Resistance and Ohm's Law

Part 3 of 23 in Cobler's Electricity Fundamentals course. New here? Start with the course map.

Let's go back to the pantry socket from Part 2. We've got the kettle on one side and a phone charger on the other, both on the same 230 V. The kettle takes 13 A, and the charger only takes about a tenth of an amp. So if the push is the same, what decides the current? Well, that's the question behind Ohm's law, and to see where the answer comes from, we need to go back inside the metal.

Resistance

Remember Part 1's picture, a lattice of copper atoms soaked in a sea of free electrons. When a voltage pushes, each free electron speeds up a little, then hits a vibrating atom, loses that extra speed and starts again. Countless collisions a second hold the electrons to the steady 0.1 mm per second drift from Part 1 (OpenStax University Physics).

Each of those collisions does two things:

• It limits the flow. The more collisions on the way, the less current a given push can drive.
• It makes heat. Every collision hands some of the electron's energy to the lattice, and a lattice that vibrates harder is a warmer metal.

How strongly something opposes the flow of charge like this is what we call its resistance, and in a metal it comes from these collisions. We measure it in ohms (Ω), and one ohm is the resistance that lets 1 amp flow when 1 volt pushes.

Ohm's law

Now imagine a short wire on an adjustable supply, kept cool so its resistance stays put. If we raise the push from 1 V to 2 V, the current doubles. If we swap in a wire with twice the resistance, the current halves. So the current through a conductor is the voltage across it divided by its resistance, and that straight-line link is what we call Ohm's law. We can write it three ways:

$$V = I \times R \qquad I = \frac{V}{R} \qquad R = \frac{V}{I}$$

(V in volts, I in amps, R in ohms.)

Let's try it on the kettle element first. From Part 2, the kettle draws 13 A from 230 V:

$$R = \frac{V}{I} = \frac{230}{13} \approx 17.7\ \Omega$$

Those 18 or so ohms are what hold the current to 13 A.

Now let's try it on a person. Your body has a resistance too, and it changes hugely with your skin. OSHA, the US workplace safety agency, uses about 100,000 ohms for dry skin and about 1,000 ohms for wet skin in its worked examples (OSHA Construction eTool). At 230 V:

• Dry: 230 ÷ 100,000 = 0.0023 A, or 2.3 milliamps (mA), which is enough to feel.
• Wet: 230 ÷ 1,000 = 0.23 A, or 230 mA. That's inside the range OSHA lists as likely to cause ventricular fibrillation, where the heart quivers instead of pumping.

That's the number behind Part 2's warning, and it's why the safety device in Part 14 is built to trip at just 30 mA.

And the phone charger? Well, its electronics decide how much current it takes (Part 15 opens one up), so it doesn't follow Ohm's law in a straight line. Seen from the socket, a charger drawing 0.09 A acts like about 2,500 ohms (230 ÷ 0.09). So Ohm's law works well for wires, busbars and heating elements, but we can't count on it for electronics.

So far, every resistance came from measuring volts and amps. But how do we go from a kettle we can plug in and measure to a cable that hasn't been laid yet? Well, what if we could work out the resistance from what the wire is made of and how big it is?

Resistivity

It turns out we can. A conductor's resistance grows with its length, shrinks as it gets thicker, and depends on its material through a number called resistivity:

$$R = \frac{\rho L}{A}$$

Here ρ (the Greek letter rho) is the material's resistivity in ohm-metres, L the length in metres and A the cross-section in square metres. A longer wire means more collisions to get through, and a thicker wire gives the electrons more lanes side by side.

Resistivity also puts numbers on Part 1's conductors and insulators. These figures are at 20 °C, from OpenStax:

So hard rubber resists current at least \(10^{20}\) times more strongly than copper, which is the real gap between a conductor and an insulator.

Now let's do a real circuit. Take Part 2's illustrative level 12 pantry circuit, with 25 m of 2.5 mm² copper cable from the distribution board to the socket. The current goes out along the live and comes back along the neutral, so it passes through 50 m of copper:

$$R = \frac{(1.68 \times 10^{-8}) \times 50}{2.5 \times 10^{-6}} \approx 0.34\ \Omega$$

That's about 7 milliohms per metre of conductor. With the 13 A kettle running, Ohm's law gives the voltage used up inside the cable, 13 × 0.34 ≈ 4.4 V. That means the kettle actually sees about 225.6 V instead of 230 V, and that loss is what we call voltage drop.

Temperature changes the answer, too. When a metal warms, its atoms vibrate harder, the electrons collide more often, and the resistance rises. For copper it climbs about 0.39% per degree Celsius (from the same OpenStax table), so a cable running at 70 °C has about 20% more resistance than at 20 °C. In a cable that's a nuisance, but in a sensor it's useful, because a part whose resistance tracks temperature is a thermometer, and Part 20 uses exactly that.

Joule heating

Now for the heat. Every amp we force through a resistance turns some electrical energy into heat. In Part 2 we saw that power is volts times amps, and across a resistance, Ohm's law says the voltage is I × R. So let's put the two together:

$$P = V \times I = (I \times R) \times I = I^2 R$$

If we swap I for V ÷ R instead, we get a second form, handy when we know the voltage:

$$P = \frac{V^2}{R}$$

This is called Joule heating, and it's really just the collisions from the first section, added up. The square is the part to remember. You'd naturally guess that twice the current makes twice the heat. But the voltage across the resistance doubles too (Ohm's law), so push times flow goes up four times, which means double the current gives four times the heat.

Let's check both forms on the kettle: 13² × 17.7 ≈ 2,990 W, and 230² ÷ 17.7 ≈ 2,990 W. That's Part 2's 3 kW, and all of it is heat we actually want, from a nichrome element chosen because it resists well and survives red heat.

Now the pantry cable from earlier:

• At 13 A: 13² × 0.34 ≈ 57 W, spread over 50 m of copper. That's about 1.1 W per metre, so it's barely warm.
• At 26 A: 26² × 0.34 ≈ 230 W, four times as much for double the current.

Joule heating is the first useful effect in this course, in kettles, water heaters and irons, and it's also the first problem, with hot cables and lost voltage. But we can turn the problem into protection, too. Say we put a short, thin metal strip in series with the circuit, chosen so that its own I²R heating melts it at a set current, before the cable's insulation gets too hot. That deliberate weak link is what we call a fuse. The thermal trip in a circuit breaker uses the same heating to bend a metal strip, as Part 14 shows.

Cable sizing and voltage drop

So how do engineers use all this? Well, every cable is sized so that its I²R heat stays inside its insulation's temperature limit, and its I × R voltage drop stays inside a set budget.

Heat comes first. PVC-insulated cable is rated for a conductor temperature of 70 °C, which is the figure the current-rating tables of IEC 60364-5-52 assume for PVC (IEC 60364-5-52 preview), the wiring standard of the International Electrotechnical Commission (IEC). So a cable has to be thick enough that its own heating, plus the room around it, never pushes it past that. Bundled cables and cables in hot ceiling voids shed heat badly, so they're allowed less current.

Voltage drop comes second. Wiring rules only let the cables use up a few percent of the supply voltage, and a long run can go over that even when the cable is thick enough for the heat. Both checks come straight out of \(R = \rho L / A\). Cable Size and Voltage Drop: Why Distance Matters works through the limits and the derating factors.

Bolted joints in the main switchboard

Now let's point a thermal camera inside a busy main switchboard, like the one we looked inside in Part 1. Most of it glows one calm colour, with bars, lugs and breakers all within a few degrees of each other. But one bolted joint on a feeder shows up 40 °C hotter than the joints beside it (an illustrative reading). Nothing has tripped, and every meter reads normal. The cause is a few thousandths of an ohm.

So what's going on in that joint? A bolted joint clamps two copper surfaces together. Even polished copper only touches at microscopic high spots, and the bolt's force crushes them into enough contact area for current to cross easily, so a healthy joint measures in millionths of an ohm. But over the years, joints loosen. The copper warms every afternoon and cools every night, vibration works at the bolt, and some bolts were never tightened properly. The contact area shrinks, the surfaces oxidise, and the resistance climbs. That makes a loose joint basically a small resistance in the path of a large current, and Ohm's law turns it into a heater.

Let's work it out with illustrative numbers for a feeder joint carrying 400 A. A healthy joint of 0.00001 Ω makes 400² × 0.00001 = 1.6 W, which is nothing. A loose joint of 0.002 Ω (2 milliohms) makes:

$$P = I^2 R = 400^2 \times 0.002 = 320\ \text{W}$$

That's about the heat of three old 100 W filament bulbs, packed into one bolt. Now you might expect a fault that hot to show up on the switchboard's meters, and to be honest that's what I'd have guessed too. But it doesn't. The voltage across the joint is only 400 × 0.002 = 0.8 V, about a third of a percent of 230 V, which is too small for any meter to flag. So the loose joint hides from every electrical reading and only shows up as heat, and that's why maintenance teams scan switchboards with thermal cameras, under load.

Load matters because of the square. If the afternoon peak lifts the feeder from 400 A to 600 A, the joint's heat rises 2.25 times, to 720 W. Heat also speeds up oxidation, which raises the resistance, which makes more heat. Left alone, the joint can run away until insulation fails or an arc starts.

The maintenance testing specification of the InterNational Electrical Testing Association (NETA) classes a temperature difference above 15 °C between similar components as a "major discrepancy; repair immediately" (IR Info). Our 40 °C joint is well past that. A competent person then isolates the board, cleans and remakes the joint, and tightens it to the maker's torque. A Trip Is a Clue covers turning findings like this into a maintenance routine.

A complete path

Now, every number in this part assumed the current already had a path, out along the live, through the element or the joint, and back along the neutral. If we break that path anywhere, there's no current at all, so V = I × R gives us nothing and the kettle never boils.

A building also needs to turn each load on and off, like the lights at 8 a.m., the kettle at the press of a button, or a fan when a room warms up. So how do we build that path, and open and close it on purpose? That's Part 4: Electric Circuits and Switches, where the switch becomes the first control in this course.

This is Part 3 of 23 in Cobler's Electricity Fundamentals course. Previous: Volts and Amps. Next: Electric Circuits and Switches.

Cobler builds CobiNeural, a platform that tracks the current on each metered feeder and flags readings that drift from normal, so to see what your switchboards are carrying, talk to us.