By the end of this chapter, you will be able to:
Mastering these skills will help you work confidently and safely with electrical systems, making you a valuable professional in the electrical trade.
Compiled by: Stephen Waweru
Electrical quantities are fundamental to the design, operation, and maintenance of agricultural machinery and systems in Kenya. Mastery of these concepts enables agricultural engineers to efficiently manage irrigation pumps, processing equipment, and storage facilities that rely on electrical power. This chapter introduces basic electrical units and measuring instruments, equipping learners with the skills to calculate and interpret electrical parameters essential for agricultural engineering applications.
Four electrical quantities describe almost every circuit on a farm: power, current, resistance and voltage. Each has its own SI unit, and each can be measured or calculated from the others, so a technician who knows the four units can read a machine nameplate, size a supply and trace a fault. The quantities are linked by two short rules: voltage equals current multiplied by resistance (V = I × R), and power equals voltage multiplied by current (P = V × I). Current and voltage are read with the instruments described under Ammeter and Voltmeter.
| Quantity | Symbol | SI unit | Unit symbol | What it describes |
|---|---|---|---|---|
| Power | P | watt | W | How fast electrical energy is converted into heat, motion or light |
| Current | I | ampere | A | How much charge flows past a point each second |
| Resistance | R | ohm | Ω | How strongly a conductor or load opposes the current |
| Voltage | V | volt | V | The electrical push that drives the current round a circuit |
The four quantities work together in every load. For example, a heating element connected across 230 V with a resistance of 46 Ω carries a current of 230 ÷ 46 = 5 A. Its power is 230 × 5 = 1150 W. The same power follows from the current and resistance alone: 5 × 5 × 46 = 1150 W.
Electrical power is the rate at which electrical energy is converted into another form, such as heat, motion or light. It is measured in watts (W). One watt is one joule of energy converted each second, so a 1000 W heater converts 1000 joules every second. On a farm, the power rating on a nameplate tells the technician what size of supply, cable and protective device a machine needs.
The power of a load is the product of the voltage across it and the current through it:
P = V × I
Where:
Power ratings of larger equipment are written in kilowatts (kW), where 1 kW = 1000 W. A rating of 3 kW is therefore 3000 W, and a calculation in watts must use 3000, not 3.
The rule P = V × I gives the true power of direct current loads and of alternating current loads that behave as pure resistance, such as heating elements and filament lamps. For an alternating current motor, the product V × I is larger than the power the motor actually uses, because the current is not fully in step with the voltage. The worked examples below therefore use heaters, lamps and direct current loads. Motors are studied under Single phase electrical machines.
Rearranging the formula gives the current and the voltage when the power is known:
I = P ÷ V
V = P ÷ I
Example 1: The heating element of an egg incubator in a hatchery runs on 230 V and draws 2.5 A. Calculate its power.
Given: V = 230 V, I = 2.5 A
P = V × I
P = 230 × 2.5
P = 575 W
Answer: 575 W (0.575 kW)
Example 2: A fish pond aerator is run from a 12 V direct current supply and draws 10 A. Find the power it takes.
Given: V = 12 V, I = 10 A
P = V × I
P = 12 × 10
P = 120 W
Answer: 120 W
Example 3: A defrost heater in a cold room for vegetables is rated at 3 kW and is supplied at 230 V. Calculate the current it draws.
Given: P = 3 kW = 3000 W, V = 230 V
I = P ÷ V
I = 3000 ÷ 230
I = 13.04 A
Answer: 13.04 A
Example 4: The lighting circuit of a poultry house carries ten filament lamps of 18 W each, supplied at 230 V. Find the current in the circuit.
Total power = 10 × 18 = 180 W
I = P ÷ V
I = 180 ÷ 230
I = 0.78 A
Answer: 0.78 A
Example 5: A group of lamps in a farm store is rated at 60 W and draws 5 A from a battery supply. Find the supply voltage.
Given: P = 60 W, I = 5 A
V = P ÷ I
V = 60 ÷ 5
V = 12 V
Answer: 12 V
Electric current is the flow of electric charge round a circuit, and its unit is the ampere (A). One ampere means that one coulomb of charge passes a point in the circuit every second, written 1 A = 1 C/s. Current is the quantity a technician reads to find out how hard a machine is working: a motor that draws more current than its nameplate allows is overloaded, and a heater that draws none has a break in its circuit. The symbol for current is I.
A useful picture is water in a pipe. The charge moving in the wire is like the water moving in the pipe, and the current is like the flow rate in litres per second. The picture helps to see that the current is the same at every point along an unbranched path, but it is only a comparison, because charge in a wire is not a liquid.
Current needs two things: a source that gives a voltage, and a complete conductive path for the charge to follow. In metal conductors the moving charges are free electrons, and the voltage of the source pushes them round the path. If the path is broken, for example by an open switch or a burnt-out element, the current stops at once. Sources used on farms include:
The size of the current depends on the voltage of the source and the resistance of the path. A higher voltage gives a larger current, and a higher resistance gives a smaller one. This relationship, written I = V ÷ R, is developed under Resistance - Ohms(Ω) and Voltage - Volts (V).
Before the electron was known, the direction of current was chosen as the direction in which positive charge would move: from the positive terminal of the source, through the external circuit, to the negative terminal. This conventional current direction is still used in all circuit diagrams, meter markings and calculations. The free electrons in a metal actually move the opposite way, from the negative terminal towards the positive terminal. The two descriptions give the same results provided one is used consistently, and the arrows on a circuit diagram always show the conventional direction.
Current is either direct or alternating, and the two are used for different jobs on a farm.
| Feature | Direct current (DC) | Alternating current (AC) |
|---|---|---|
| Direction | Flows one way round the circuit | Reverses direction regularly |
| Typical sources | Batteries, photovoltaic panels, rectified supplies | Mains supply, alternators and generators |
| Typical farm uses | Vehicle systems, sensors, controllers, battery-backed lights | Heaters, lamps, motors and machine tools |
| Supply details | Steady voltage that keeps its polarity | In Kenya the supply changes direction 100 times a second, which is 50 cycles per second (50 hertz) |
The type of current decides which ammeter is suitable, as described under Ammeter.
A current has effects that technicians use and effects they must guard against:
Many sensors and electronic controllers on a farm draw only a small fraction of an ampere, so the milliampere (mA) is used. One ampere equals 1000 milliamperes, so 40 mA = 40 ÷ 1000 = 0.04 A. A calculation using the formulas must use amperes, so a value in milliamperes is converted first.
Current is read with an ammeter or a clamp meter, as described under Ammeter. A technician compares the reading with the rated current on the machine nameplate, as shown in the next section.
The rated current on a nameplate is the current the machine is designed to carry continuously. The measured current shows how heavily the machine is loaded:
Load (%) = measured current ÷ rated current × 100
A reading well above the rated value, called overload, shows a fault or too much work, such as a jammed blade or a worn bearing. The extra current heats the windings and can damage them if it continues.
Overload (%) = (measured current − rated current) ÷ rated current × 100
Example 1: A rain gauge logger at a field weather station draws 40 mA from a 12 V direct current supply. Express the current in amperes and find the power drawn.
Given: I = 40 mA, V = 12 V
I = 40 ÷ 1000 = 0.04 A
P = V × I
P = 12 × 0.04
P = 0.48 W
Answer: 0.04 A, which is a power of 0.48 W
Example 2: A 2 kW immersion heater in the parts-washing tank of a farm machinery workshop runs on 230 V. Calculate the current it draws.
Given: P = 2 kW = 2000 W, V = 230 V
I = P ÷ V
I = 2000 ÷ 230
I = 8.70 A
Answer: 8.70 A
Example 3: A bale elevator motor in a hay barn has a rated current of 14 A. A clamp meter reads 11.2 A while it runs. Find the load as a percentage of the rated current.
Given: rated current = 14 A, measured current = 11.2 A
Load = 11.2 ÷ 14 × 100
Load = 80%
Answer: 80% load, so the motor is working within its rating
Example 4: A chaff cutter motor has a rated current of 9 A. When the blades jam, a clamp meter reads 12 A. Find the overload as a percentage of the rated current.
Given: rated current = 9 A, measured current = 12 A
Excess current = 12 − 9 = 3 A
Overload = 3 ÷ 9 × 100
Overload = 33.33%
Answer: 33.33% above the rated current, so the machine is switched off and the blockage cleared
Example 5: A farm gatehouse is supplied from one 12 V battery. A light takes 1.5 A, a camera takes 0.5 A and a gate controller takes 2 A. Find the total current taken from the battery and the total power.
Total current = 1.5 + 0.5 + 2 = 4 A
P = V × I
P = 12 × 4
P = 48 W
Answer: 4 A in total, and 48 W
Resistance is the opposition that a conductor or a load offers to the flow of electric current. Its unit is the ohm (Ω). One ohm is the resistance through which a voltage of one volt drives a current of one ampere, so 1 Ω = 1 V ÷ 1 A. A technician meets resistance in heating elements, which are made to have it, and in cables and joints, where it is unwanted because it wastes energy as heat and lowers the voltage at the load.
Resistance is related to voltage and current by Ohm's law:
R = V ÷ I
Where R is the resistance in ohms (Ω), V is the voltage across the conductor in volts (V) and I is the current through it in amperes (A). Ohm's law holds for a conductor whose temperature stays steady. The same relationship gives the other two quantities:
V = I × R
I = V ÷ R
Four things decide how much resistance a conductor has. The first is length: the resistance is proportional to the length, so doubling the length of a cable doubles its resistance. The second is cross-sectional area: a thicker conductor has less resistance, so a cable with twice the area has half the resistance. The third is material: copper and aluminium conduct well and are used for cables, while alloys such as nichrome have a much higher resistance and are used for heating elements. The fourth is temperature: the resistance of most metals rises as they get hotter, so a warm element has a higher resistance than the same element when cold.
These four points explain why a long, thin extension cable gives a poorer supply at its far end than a short, thick one carrying the same current.
Heating elements are made to have a chosen resistance so that they convert the supply voltage into the required heat. Cables are chosen with enough area to keep their resistance small, so that little voltage is lost along them. Resistors in sensors and control circuits limit the current to a safe value for the device. A technician finds a fault by comparing the calculated resistance of a load with its normal value: a much higher value suggests a poor joint or a partly broken element.
Example 1: A propagation heating mat in a seedling nursery is connected to 230 V and draws 0.46 A. Calculate its resistance.
Given: V = 230 V, I = 0.46 A
R = V ÷ I
R = 230 ÷ 0.46
R = 500 Ω
Answer: 500 Ω
Example 2: The level sensor circuit of a fish hatchery tank is supplied with 5 V and carries 25 mA. Find the resistance of the circuit.
Given: V = 5 V, I = 25 mA = 25 ÷ 1000 = 0.025 A
R = V ÷ I
R = 5 ÷ 0.025
R = 200 Ω
Answer: 200 Ω
Example 3: A cable feeding a machinery shed carries 20 A, and the voltage drop along it is 6 V. Find the resistance of the cable. Then find the resistance of 2.5 times that length of the same cable.
Given: V = 6 V, I = 20 A
R = V ÷ I
R = 6 ÷ 20
R = 0.3 Ω
Resistance is proportional to length, so for 2.5 times the length:
R = 0.3 × 2.5 = 0.75 Ω
Answer: 0.3 Ω, and 0.75 Ω for 2.5 times the length
Example 4: A heating element in a seed germination cabinet has a resistance of 57.5 Ω and is connected to 230 V. Find the current it draws.
Given: R = 57.5 Ω, V = 230 V
I = V ÷ R
I = 230 ÷ 57.5
I = 4 A
Answer: 4 A
Example 5: On a test bench, a current of 1.5 A flows through an 8 Ω resistor that loads a battery charge controller. Find the voltage across the resistor.
Given: I = 1.5 A, R = 8 Ω
V = I × R
V = 1.5 × 8
V = 12 V
Answer: 12 V
Voltage, also called potential difference, is the electrical push that drives current round a circuit. Its unit is the volt (V). One volt means that one joule of energy is given to each coulomb of charge that passes, written 1 V = 1 J/C. It is also the potential difference that drives a current of one ampere through a resistance of one ohm. A battery, a photovoltaic panel and a generator are all sources of voltage, and a voltage exists between two points even when no current flows.
The voltage across a load is linked to the current through it and its resistance by Ohm's law:
V = I × R
Where V is the voltage in volts (V), I is the current in amperes (A) and R is the resistance in ohms (Ω). Rearranged, I = V ÷ R and R = V ÷ I.
The voltage a source produces with nothing connected is called its electromotive force (e.m.f.), and the voltage at its terminals falls a little when it supplies current, because the source has some resistance of its own. The way an e.m.f. is produced in a cell is treated under Determination of e.m.f. in cells. The voltage across a load is the potential difference between its two ends. A load connected across a source receives nearly all of the source voltage when the cables joining them have little resistance.
A farm uses several voltage levels, and a technician must know which a machine needs:
The value on a machine nameplate must match the supply. A load connected to a voltage far above its rating overheats or fails, and a load on a voltage far below its rating performs poorly.
When current flows through a cable, some voltage is lost along it, because the cable has resistance. The voltage drop equals the current multiplied by the cable resistance (V = I × R), and the voltage at the load is the supply voltage minus the drop. A long cable carrying a large current gives a larger drop, and loads at its end may heat poorly or run slowly.
Voltage is read with a voltmeter, as described under Voltmeter. A higher voltage is more dangerous to a person, so the supply is isolated before any repair work on a machine.
Example 1: A work lamp circuit in a produce pack house has a resistance of 6 Ω and carries 2 A. Calculate the voltage across it.
Given: I = 2 A, R = 6 Ω
V = I × R
V = 2 × 6
V = 12 V
Answer: 12 V
Example 2: A cable feeds a cold room from a distribution board that supplies 230 V. The cable has a resistance of 0.45 Ω and carries 16 A. Find the voltage drop, the voltage at the cold room and the drop as a percentage of the supply.
Given: R = 0.45 Ω, I = 16 A, supply = 230 V
Voltage drop = I × R = 16 × 0.45 = 7.2 V
Voltage at the cold room = 230 − 7.2 = 222.8 V
Percentage drop = 7.2 ÷ 230 × 100 = 3.13%
Answer: a drop of 7.2 V, 222.8 V at the cold room, and a drop of 3.13% of the supply
Example 3: A 24 V control circuit in a hatchery has a load resistance of 120 Ω. Calculate the current in the circuit.
Given: V = 24 V, R = 120 Ω
I = V ÷ R
I = 24 ÷ 120
I = 0.2 A
Answer: 0.2 A (200 mA)
Example 4: A handheld livestock tag reader runs on a 6 V battery and draws 0.15 A. Find its resistance.
Given: V = 6 V, I = 0.15 A
R = V ÷ I
R = 6 ÷ 0.15
R = 40 Ω
Answer: 40 Ω
Example 5: A soil moisture probe in a seedbed gives 2.4 V across a circuit of 1200 Ω. Find the current in milliamperes.
Given: V = 2.4 V, R = 1200 Ω
I = V ÷ R
I = 2.4 ÷ 1200
I = 0.002 A
I = 0.002 × 1000 = 2 mA
Answer: 0.002 A, which is 2 mA
Kenyan Context Example:
The Kenya Electricity Generating Company (KenGen) supplies power to agricultural processing plants. For example, a maize milling plant in Eldoret operated by Unga Group Limited uses electric motors rated at 240 V and 12 A. Calculating the power consumption helps engineers at KenGen and Unga Group optimize energy use and reduce operational costs.
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Create a free accountThis chapter introduced the fundamental electrical quantities and their corresponding SI units, starting with power measured in Watts, current in Amperes, resistance in Ohms, and voltage in Volts. It explained how these basic units describe essential electrical properties in circuits. The use and function of measuring instruments were covered, with particular emphasis on the voltmeter for measuring electrical potential difference. The concept of derived units was explored, showing how combinations of basic units form new electrical units relevant to practical applications. Various units of electrical quantities were discussed to establish a clear understanding of their relationships. The chapter concluded with calculations involving charge, force, work, and power, demonstrating how these quantities are interrelated and applied in electrical engineering contexts. This foundation prepares students for more complex electrical analyses and practical measurement techniques.
Kenyan Context Example:
Throughout the chapter, examples from a farm irrigation scheme, a dairy farm, a poultry house, a grain store and a feed mill illustrate the practical application of electrical quantity concepts in real agricultural engineering scenarios.
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