Ohm and Kirchhoff's laws in Year 3

If you are in Year 3 of a vocational school (istituto professionale) with the Maintenance and technical assistance track (or similar, with an electrical focus), the electrical/electronic technology subject of your track is starting these days. The exact name changes from school to school, but the first topics are almost always the same: voltage, current, resistance, Ohm’s law, and series and parallel circuits. The first test arrives in 5 to 8 weeks.

In Year 3 the heavy specialist subjects appear, and it is the jump in technical difficulty that students feel most sharply: even before the calculation, the type of request changes. Knowing the formula is no longer enough: you have to be able to apply it inside a diagram, with the right units, and often the same skill comes back in the laboratory, assessed with a grid of practical tasks instead of a single grade. With the school year split into two terms and the school report at the end of January, this first test weighs quite a lot on the first term.

Here I find it useful to be direct: what follows is tuned to the practical approach of vocational schools. No complex network theorems, no Thévenin, only what you really need to read a diagram, calculate what it asks and not get the conversions wrong.

Voltage, current, resistance: what they really are

The analogy with water in a pipe is imprecise in some details, but to get your bearings in the first days it works:

Voltage (or potential difference, $V$ , in volts) is what “pushes” the charges to move, like the pressure that pushes water through a pipe.
Current ( $i$ , in amperes) is the amount of charge that crosses a section of the circuit per unit of time, like the flow rate of water.
Resistance ( $R$ , in ohms, symbol $Ω$ ) is how much the material opposes the passage of current, like the friction inside the pipe, or a narrowing.

The analogy stops here: there is no need (and it is not entirely correct) to push it further, but it is enough to fix the role of the three quantities before bringing in the formulas.

The first mistake: units and their multiples and submultiples

Even before Ohm’s law, the place where the most marks are lost is the conversion between multiples and submultiples:

1 mA = 1 0^{- 3} A, 1 kΩ = 1 0^{3} Ω, 1 MΩ = 1 0^{6} Ω

A resistor of $4.7 kΩ$ is $4700 Ω$ , not $4.7 Ω$ . A current of $150 mA$ is $0.150 A$ . Rule of thumb: convert everything to base units (V, A, $Ω$ ) before touching any formula. Anyone who calculates with the submultiples still in gets the result wrong by a factor of 1000 and often does not notice, because the sum “works” numerically anyway.

Ohm’s law

V = R i

The voltage across a component equals the resistance times the current flowing through it. From here you get the other two forms, $i = V / R$ and $R = V / i$ , but they should not be learned as three separate formulas: it is a single relation between three quantities, and which one you isolate depends only on what the exercise asks.

Power and the Joule effect

A component carrying current dissipates energy as heat. This is the Joule effect, the same principle behind an electric heater or the filament of an old light bulb. The power dissipated:

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

Three forms of the same formula: you use the one that contains the data you have, without first having to work out all three.

Resistors in series and in parallel

The criterion for recognising them in a diagram, before even looking at the values:

in series: the same electrons pass through them one after the other, so the current is the same in both; there is no node between them.
in parallel: they are connected to the same two points (nodes) of the circuit, so the voltage across them is the same.

The formulas for the equivalent resistance:

series: R_{eq} = R_{1} + R_{2} + \dots

parallel: \frac{1}{R _{eq}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \dots

For just two resistors in parallel there is a widely used shortcut that avoids going through the reciprocals:

R_{eq} = \frac{R _{1} \cdot R _{2}}{R _{1} + R _{2}}

Two plausibility checks that almost always save you: the equivalent resistance of a series is always greater than the largest in the group; that of a parallel is always smaller than the smallest. If your result does not respect this, you have inverted a formula or forgotten to take the reciprocal at the end.

The voltage divider

Two resistors in series connected to a generator divide the total voltage in proportion to their values. With $R_{1}$ and $R_{2}$ in series across a generator $V_{s}$ , the voltage across $R_{2}$ is:

V_{R_{2}} = V_{s} \cdot \frac{R _{2}}{R _{1} + R _{2}}

It is the most used set-up in the laboratory to obtain a voltage lower than the one available, and one of the most frequent questions in the test because it requires reasoning in proportion, not just applying Ohm once.

Kirchhoff’s two laws

You need them when a circuit cannot be reduced entirely to simple series and parallels, or when there are several generators.

First law (nodes). At a node, a point where three or more conductors meet, the sum of the currents flowing in equals the sum of the currents flowing out. No charge accumulates at a node: what goes in, comes out.

\sum i_{in} = \sum i_{out}

Second law (loops). Going round a closed loop (a ring of the circuit) in a direction of your choice, the algebraic sum of the voltages (generators and drops across resistors) is zero. The sign convention: assign a direction of travel to the loop; each generator contributes $+ E$ if you cross it from the $-$ terminal to the $+$ terminal, and $- E$ in the opposite direction; each resistor contributes $- R i$ if you go through it in the same direction as the current flowing through it, and $+ R i$ in the opposite direction.

\sum E - \sum R i = 0

The most delicate point is not the formula, it is the consistency of the signs: choose a direction for the loop and for every unknown current before writing the equation, and do not change them halfway.

In the laboratory: how to connect the instruments

This is where theory and practical tests meet, and it is material for the assessment grid as much as for the written test.

The voltmeter measures a voltage, so it must be connected in parallel with the component whose voltage drop you want to read: it must “see” the same potential difference, without diverting the current of the branch.
The ammeter measures a current, so it must be connected in series in the branch you want to measure: the current must flow through it entirely, otherwise it does not measure all of it.
The ohmmeter is always used with the circuit switched off (and, if possible, with the component disconnected from the rest): it measures resistance by sending a small test current itself, and if the circuit is powered that current adds to the one already present, giving a wrong reading and, in some cases, damaging the instrument.

Memorising them is not enough if you then swap them in the lab: repeat them while looking at the diagram you are wiring up, not just reading them on a sheet of paper.

Four worked exercises

1. Mixed circuit: currents in every branch

A generator of $V = 24 V$ supplies a resistor $R_{1} = 4.0 Ω$ in series with the parallel of $R_{2} = 6.0 Ω$ and $R_{3} = 12 Ω$ . Calculate the equivalent resistance, the total current and the current in each branch.

First the innermost block, the parallel:

R_{23} = \frac{R _{2} \cdot R _{3}}{R _{2} + R _{3}} = \frac{6.0 \cdot 12}{18} = 4.0 Ω

Check: $4.0 Ω$ is smaller than the smallest, $6.0 Ω$ . Now the series:

R_{eq} = R_{1} + R_{23} = 4.0 + 4.0 = 8.0 Ω

i_{tot} = \frac{V}{R _{eq}} = \frac{24}{8.0} = 3.0 A

This current flows through all of $R_{1}$ (it is in series). The voltage across $R_{1}$ :

V_{1} = R_{1} i_{tot} = 4.0 \cdot 3.0 = 12 V

Across the parallel there remains $24 - 12 = 12 V$ , the same across $R_{2}$ and $R_{3}$ :

i_{2} = \frac{12}{6.0} = 2.0 A, i_{3} = \frac{12}{12} = 1.0 A

Check: $i_{2} + i_{3} = 3.0 A$ , exactly the total current: Kirchhoff’s first law applied at the node where the parallel rejoins.

2. Power dissipated in the same circuit

With the values of the previous exercise, calculate the power dissipated in each resistor and check the balance against the power supplied by the generator.

P_{1} = V_{1} i_{tot} = 12 \cdot 3.0 = 36 W

P_{2} = 12 \cdot 2.0 = 24 W, P_{3} = 12 \cdot 1.0 = 12 W

P_{1} + P_{2} + P_{3} = 36 + 24 + 12 = 72 W

The power supplied by the generator, $P_{gen} = V i_{tot} = 24 \cdot 3.0 = 72 W$ , matches: all the electrical energy supplied is dissipated in the three resistors, nothing is lost and nothing is created. It is the quickest way to check you have not made a slip along the way.

3. Voltage divider

A generator of $V_{s} = 12 V$ supplies $R_{1} = 2.2 kΩ$ and $R_{2} = 3.3 kΩ$ in series. Calculate the voltage across $R_{2}$ and the current in the circuit.

Conversion first of all: $R_{1} = 2200 Ω$ , $R_{2} = 3300 Ω$ .

V_{R_{2}} = V_{s} \cdot \frac{R _{2}}{R _{1} + R _{2}} = 12 \cdot \frac{3300}{5500} = 7.2 V

i = \frac{V _{s}}{R _{1} + R _{2}} = \frac{12}{5500} \approx 2.2 mA

Note the current in milliamperes: in circuits with resistors in kilohms the typical currents are small, of the order of milliamperes. It is a good plausibility check while you work.

4. A loop with two generators: Kirchhoff’s second law

In a closed loop there are two generators, $E_{1} = 9.0 V$ and $E_{2} = 3.0 V$ , connected with opposite polarity, and two resistors in series $R_{1} = 100 Ω$ and $R_{2} = 50 Ω$ . Calculate the current in the loop and the voltage drop across each resistor.

Choose a direction of travel in agreement with $E_{1}$ : going round the loop in that direction, $E_{1}$ contributes $+ E_{1}$ (crossed from $-$ to $+$ ), while $E_{2}$ , having opposite polarity, contributes $- E_{2}$ . The two resistors, traversed in the same direction as the unknown current $i$ , give $- R_{1} i$ and $- R_{2} i$ :

E_{1} - E_{2} - R_{1} i - R_{2} i = 0

i = \frac{E _{1} - E _{2}}{R _{1} + R _{2}} = \frac{9.0 - 3.0}{100 + 50} = \frac{6.0}{150} = 0.040 A = 40 mA

The voltage drops:

V_{R_{1}} = R_{1} i = 100 \cdot 0.040 = 4.0 V, V_{R_{2}} = 50 \cdot 0.040 = 2.0 V

Check with the second law: $V_{R_{1}} + V_{R_{2}} = 6.0 V = E_{1} - E_{2}$ . The negative sign of $E_{2}$ in the equation is not a whim: it comes from the direction in which you crossed it relative to the chosen direction of travel, and if you change that, the sign with which it enters the final balance must change consistently too.

The mistakes that cost the most marks

Leaving submultiples unconverted. $kΩ$ , $mA$ , $μ A$ : convert them to base units before applying any formula, not after.
Confusing series and parallel by looking at the drawing instead of the structure. The criterion is “same current” or “same voltage”, not “how they are drawn on the page”: a diagram drawn in an unusual way can hide a parallel that looks like a series.
Forgetting to take the reciprocal at the end when calculating a parallel. With $1/ R_{eq} = 1/6 + 1/12$ the result is $1/ R_{eq} = 0.25$ , so $R_{eq} = 4.0 Ω$ , not $0.25 Ω$ . The “smaller than the smallest” check always catches it.
Getting the sign convention wrong in Kirchhoff because the direction of travel was not fixed before writing the equation.
Using the ohmmeter on a live circuit, or swapping the series and parallel connections of ammeter and voltmeter in the lab: here the mistake costs more than a mark, it can damage the instrument.

How to tell whether you are ready

The quickest way is not to reread your notes: it is to take any diagram from the book, cover the text of the exercise and try to answer two questions before calculating anything: “which resistors are in series and which in parallel?” and “what direction does the current have in each branch?”. If you answer those two without hesitating, the calculation almost does itself.

For vocational-school electrical technology there is not yet a dedicated practice test on the site: you will find the catalogue of practice tests in the practice tests for high school section (questions in Italian), useful as a reference for the format even though the programme is not identical. If instead you want to see more worked exercises on the same topics (resistors, Ohm’s law, circuits) from a science high school angle, with electrostatics added, you will find them in electrostatics and circuits: the Year 4 test: it is aimed at the liceo, so it also has a part on charges and potential that you do not do in Year 3 of a vocational school, but the section on direct-current circuits is exactly the same topic.

When the circuit has several generators and you only want to know what happens across one load, the next step is Thévenin and Norton with worked exercises. The set-up mistakes that come back throughout physics are collected in typical physics mistakes in high school.

If you want a hand

On this chapter the useful work is almost always concentrated: one or two lessons to fix multiples and submultiples and the series/parallel criterion properly, then mixed exercises with Kirchhoff until you can read the diagram at a glance. I work online, one to one. You will find the overall picture on the physics tutoring page, and to start from a diagnosis you can book a free first chat through the contact page: send me the programme covered so far and the date of the test, and I will tell you honestly what needs doing between now and then.

Ohm's law and Kirchhoff's laws in Year 3