Electric current
From charge in motion to solving a complete circuit. This is where Kirchhoff's laws appear: the tool used to analyze any network, however complicated, without needing intuition.
01What current is
Electric current is charge in orderly motion. Its magnitude is the charge that crosses a cross-section of the conductor per unit of time.
In metals, the charges that move are the electrons, from negative to positive. But when the direction of current was defined, in the 18th century, this was not known: it was chosen to be from positive to negative. That convention is still in use today.
In practice nothing changes: all the formulas, all the symbols and all the circuits use the conventional direction, and you must always work with it. The actual direction only matters when explaining the internal operation of a semiconductor.
A surprising fact: electrons drift extremely slowly (a few millimeters per second in an ordinary copper wire). What travels fast is not the electrons but the electric field, at nearly the speed of light. That is why the lamp lights instantly even though the electron that left the switch takes hours to arrive.
Conditions for current to flow
- Free carriers: a conducting material.
- A potential difference to push them: a source.
- A closed circuit: if the path is open, nothing flows even though there is voltage.
02Resistance of a conductor
The resistance of a length of conductor is not an arbitrary figure: it depends on the material and on the geometry.
| Material | ρ (Ω·mm²/m) at 20 °C | α (1/°C) | Use |
|---|---|---|---|
| Silver | 0.0159 | 0.0038 | The best conductor. Relay contacts. |
| Copper | 0.0172 | 0.0039 | Wires, windings, PCB traces. |
| Aluminum | 0.0282 | 0.0039 | Overhead lines: more resistive but much lighter. |
| Tungsten | 0.0560 | 0.0045 | Lamp filaments. |
| Nickel-chromium | 1.10 | 0.0004 | Heating elements. |
| Constantan | 0.49 | 0.00002 | Standard resistors: almost no variation with temperature. |
| Carbon | 40 | −0.0005 | Ordinary resistors. Negative coefficient. |
Variation with temperature
A 2.5 mm² copper cable feeds an appliance 30 m away (60 m of conductor, out and back) with a current of 10 A.
- R = 0.0172 × 60 / 2.5 = 0.41 Ω
- Voltage drop: V = 10 × 0.41 = 4.1 V (1.9 % of 220 V: acceptable, the regulatory limit is 3 %)
- Power lost in the cable: P = I²·R = 100 × 0.41 = 41 W, all of it converted into heat
Had 1.5 mm² cable been used, the resistance would be 0.69 Ω, the drop 6.9 V (3.1 %, outside the standard) and the loss 69 W. This calculation is what determines the cross-section of the cables in an installation, and it is taken up in depth in Industrial Installations, in Year 6.
Tungsten has α = 0.0045 and an incandescent lamp operates at about 2500 °C. Its hot resistance is about 15 times higher than when cold. On switch-on, during the first milliseconds, a current 15 times the rated value flows — and that spike is what eventually breaks the filament. The same thing, for a different reason, explains why starting a motor draws several times its rated current.
03Kirchhoff's laws
Ohm's law handles a single resistor. For a network with several sources and several paths, two more principles are needed, which are really the conservation of charge and of energy applied to circuits.
At every node, the sum of the currents flowing in equals the sum of those flowing out.
Charge does not accumulate at any point in the circuit.
In every closed loop, the algebraic sum of the voltages is zero.
Going once around the loop brings you back to the same potential: what the source delivers, the loads consume.
04How to solve a circuit
The mesh method is systematic: it is always applied the same way and requires no intuition.
- Draw the circuit neatly and identify the nodes and the independent meshes.
- Assign a direction to the current in each mesh. It does not matter if you guess right: if the chosen direction is backwards, the result comes out negative and that is already the answer.
- Go around each mesh in a fixed direction and write its equation using these conventions:
- Source traversed from − to + (entering at the negative terminal): adds.
- Source traversed from + to −: subtracts.
- Resistor traversed along the assumed current: −I·R.
- Resistor traversed against it: +I·R.
- Solve the system of equations.
- Check with the first law at some node and with the power balance: the power delivered by the sources must equal the power dissipated by the resistors.
Left mesh: source E1 = 12 V with R1 = 2 Ω. Right mesh: source E2 = 6 V with R2 = 3 Ω. Shared central branch: R3 = 6 Ω. The three currents are to be found.
Top node: I₁ + I₂ = I₃
Left mesh: 12 = 2·I₁ + 6·I₃
Right mesh: 6 = 3·I₂ + 6·I₃
Substituting I₃ = I₁ + I₂ into the last two:
- 12 = 2·I₁ + 6·I₁ + 6·I₂ = 8·I₁ + 6·I₂
- 6 = 3·I₂ + 6·I₁ + 6·I₂ = 6·I₁ + 9·I₂
From the first, I₁ = 1.5 − 0.75·I₂. Substituting into the second: 6 = 9 + 4.5·I₂, hence
I₁ = 2.00 A · I₂ = −0.67 A · I₃ = 1.33 A
Check on the meshes: 2×2 + 6×1.33 = 4 + 8 = 12 ✓ and 3×(−0.67) + 6×1.33 = −2 + 8 = 6 ✓.
The negative sign of I₂ means that current flows the opposite way to what was assumed: the 6 V source is not delivering energy but receiving it — it is being charged, if it were a battery. That result is correct, and it is exactly what happens when a charger is connected to a battery.
Check by power: E1 delivers 12 × 2 = 24 W; E2 receives 6 × 0.67 = 4 W. The resistors dissipate 2×2² + 3×0.67² + 6×1.33² = 8 + 1.3 + 10.7 = 20 W. Balance: 24 W = 4 W + 20 W ✓.
05The Wheatstone bridge
It is the most elegant application of Kirchhoff's laws and is still at the heart of almost all industrial sensors.
It is a null method: no quantity is measured; you adjust until a difference vanishes. Null methods are much more accurate than deflection methods because they do not depend on the calibration of the indicator. Today the unbalanced bridge is used, measuring the voltage across the diagonal, in almost all resistive sensors: strain gauges in a scale, Pt100 temperature sensors, pressure sensors. The subject returns in Industrial Electronics in Year 6, combined with an instrumentation amplifier.
06Power, Joule's law and efficiency
The form P = I²·R is the most revealing for installations: the losses in a cable grow with the square of the current. Doubling the current quadruples the losses. This is the reason electrical energy is transmitted at high voltage: for the same power, more voltage means less current, and the losses fall with the square.
07Real generators
A real source is not an ideal voltage source: it has an internal resistance that makes the terminal voltage drop when current is drawn from it.
A new battery has r ≈ 0.2 Ω; a worn-out one can reach 5 Ω or more.
- With no load, the voltmeter (which draws almost no current) reads 1.5 V on both.
- With a 10 Ω load: the new one delivers 1.5 × 10/10.2 = 1.47 V; the worn-out one, 1.5 × 10/15 = 1.00 V.
That is why a battery is tested under load, never open-circuit. It is also the explanation for why a car's lights dim when starting: the starter motor draws 200 A and the internal drop in the battery pulls down the voltage of the whole system.
Maximum power transfer
An important and counterintuitive result: the power delivered to the load is a maximum when RL = r, that is, when the load equals the internal resistance of the generator.
Under that condition, half of the power is dissipated inside the generator: the efficiency is 50 %. Impedance matching is useful for signals (an antenna, a speaker, a transmission line), where what matters is transferring the largest possible signal. In power applications (an installation, a power supply) the opposite is sought: the internal resistance should be as small as possible compared with the load, to get good efficiency.
08In the lab
Build the circuit from the worked example (12 V with 2 Ω, 6 V with 3 Ω, common branch of 6 Ω; you can use resistances 100 times larger and a dual supply). Measure the three currents by inserting the ammeter, and the voltages across each resistor. Verify the 1st law at the node and the 2nd in the two meshes. Compare with the calculated values and explain the negative current.
With a coil of enameled copper wire of known cross-section: measure its resistance with the ohmmeter (or with the voltmeter-ammeter method if it is very low), measure the length, and calculate ρ by solving R = ρ·L/A. Compare with the table value. Repeat with a piece of the same wire of half the length and verify that the resistance is halved.
Measure the open-circuit voltage (E) of a battery. Then connect a known load RL and measure the terminal voltage V. Solve for the internal resistance:
Repeat with a new battery and a worn-out one and compare. This is the lab that explains once and for all why batteries are tested under load.
With three known resistors and a resistance decade box (or a multiturn potentiometer), build the bridge and adjust until the galvanometer reads zero. Calculate Rx and compare it with the direct ohmmeter measurement. The accuracy of the bridge is usually better. Then unbalance it on purpose by replacing one resistor with a thermistor and warming it with your hand: you see the voltage appear across the diagonal, which is the principle of every bridge sensor.
09Common mistakes
| Symptom | Usual cause |
|---|---|
| The system of equations does not balance | Signs applied incorrectly when traversing the meshes. Fix one direction of travel and stick to it around the whole mesh. |
| A current comes out negative and looks like a mistake | It is not: it means the current flows opposite to the assumed direction. The magnitude is correct. |
| The voltage measured at the source is lower than the rated one | Drop across the internal resistance, or wires that are too thin. |
| A distant appliance works poorly | Voltage drop in the line. Recalculate with R = ρL/A and increase the cross-section. |
| The calculated drop is half of what is measured | The out-and-back conductor was forgotten: the length is twice the distance. |
| The bridge never reaches zero | A branch is open, the source is disconnected, or the range of the decade box is not enough for that Rx. |
| The measured resistance changes while measuring | The component is heating up from the instrument's own current, or has a high temperature coefficient. |
10Self-assessment
Why is the conventional direction of current opposite to the actual motion of the electrons?
Because of a historical convention that predates the discovery of the electron. It is kept because all the formulas and symbols are built on it, and because for circuit analysis the result is identical.
A copper cable 100 m long with a 4 mm² cross-section: what is its resistance?
R = 0.0172 × 100 / 4 = 0.43 Ω. If it were to feed an appliance 100 m away, 200 m of conductor would be needed and the resistance would be 0.86 Ω.
At a node, 8 A and 3 A flow in, and one branch carries 4 A out. How much flows in the fourth branch?
8 + 3 = 4 + I → I = 7 A, leaving the node.
When solving a circuit, a current comes out as −2 A. What does it mean?
That its magnitude is 2 A but it flows in the opposite direction to the one assumed when setting up the equations. It is not an error: it is information. If that branch has a source, it means it is receiving energy instead of delivering it.
Why is electrical energy transmitted at high voltage?
Because the losses in the line are P = I²·R. To transmit the same power at a higher voltage, less current is needed, and since the losses depend on the square of the current, they fall drastically. Raising the voltage 10 times reduces the losses 100 times.
A Wheatstone bridge balanced with R1 = 100 Ω, R2 = 470 Ω and R3 = 220 Ω. What is the value of Rx?
Rx = R3 × R2/R1 = 220 × 470/100 = 1034 Ω.
A 9 V battery with r = 2 Ω feeds an 18 Ω load. What are the terminal voltage and the power in the load?
I = 9/(18+2) = 0.45 A. Vterm = 9 − 0.45×2 = 8.1 V. Pload = 8.1 × 0.45 = 3.65 W. The battery dissipates internally 0.45²×2 = 0.4 W.
When is impedance matching worthwhile, and when is it not?
It is worthwhile with signals (antennas, transmission lines, speakers), where the aim is to transfer the maximum possible power even if the efficiency is 50 %. It is not worthwhile in power distribution or in power supplies, where the internal resistance should be much lower than the load to get good efficiency.
Why does an incandescent lamp almost always burn out when it is switched on?
Because cold tungsten has a resistance about 15 times lower than at operating temperature. At the instant of switch-on a huge current spike flows, and that thermal and mechanical stress is what eventually breaks the filament.