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Part 2 of 6

Back to wires: Direct current

May 29, 2026 electrical

What is moving inside a conductor

Direct current flows steadily in one direction, the kind a battery, a solar panel, or a phone charger delivers. Its counterpart, alternating current, reverses many times a second, and we get to it in a later part. Five quantities carry the whole of direct current, charge, current, voltage, resistance, and power, with one law tying the first four together, and they all begin with what moves inside the wire.

A conductor can be copper, aluminium, or gold; what they share is a particular kind of bonding. In an insulator like rubber or plastic, each electron is locked in a bond between two specific atoms and goes nowhere. In a metal, the outer electrons are not owned by any one atom; they belong to the whole lattice and drift freely through it. So a metal wire is packed with electrons wandering every which way, going nowhere in particular because there is nothing to make them all go the same way.

Each electron carries a fixed amount of charge, the property that makes particles push and pull on each other. Charge is not a substance; it is a property of the electron, the way mass is a property of any object. We write it Q and measure it in coulombs. One coulomb is the combined charge of about 6 × 10¹⁸ electrons. That large unit exists because a single electron's charge is far too small to work with; the coulomb is just a pile big enough to be useful.

The rate at which charge flows

Get all those free electrons moving the same way and you have a current. Current is not the electrons themselves; it is the rate at which charge passes a point. Stand at one cross-section of the wire and measure how many coulombs go past each second. That is the current, I, measured in amperes. One ampere is one coulomb per second. Turned around: multiply the current by the time and you get the charge that passed.

\[I = \frac{Q}{t} \qquad \text{1 A} = \text{1 C/s}\]

One convention needs settling before we go further. In every diagram and formula, current is drawn flowing out of the positive terminal, around the circuit, and into the negative one, as if positive charge were moving. In a metal it is really the electrons going the other way: the source pushes an excess of electrons onto the negative terminal and leaves a deficit at the positive one, so electrons flow from the full side toward the empty one, from minus to plus. The convention was fixed in the 1700s before electrons were discovered, and by the time the mistake was found, every instrument and textbook already used it. The arithmetic works out the same either way, so we use the convention and move on.

One more thing that puzzles people. If electrons only creep a few millimetres a second, why does a lamp light the instant you flip the switch? Because no electron needs to travel from the switch to the lamp. The wire is already packed with electrons end to end. Push one in at one end and one pops out the other at once, the way a marble pushed into a full tube sends another out the far end immediately. The push travels at close to the speed of light; the electrons barely shuffle.

What drives the charge

The charge is already in the wire; every free electron carries it. The source does not add electrons; it creates a pressure difference between the two ends of the conductor, and that pressure gets the existing electrons all moving the same way. We call that pressure voltage, write it V, and measure it in volts.

Voltage is always a difference between two points, never a property of one point alone. A 12 V battery means there are 12 volts between its terminals. A 230 V socket means 230 V between live and neutral. You measure it across a component, across two ends, because "the voltage at one point" is as meaningless as "the height of a place" without saying height above what.

The volt is defined by energy. One volt means one joule is exchanged per coulomb of charge moved between the two points. A 12 V battery gives each coulomb 12 joules. A 230 V socket gives each coulomb 230 joules. That is the energy each coulomb picks up from the source and will spend in the circuit.

In real life: a phone charger runs at 5 V, a car battery at 12 V, household mains at 230 V in Europe. What actually harms you is not voltage directly but the current it drives through your body. Your skin has a high resistance, and at 12 V that resistance limits the current to a harmless fraction of a milliamp. At 230 V, the same resistance allows enough current to be fatal. More voltage means more current through the same resistance, which is the next thing to look at.

What opposes the flow

Voltage is the push; resistance is what pushes back. Every material opposes the movement of electrons to some degree. As electrons drift through the lattice they collide with atoms and lose a little energy at each collision. That opposition is resistance, written R and measured in ohms, symbol Ω.

The water analogy holds well here. A wide, short pipe passes a lot of flow for a given pressure. A long, narrow one passes far less. A conductor works the same way: short and thick means low resistance, long and thin means high. A long thin extension lead running a heavy load gets warm and runs the load feebly, because its resistance limits the current and the energy lost to collisions turns into heat in the wire itself rather than being delivered to the load. A short fat cable does the same job easily.

Resistance depends on three things: the material, the length, and the cross-section. Copper has lower resistance than aluminium for the same dimensions. Longer wire has more resistance; wider wire has less. The cable sizing rules you already know from the trade follow directly from this: too thin a cable means too much resistance, too much heat, and not enough current reaching the load.

A short circuit is the extreme case: a wire bridging two points with no load, resistance near zero. The current is not limited by a useful load but only by the wire itself, which is why it reaches thousands of amps and the wire melts or the fuse blows. A starter motor drawing 150 A is not a short because its resistance, though small at 0.08 Ω, is doing real work: converting electrical energy into mechanical torque.

The law that ties all three together

Three quantities, one relationship. Ohm's law says the current through a conductor equals the voltage across it divided by its resistance.

\[I = \frac{V}{R}\]

More voltage drives more current; more resistance allows less. You will most often use it rearranged as V = IR, which tells you the voltage drop across a component given the current through it and its resistance. The third form, R = V/I, finds the resistance from a measurement.

A word on measuring. Voltage is measured across a component: one probe on each terminal, from outside. The voltmeter has very high internal resistance, so it draws almost no current out of the circuit and leaves the voltage it is reading essentially undisturbed. Current is measured through a component: the meter must be in the current path. You break the circuit at one side of the component, insert the ammeter in the gap, and current flows through the meter as it does through the component. The ammeter has very low internal resistance, so it adds almost no resistance to the circuit. In practice, a clamp meter measures the magnetic field around the wire instead, so you never need to break the circuit at all.

Ohm's law explorer

Drag the levers below. The line shows I = V/R for the resistance you choose, and the dot marks the operating point at the chosen voltage.

Raising R tilts the line flatter, so the same voltage produces less current.

What the circuit does with the energy

Each coulomb picks up energy from the source and spends it in the circuit. Power is the rate at which that happens. Voltage is joules per coulomb; current is coulombs per second. Multiply them and the coulombs cancel, leaving joules per second, which is power, P, measured in watts.

\[P = V\,I \qquad \text{1 W} = \text{1 J/s}\]

One watt is one joule per second. A 2 kW heater uses 2000 joules every second, roughly the output of ten people cycling hard. Your body at rest produces about 80 W of heat, which is why a packed room gets warm.

Because Ohm's law locks V, I, and R together, you can write power in two other ways when you only know two of the three quantities.

\[P = V I = I^2 R = \frac{V^2}{R}\]

The I²R form is the one that matters most on the job. Power rises with the square of the current: double the current and the power quadruples. The reason is that raising the current does two things at once: more coulombs arrive each second, and each coulomb falls through a larger voltage drop, since that drop is V = IR. The two effects multiply, giving the square. This is why an overloaded cable heats fiercely and why cable sizing is a safety matter, not tidiness.

Power is the rate; energy is the running total. Run P watts for time t and the energy adds up as W = Pt. Joules are too small for household quantities, so the meter counts kilowatt-hours instead: one kilowatt-hour is a thousand watts held for one hour, the unit on your bill.

\[W_{\text{kWh}} = \frac{P_{\text{W}}}{1000}\times t_{\text{hours}}\]

Why a real source sags under load

So far the battery has been a perfect push that never weakens. A perfect source, by definition, has no internal resistance, so its terminal voltage equals its EMF regardless of how much current you draw. Real sources are less generous.

A real battery has a small resistance inside it, in the materials and connections of the cell. We call it internal resistance, written r. It sits in series with the rest of the circuit, so every ampere drawn must push through it first. The battery's true voltage with nothing connected and no current flowing is the electromotive force, or EMF, written ℰ. Draw current I and Ohm's law takes its cut: a drop of Ir is lost inside the battery and never reaches the terminals. What remains is the terminal voltage U.

\[U = \mathcal{E} - I\,r\]

A healthy car battery has r around 0.02 Ω. When the starter pulls 150 A, the internal drop is 150 × 0.02 = 3 V, pulling the terminal voltage from 12 V to 9 V. Every branch in the circuit, including the headlights, now sees 9 V instead of 12 V, so the headlights dim. Without internal resistance a perfect source would hold 12 V across all branches no matter what the starter drew; the dimming would not happen. The dimming is the signature of r.

An old battery with r = 0.4 Ω cannot deliver 100 A at all: the formula gives a negative terminal voltage, which tells you the model has broken down and the battery collapses before reaching that current. High internal resistance is why a weak battery fails on a cold morning.

Two ways to wire components together

There are only two basic ways to connect components, and anything more complex is a mix of the two.

Series means end to end on a single path. The current has nowhere else to go, so it passes through every component in turn. The same current flows everywhere in the loop. Each component takes a share of the voltage proportional to its resistance; the higher the resistance, the larger its share. Because each component adds one more obstacle to the only path, resistances simply add.

\[R_\text{series} = R_1 + R_2\]

Parallel means side by side, both ends connected to the same two points. The current now has a choice of routes and divides between them. Because both components connect between the same two points, both see the same voltage. Each branch draws its own current independently: I₁ = V/R₁ and I₂ = V/R₂. The total current is their sum, I = V/R₁ + V/R₂, which means the equivalent resistance satisfies:

\[\frac{1}{R_\text{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} \qquad R_\text{parallel} = \frac{R_1 R_2}{R_1 + R_2}\]

The result that catches people out: a parallel combination is always lower than the smallest resistor in it. The reason is that every additional branch adds another route for current. More routes means more total current for the same voltage, which means lower resistance. Two 100 Ω resistors in parallel give 50 Ω; add a third 100 Ω branch and it drops to 33 Ω.

A circuit you can watch

Here a real source with EMF ℰ and internal resistance r drives two resistors, switchable between series and parallel. The moving dots are conventional current, their speed proportional to the current in each wire. In series the same current runs through both resistors; in parallel it splits at the node, with more flowing through the smaller resistance. Raise the internal resistance and the terminal voltage droops further below the EMF.

What it costs to run

Power is instantaneous, but the bill depends on energy used over time. Choose a load and its hours per day to see the monthly cost at €0.22 per kWh, the approximate European residential tariff.

Combine the resistors yourself

Set the two resistors and switch between series and parallel. A series combination always exceeds either resistor alone; a parallel one is always smaller than the smaller of the two.

A worked example, start to finish

A 12 V battery with internal resistance r = 0.5 Ω feeds a 5.5 Ω heating element for 2 hours. Find the current, the terminal voltage, the power delivered to the element, and the energy used.

The internal resistance and the element sit in series, so the EMF drives current through their combined resistance.

\[I = \frac{\mathcal{E}}{R + r} = \frac{12}{5.5 + 0.5} = 2\ \text{A}\]

The terminal voltage is what the element actually sees: the EMF minus the drop lost inside the battery.

\[U = \mathcal{E} - I r = 12 - (2 \times 0.5) = 11\ \text{V}\]

Use I²R for the power delivered to the element, since we know the current and the element's resistance.

\[P = I^2 R = 2^2 \times 5.5 = 22\ \text{W}\]

Energy is power times time. Express it in Wh first, then convert if needed.

\[W = P\,t = 22 \times 2 = 44\ \text{Wh} = 0.044\ \text{kWh}\]

The one weakness of direct current

DC electrons drift slowly in one direction, always. That steady drift carries metal ions with it. A metal ion that moves leaves a gap where it came from and deposits where it arrives. The end where current leaves the metal into the surrounding medium, soil or water, corrodes. On a grid carrying DC for decades, this acts along every buried conductor.

Alternating current reverses direction fifty times a second in Europe. The electrons jiggle back and forth instead of marching. Metal ions migrate one way for half a cycle, then back the other way for the next half, so the net migration over a full cycle is nearly zero. The corrosion nearly cancels.

The electrons in a house wire sway by about the width of a bacterium. Yet full kilowatts arrive instantly. The energy is not carried by the electrons; it travels in the electric and magnetic fields surrounding the conductor, which propagate at close to the speed of light. The electrons are the medium, not the messenger, the way air is the medium for sound without itself travelling from speaker to ear. That is why so little electron movement delivers so much power.

The cancellation of corrosion is one reason the world's grids run on alternating current. The rest of the story comes in a later part.

Where this comes from

  • Ohm's law, the power relations, and series and parallel combinations: Alexander and Sadiku, Fundamentals of Electric Circuits, ch. 1 to 2.
  • The definitions of charge, current and EMF, with worked DC examples: Hughes, Electrical and Electronic Technology.

Check yourself

1. A 9 V supply is connected across a 3 Ω resistor. What current flows?

I = V/R = 9/3 = 3 A.

2. Two 100 Ω resistors are connected in parallel. What is the equivalent resistance?

Rₑq = (1/100 + 1/100)⁻¹ = 50 Ω, half of one, because a parallel combination is always smaller than the smallest resistor in it.

3. A real 12 V source has internal resistance 1 Ω and delivers 3 A. What is its terminal voltage?

U = ℰ − Ir = 12 − 3 × 1 = 9 V.

4. A 20 Ω and a 30 Ω resistor are connected in series across a 10 V supply. What current flows, and what is the voltage across the 30 Ω resistor?

Rₑq = 20 + 30 = 50 Ω, so I = 10/50 = 0.2 A. The voltage across the 30 Ω resistor is V = 0.2 × 30 = 6 V. The remaining 4 V sits across the 20 Ω resistor, and the two drops add back to 10 V as expected.

5. A 4 Ω resistor carries 3 A. What power does it dissipate?

P = I²R = 3² × 4 = 36 W.

6. A 2 kW heater runs for 3 hours a day. How much energy does it use per day in kWh, and what does that cost at €0.22 per kWh?

W = 2 × 3 = 6 kWh. Cost = 6 × 0.22 = €1.32.

7. A current of 2 A flows for 5 minutes. How much charge passes a given point?

Q = I × t = 2 × 300 = 600 C.

8. A lamp draws 0.26 A from a 230 V supply. What is its power rating?

P = VI = 230 × 0.26 ≈ 60 W.

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