone learns once:
The capacitor reaches 63% of the way in one τ, 95% in three, 99% in five. A 10 kΩ resistor with 100 µF has τ = 1 second. A capacitor charged to 5 V stores ½CV² = 1.25 mJ, and the resistor turns exactly as much into heat while charging it, whatever the resistance. That half-and-half loss is also why switching chips cost power (see Electronics, 34/35).
Feed the same RC a square wave and it becomes a filter. When τ is much shorter than a half-period the output follows the wave. When τ is much longer the capacitor never catches up and you get a nearly flat level: slow changes pass, fast ones are averaged away.
Input is 5 V. In square-wave mode the output settles to a steady swing between V·a/(1+a) and V/(1+a), where a = e−half period ÷ τ.
The mains and every radio signal are sine waves, and a sine wave through an RLC circuit comes out a sine wave of the same frequency, just scaled and shifted in time. That changes the problem: you no longer need a differential equation, only the size and the shift. Engineers write that pair as a single arrow, a phasor, and the arrow's behaviour has a name, impedance.
Every resistor, coil and capacitor now obeys Ohm's law with a complex number. A coil's reactance ωL grows with frequency; a capacitor's 1 ÷ ωC shrinks. In a series RLC they point in opposite directions, so at one frequency they cancel completely. There the impedance is only R, the current peaks, and the circuit rings at its natural frequency:
With L = 10 mH and C = 1 µF, f0 is 1,592 Hz. Make R only 10 Ω and Q is 10: the circuit answers loudly within about 160 Hz of that note and ignores everything else. This is a radio tuner, and it is also a filter: take the output across the resistor and you have a band-pass (the tuner); across the capacitor, a low-pass; across the coil, a high-pass. It is the mechanical mass-and-spring of Vibrations (34/17) in another medium, with Q playing the part of the damping ratio.
Fixed: L = 10 mH, C = 1 µF. In the phasor picture VR points right (in phase with the current), VL up, VC down; the dashed arrow is the source. At f0 the up and down arrows are equal and the dashed arrow lies flat.
Circuit analysis is bookkeeping with two laws: charge does not pile up at a node, and voltage adds to zero around a loop. From them come dividers, node equations and the Thévenin trick that reduces any two-terminal mess to a source and a resistor. Heat grows with the square of the current, which is why power moves at high voltage and why the lights dim when the kettle starts. Add a capacitor and the answer is an exponential with a time constant; add a coil and a sine wave and it is impedance, resonance and filters.
Try this: next time a charger or power brick feels warm, touch the cable too. Then watch a lamp on the same circuit as a kettle, a hair dryer or a toaster when it switches on. If you can see a dip, you have just measured a Thévenin resistance with your eyes.
Sources. Ohm's law, Kirchhoff's laws (1845), node analysis, Thévenin's and Norton's theorems, RC and RLC behaviour are standard textbook material (Nilsson and Riedel; Alexander and Sadiku; MIT 6.002). The usual textbook attributions are used: Ohm 1827, Kirchhoff 1845, Helmholtz 1853, Thévenin 1883. The supply reference impedance (about 0.4 Ω + j0.25 Ω for a 230 V single-phase test) is from IEC 61000-3-3 as listed by test-equipment makers (Kikusui and Pacific Power datasheets); other documents list different network values, so the 0.5 Ω used here is illustrative. Computed in node: every number in the readouts and worked examples (the 28.8 W and 1.8 W cable losses, the 2.5 V and 1.67 V divider, node voltages 6.909 V and 3.636 V, the 225 V kettle, maximum power and 50% efficiency, 3.11 V and 1.89 V RC swing, f0 = 1,591.5 Hz and Q = 10).