Aandrijftechn.Hf.1.pdf

# Units and fundamental concepts This section introduces the foundational concepts of physical units, focusing on the SI system, base and derived quantities, and prefixes, and then delves into fundamental concepts in mechanics. ### 1.1 Units and the SI system Units are crucial for building and remembering physical formulas. The International System of Units (SI) is the standard system used for measurements [4](#page=4). #### 1.1.1 The SI system and base units The SI system is built upon a set of base units that represent fundamental physical quantities. There are seven base units in the SI system, with five being particularly important and commonly used in physics [5](#page=5). * **Length**: The base unit is the meter, symbolized by $m$ [5](#page=5) [6](#page=6). * **Mass**: The base unit is the kilogram, symbolized by $kg$ [5](#page=5) [6](#page=6). * **Time**: The base unit is the second, symbolized by $s$ [5](#page=5) [6](#page=6). * **Electric current**: The base unit is the ampere, symbolized by $A$. Note that the symbol for electric current intensity is a capital 'I' [5](#page=5) [6](#page=6). * **Thermodynamic temperature**: The base unit is the kelvin, symbolized by $K$. Celsius ($°C$) is also frequently used [5](#page=5) [6](#page=6). Other commonly used units, which are not SI base units, include: * 1 inch = 2.54 cm [5](#page=5). * 1 horsepower (pk) = 736 W = 0.736 kW [5](#page=5). * 1 bar = 100,000 Pa = $10^5$ Pa [5](#page=5). #### 1.1.2 Derived quantities and units Derived quantities are those that can be expressed in terms of base quantities through mathematical relationships. Their units are consequently derived from the base units. It is important to be familiar with these [7](#page=7). * **Force (F)**: The SI unit is the newton ($N$), which is equivalent to $kg \\cdot m \\cdot s^{-2}$ [7](#page=7). * **Pressure (p)**: The SI unit is the pascal ($Pa$) [7](#page=7). * **Work or energy (W)**: The SI unit is the joule ($J$) [7](#page=7). * **Power (P)**: The SI unit is the watt ($W$) [7](#page=7). * **Velocity (v)**: The unit is typically meters per second ($m/s$) [7](#page=7). #### 1.1.3 Prefixes Prefixes are used to modify the magnitude of SI units, making them larger or smaller by powers of ten [8](#page=8). **Multiples:** * tera ($T$) = $10^{12}$ * giga ($G$) = $10^9$ * mega ($M$) = $10^6$ * kilo ($k$) = $10^3$ * hecto ($h$) = $10^2$ * deka ($da$) = $10^1$ **Submultiples:** * deci ($d$) = $10^{-1}$ * centi ($c$) = $10^{-2}$ * milli ($m$) = $10^{-3}$ * micro ($\\mu$) = $10^{-6}$ * nano ($n$) = $10^{-9}$ * pico ($p$) = $10^{-12}$ > **Tip:** Understanding and correctly applying these prefixes is essential for accurate calculations and clear communication of measurements. #### 1.1.4 Unit conversion examples Converting between units is a common task. For example, to convert meters per second ($m/s$) to kilometers per hour ($km/h$): $10 , m/s = 10 \\times \\frac{3600}{1000} , km/h = 10 \\times 3.6 , km/h = 36 , km/h$ [4](#page=4). > **Example:** If a car travels at $20 , m/s$, its speed in $km/h$ is $20 \\times 3.6 = 72 , km/h$. ### 1.2 Fundamental concepts in mechanics Mechanics deals with the study of motion and the forces that cause it. Key fundamental concepts include: * **Mass**: A measure of inertia or the amount of matter in an object [6](#page=6). * **Force**: An interaction that can change the motion of an object [7](#page=7). * **Pressure**: Force applied over a unit area [7](#page=7). * **Work**: The energy transferred when a force moves an object over a distance [7](#page=7). * **Power**: The rate at which work is done or energy is transferred [7](#page=7). > **Studeertip:** Thoroughly understand the quantities, SI units, and formulas associated with mechanics. Practice applying these concepts through exercises to solidify your knowledge [10](#page=10). * * * # Thermodynamics and refrigeration principles This section explores fundamental concepts in thermodynamics and refrigeration, including Kelvin temperature, heat, specific heat, phase changes, and the relationship between evaporation pressure and temperature. ### 2.1 Kelvin temperature The Kelvin scale is the SI unit for thermodynamic temperature, with absolute zero defined as 0 K, which is equivalent to -273 °C. Temperature differences ($\\Delta T$) are expressed in Kelvin [20](#page=20). ### 2.2 Heat Heat is a form of energy or work. Historically, the calorie was used as a unit of heat, where 1 calorie is equal to 4.19 joules (J). The SI unit for heat (and cold) is the joule (J). While the symbol 'W' can also represent heat, 'Q' is more commonly used in this context, distinct from the flow rate 'Q' in hydraulics [21](#page=21). ### 2.3 Heat quantity and specific heat The quantity of heat ($Q$) or work ($W$) required to change the temperature of a substance is given by the formula: $$Q = c \\cdot m \\cdot \\Delta T$$ [22](#page=22). Where: * $Q$ is the heat quantity (in joules). * $c$ is the specific heat capacity of the substance (in J/kg.K). * $m$ is the mass of the substance (in kg). * $\\Delta T$ is the change in temperature (in K or °C). The specific heat capacity ($c$) of a substance is defined as the amount of energy (in kilojoules) required to raise the temperature of 1 kilogram (or 1 gram) of that substance by 1 Kelvin (or 1 degree Celsius) [22](#page=22). Examples of specific heat capacities include: * Water ($c\_w$): 4190 J/kg.K or 4.19 kJ/kg.K [22](#page=22). * Oil ($c\_{olie}$): 2500 J/kg.K [22](#page=22) [23](#page=23). * Steel (Fe) ($c\_{staal(Fe)}$): 460 J/kg.K [22](#page=22) [23](#page=23). #### 2.3.1 Example calculation of heat and power **Problem:** A tractor with a steel gearbox (transmission) weighing 1200 kg contains 90 liters of oil. 1 liter of oil weighs 820 grams. Calculate the heat or work ($W$) produced to warm this transmission from 10 °C to 80 °C in 10 minutes. Also, calculate the required power ($P$) in kW. **Solution:** 1. **Heat required to warm the steel transmission:**$W\_{Fe} = Q\_{Fe} = c\_{Fe} \\cdot m\_{Fe} \\cdot \\Delta T$$W\_{Fe} = 460 , \\text{J/kg.K} \\cdot 1200 , \\text{kg} \\cdot (80 - 10) , \\text{°C}$$W\_{Fe} = 460 \\cdot 1200 \\cdot 70 = 38,640,000 , \\text{J} = 38,640 , \\text{kJ}$ [23](#page=23). 2. **Heat required to warm the transmission oil:** Mass of oil ($m\_{olie}$) = 90 l $\\cdot$ 0.820 kg/l = 73.8 kg $W\_{olie} = Q\_{olie} = c\_{olie} \\cdot m\_{olie} \\cdot \\Delta T$$W\_{olie} = 2500 , \\text{J/kg.K} \\cdot 73.8 , \\text{kg} \\cdot (80 - 10) , \\text{°C}$$W\_{olie} = 2500 \\cdot 73.8 \\cdot 70 = 12,915,000 , \\text{J} = 12,915 , \\text{kJ}$ [24](#page=24). 3. **Total heat required:**$W\_{total} = Q\_{total} = W\_{Fe} + W\_{olie}$$W\_{total} = 38,640 , \\text{kJ} + 12,915 , \\text{kJ} = 51,555 , \\text{kJ}$ [24](#page=24). 4. **Calculate total power:** Time ($t$) = 10 minutes = 10 $\\cdot$ 60 seconds = 600 seconds $P = W\_{total} / t$$P = 51,555 , \\text{kJ} / 600 , \\text{s}$$P = 85.925 , \\text{kW}$ [24](#page=24). ### 2.4 Phase changes of a substance Substances exist in three states of aggregation: solid, liquid, and gaseous. Transitions between these states occur at a constant temperature. These transitions include [25](#page=25): * **Freezing (Stollen):** Transition from liquid to solid, involving latent heat of fusion (Stollingswarmte) [25](#page=25). * **Melting (Smelten):** Transition from solid to liquid [25](#page=25). * **Evaporation (Verdampen):** Transition from liquid to gas [25](#page=25). * **Condensation (Condenseren):** Transition from gas to liquid [25](#page=25). The specific heat capacity can vary between different phases of the same substance; for example, $c\_w$ (water) is 4.19 kJ/kg.K, while $c\_w$ (ice) is 2.1 kJ/kg.K [25](#page=25). ### 2.5 Evaporation pressure and temperature Pressure plays a critical role in the change of a substance's state of aggregation [26](#page=26). * An increase in pressure leads to a higher boiling point [26](#page=26). * A decrease in pressure leads to a lower boiling point [26](#page=26). For every pressure, there is a specific boiling temperature. The pressure corresponding to the boiling point is known as the saturated vapor pressure. For instance, at a pressure of 0.9 bar, the boiling temperature of water is 95 °C [26](#page=26). * * * # Hydraulics basics This section covers fundamental concepts in hydraulics, focusing on pressure, Pascal's law, flow rate, and hydraulic power, which are essential for understanding fluid power systems [28](#page=28). ### 3.1 Pressure Pressure ($p$) is defined as a force ($F$) distributed evenly over an area ($A$). The SI unit for pressure is Newtons per square meter ($N.m^{-2}$), which is equivalent to the Pascal ($Pa$). A common unit used in practice is the bar, where 1 bar equals $10^5 Pa$ [30](#page=30). Gases are compressible, meaning their volume can change significantly under pressure. In contrast, liquids are virtually incompressible, which makes them ideal for use in hydraulic systems as they can transmit force efficiently. Hydraulics, in essence, is the study of fluids, particularly their behavior under pressure and in motion [30](#page=30). ### 3.2 Pascal's law Pascal's law is a cornerstone principle in hydraulics. It states that any pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel, in all directions. This means that pressure at one point in a confined fluid will be the same at all other points at the same horizontal level, regardless of the shape of the container [31](#page=31). > **Tip:** Understanding Pascal's law is crucial for comprehending how hydraulic systems can generate large output forces from smaller input forces, as seen in hydraulic jacks and presses. ### 3.3 Flow rate (debit) Flow rate, denoted by $Q$, quantifies the volume of fluid or gas that moves through a system per unit of time. The SI unit for flow rate is cubic meters per second ($m^3/s$ or $m^3 s^{-1}$). However, other units like liters per minute ($l/min$) are also commonly used and may require conversion in calculations [32](#page=32). In hydraulic pumps, the displacement volume, often expressed in cubic centimeters per revolution ($cm^3/rev$) or milliliters per revolution ($ml/rev$), is a key parameter. When combined with the pump's rotational speed (revolutions per minute, $rpm$), the flow rate can be calculated [32](#page=32). The flow rate ($Q$) is also related to the cross-sectional area ($A$) of a conduit or pipe and the fluid's velocity ($v$) within it, through the formula: $$Q = A \\cdot v$$ This relationship allows for the calculation of fluid velocity or the extension speed of a hydraulic cylinder if the flow rate and the cross-sectional area are known. The SI unit for velocity is meters per second ($m/s$ or $m s^{-1}$) [32](#page=32). ### 3.4 Hydraulic power Power is generally defined as the rate at which work is done, or energy is transferred, over time. The base formula for power ($P$) is [33](#page=33): $$P = \\frac{W}{t}$$ Where $W$ is work and $t$ is time. The SI unit for power is the Watt ($W$), which can be expressed in fundamental units as $J/s$, $N.m/s$, or $kg.m^2/s^3$ [33](#page=33). In hydraulics, hydraulic power is directly related to pressure and flow rate. The formula for hydraulic power is: $$P = p \\cdot Q$$ Using the SI units, this translates to Watts ($W$) when pressure is in Pascals ($Pa$) and flow rate is in cubic meters per second ($m^3/s$). The derivation confirms the unit consistency [33](#page=33): $$W = Pa \\cdot \\frac{m^3}{s} = \\frac{N}{m^2} \\cdot \\frac{m^3}{s} = \\frac{N \\cdot m}{s} = \\frac{J}{s} = W$$ Further expansion using base SI units: $$W = \\left(\\frac{kg \\cdot m}{s^2 \\cdot m^2}\\right) \\cdot \\frac{m^3}{s} = \\frac{kg \\cdot m^4}{m^2 \\cdot s^3} = \\frac{kg \\cdot m^2}{s^3}$$ #### 3.4.1 Example calculation of hydraulic power **Exercise 1:** A hydraulic pump has a flow rate of 40 liters per minute. The inner diameter of the cylinder is 5 cm, and the load to be moved is 3000 kg. Calculate the hydraulic power of the oil pump in kilowatts (kW) [34](#page=34). **Tips for solving:** 1. Convert the flow rate ($Q$) to the SI unit $m^3/s$. 2. Use the formula $P = p \\cdot Q$. To calculate pressure ($p$), you will need the force ($F$) and the area ($A$). 3. Calculate force $F$ using $F = m \\cdot a$, where $m$ is mass and $a$ is acceleration (for gravitational force, $a \\approx 9.81 m/s^2$). 4. Calculate the area $A$ of the cylinder piston using $A = \\pi r^2$ or $A = \\frac{\\pi D^2}{4}$, where $r$ is the radius and $D$ is the inner diameter of the piston. 5. Ensure all units are consistent before performing calculations. #### 3.4.2 Exercise with displacement volume and speed **Exercise 2:** A hydraulic pump has a displacement volume of 90 $cm^3$ per revolution, and the tractor's engine rotates at 2000 revolutions per minute ($rpm$). There is a gear reduction ratio of 24/48 applied to the oil pump. The load to be moved is 2000 kg. The cylinder has an inner diameter of 20 cm and a stroke length of 1 meter. Calculate the flow rate, pressure, and hydraulic power of the oil pump. Also, calculate the extension speed of the piston [35](#page=35). **Key considerations:** * The stroke length of the cylinder (1 meter) is not directly needed for calculating flow rate, pressure, or power but would be used to determine the time taken for the piston to extend [35](#page=35). * The gear reduction ratio needs to be accounted for when calculating the actual pump speed. * * * **Summary of Key Formulas:** * **Pressure:** $p = \\frac{F}{A}$ [30](#page=30). * **Flow Rate:** $Q = \\frac{V}{t}$ [32](#page=32). * **Flow Rate & Velocity:** $Q = A \\cdot v \\implies v = \\frac{Q}{A}$ [32](#page=32). * **Hydraulic Power:** $P = p \\cdot Q$ [29](#page=29) [33](#page=33). * **Power (General):** $P = \\frac{W}{t}$ [33](#page=33). **Important Units:** * Pressure: Pascal ($Pa$), bar [30](#page=30). * Flow Rate: $m^3/s$, $l/min$ [32](#page=32). * Power: Watt ($W$), Kilowatt ($kW$) [33](#page=33). * Displacement Volume: $cm^3/rev$, $ml/rev$ [32](#page=32). * Speed: $m/s$, $rpm$ [32](#page=32). > **Tip:** Always pay close attention to unit conversions, especially between metric units (like liters and cubic meters) and time units (like minutes and seconds), as these are common sources of error in calculations. Ensure you are using the correct internal diameter for area calculations in cylinder applications. * * * # Electricity fundamentals and circuit analysis This section delves into the foundational concepts of electricity, including its definition, chemical basis, and the fundamental properties of charge, voltage, current, and resistance, along with methods for analyzing circuits with various resistor configurations [37](#page=37). ### 4.1 Definition and chemical background of electricity Electricity is fundamentally defined as the **displacement of electrons** from one atom to another, constituting an electron flow. This phenomenon has its roots in the atomic structure of matter. Substances are composed of molecules, which in turn are made up of atoms. An atom, like hydrogen (H), can be conceptualized with a positively charged nucleus at its center and negatively charged electrons orbiting it, similar to a solar system [39](#page=39). #### 4.1.1 Atomic structure and electron behavior In an atom, such as copper (Cu), the nucleus contains protons (positive charges), and electrons (negative charges) orbit the nucleus. A copper atom has 29 protons and 29 electrons. Some electrons are tightly bound to the atom, while others, particularly those in the outermost shell, are considered free electrons. The "jumping" or transfer of these free electrons from one atom to another is the essence of electrical current. (#page=40, 41) [40](#page=40) [41](#page=41). #### 4.1.2 Conductors and insulators The ease with which free electrons can "jump" from one atom to another determines a material's conductivity. Materials with fewer than four electrons in their outermost shell are generally good conductors because their electrons are more readily available to move. Copper, with only one electron in its outer shell, is an excellent conductor. Conversely, insulators have more than four electrons in their outer shell, making electron transfer more difficult [41](#page=41). ### 4.2 Fundamental electrical quantities #### 4.2.1 Electric charge Atoms are electrically neutral when the number of protons (positive charges) equals the number of electrons (negative charges). An imbalance, such as an excess or deficiency of electrons around atoms, can lead to electrical charge, often caused by friction. Initially, this may manifest as static electricity, but when this charge flows through a conductor, it becomes dynamic electricity. An electrical source is necessary to maintain these positive and negative charges at different points, creating a potential difference that drives the electron flow from the negative to the positive terminal [42](#page=42). #### 4.2.2 Potential difference and voltage (U) For an electron current to flow through a conductor, there must be a difference in charge between its two ends. This concept is analogous to the height difference between two water vessels, which allows water to flow. The difference in charge between the ends of a conductor creates a **voltage**, measured in volts (V). This electrical potential difference is comparable to the pressure difference ($\\Delta p$) in hydraulics. Voltage can be generated chemically or mechanically [43](#page=43) [44](#page=44). > **Tip:** The water analogy is a useful way to visualize electrical concepts: pressure difference ($\\Delta p$) is analogous to voltage (U), water flow rate (Q) is analogous to current (I), and the resistance of a pipe is analogous to electrical resistance (R). (#page=43, 44) [43](#page=43) [44](#page=44). #### 4.2.3 Electric current (I) Electric current (I) is the rate of flow of electric charge, analogous to the flow rate of water in a pipe. The SI unit for current is the **ampere (A)** [45](#page=45). * **AC (Alternating Current)** refers to current that periodically reverses direction [45](#page=45). * **DC (Direct Current)** refers to current that flows in only one direction [45](#page=45). #### 4.2.4 Resistance (R) Electrical resistance (R) is the opposition to the flow of electric current, similar to the friction or obstruction a water current encounters in a pipe. The SI unit for resistance is the **ohm ($\\Omega$)**. The resistance of an electrical cable depends on three factors [46](#page=46): * **Material type (resistivity, $\\rho$)**: Different materials have inherent resistance properties. For copper (Cu) at 15°C, the resistivity ($\\rho\_{15°C}$) is 0.0175 $\\Omega \\cdot \\text{mm}^2/\\text{m}$ [46](#page=46) [47](#page=47). * **Cross-sectional area (A)**: A larger cross-sectional area offers less resistance [46](#page=46). * **Length (l)**: A longer conductor has higher resistance [46](#page=46). ##### 4.2.4.1 Formulas for resistance Two key formulas are used to calculate resistance: 1. **Pouillet's Law**: This law relates resistance to the material's resistivity, length, and cross-sectional area. $$R = \\rho \\frac{l}{A}$$ [47](#page=47). 2. **Ohm's Law**: This fundamental law relates voltage, current, and resistance in a circuit. $$U = I \\cdot R$$ or $$I = \\frac{U}{R}$$ (#page=44, 47) [44](#page=44) [47](#page=47). > **Example:** Calculate the resistance of a copper wire measuring 20 meters in length with a cross-sectional area of 1.5 mm$^2$. Since the current flows both ways in a connected circuit, the total length of the wire is 2 \* 20 m = 40 m. > > Using Pouillet's Law: $R = 0.0175 , \\Omega \\cdot \\text{mm}^2/\\text{m} \\times \\frac{40 , \\text{m}}{1.5 , \\text{mm}^2} = 0.4667 , \\Omega$ [48](#page=48). > **Example:** A copper wire with a cross-sectional area of 2.5 mm$^2$ has a resistance of 7 $\\Omega$. Calculate the length of the wire. > > Rearranging Pouillet's Law for length: $l = \\frac{R \\cdot A}{\\rho}$$l = \\frac{7 , \\Omega \\times 2.5 , \\text{mm}^2}{0.0175 , \\Omega \\cdot \\text{mm}^2/\\text{m}} = 1000 , \\text{m}$ [48](#page=48). > **Example:** A copper wire 50 meters long has a resistance of 1.75 $\\Omega$. Calculate its cross-sectional area (in mm$^2$) and diameter (in mm). > > First, calculate the area using Pouillet's Law rearranged for A: $A = \\frac{\\rho \\cdot l}{R}$$A = \\frac{0.0175 , \\Omega \\cdot \\text{mm}^2/\\text{m} \\times 50 , \\text{m}}{1.75 , \\Omega} = 0.5 , \\text{mm}^2$ > > To find the diameter, use the formula for the area of a circle, $A = \\pi r^2 = \\frac{\\pi D^2}{4}$, where $D$ is the diameter and $r$ is the radius. $r^2 = \\frac{A}{\\pi} \\implies r = \\sqrt{\\frac{A}{\\pi}} = \\sqrt{\\frac{0.5 , \\text{mm}^2}{\\pi}} \\approx 0.4 , \\text{mm}$$D = 2r = 2 \\times 0.4 , \\text{mm} = 0.8 , \\text{mm}$ [49](#page=49). ### 4.3 Circuit analysis: switching resistors #### 4.3.1 Series circuits In a series circuit, components are connected end-to-end, so the **current (I) is the same through all components** [52](#page=52). * The **total resistance (R$\_{tot}$)** is the sum of the individual resistances: $$R\_{\\text{tot}} = R\_1 + R\_2 + R\_3 + \\dots$$ [52](#page=52). * The **total voltage (U$\_{tot}$)** is the sum of the voltages across each component: $$U\_{\\text{tot}} = U\_1 + U\_2 + U\_3 + \\dots$$ [52](#page=52). > **Example:** In a series circuit with resistances of 5 $\\Omega$, 8 $\\Omega$, 20 $\\Omega$, and 3 $\\Omega$, connected to a 12 V source: $R\_{\\text{tot}} = 5 + 8 + 20 + 3 = 36 , \\Omega$ [52](#page=52). $I\_{\\text{tot}} = \\frac{U\_{\\text{tot}}}{R\_{\\text{tot}}} = \\frac{12 , \\text{V}}{36 , \\Omega} = \\frac{1}{3} , \\text{A} \\approx 0.33 , \\text{A}$ [53](#page=53). Individual voltages can then be calculated using Ohm's Law: $U\_1 = I\_{\\text{tot}} \\cdot R\_1 = 0.33 , \\text{A} \\cdot 5 , \\Omega = 1.65 , \\text{V}$ [53](#page=53). $U\_2 = 0.33 , \\text{A} \\cdot 8 , \\Omega = 2.64 , \\text{V}$ [53](#page=53). $U\_3 = 0.33 , \\text{A} \\cdot 20 , \\Omega = 6.60 , \\text{V}$ [53](#page=53). $U\_4 = 0.33 , \\text{A} \\cdot 3 , \\Omega = 0.99 , \\text{V}$ [53](#page=53). #### 4.3.2 Parallel circuits In a parallel circuit, components are connected across each other, so the **voltage (U) is the same across all components** [54](#page=54). * The **total current (I$\_{tot}$)** is the sum of the individual currents: $$I\_{\\text{tot}} = I\_1 + I\_2 + I\_3 + \\dots$$ (#page=54, 55) [54](#page=54) [55](#page=55). * The **total resistance (R$\_{tot}$)** is calculated using the reciprocal of the sum of the reciprocals of the individual resistances: $$\\frac{1}{R\_{\\text{tot}}} = \\frac{1}{R\_1} + \\frac{1}{R\_2} + \\frac{1}{R\_3} + \\dots$$$$R\_{\\text{tot}} = \\frac{1}{\\frac{1}{R\_1} + \\frac{1}{R\_2} + \\frac{1}{R\_3} + \\dots}$$ [54](#page=54). > **Example:** In a parallel circuit with resistances of 5 $\\Omega$, 8 $\\Omega$, 20 $\\Omega$, and 3 $\\Omega$, connected to a 12 V source: $\\frac{1}{R\_{\\text{tot}}} = \\frac{1}{5} + \\frac{1}{8} + \\frac{1}{20} + \\frac{1}{3} = \\frac{24 + 15 + 6 + 40}{120} = \\frac{85}{120}$$R\_{\\text{tot}} = \\frac{120}{85} \\approx 1.41 , \\Omega$ [54](#page=54). Since the voltage is the same across all components: $U\_{\\text{tot}} = U\_1 = U\_2 = U\_3 = U\_4 = 12 , \\text{V}$. (#page=54, 55) [54](#page=54) [55](#page=55). Individual currents can then be calculated using Ohm's Law: $I\_1 = \\frac{U\_1}{R\_1} = \\frac{12 , \\text{V}}{5 , \\Omega} = 2.4 , \\text{A}$ [55](#page=55). $I\_2 = \\frac{U\_2}{R\_2} = \\frac{12 , \\text{V}}{8 , \\Omega} = 1.5 , \\text{A}$ [55](#page=55). $I\_3 = \\frac{U\_3}{R\_3} = \\frac{12 , \\text{V}}{20 , \\Omega} = 0.60 , \\text{A}$ [55](#page=55). $I\_4 = \\frac{U\_4}{R\_4} = \\frac{12 , \\text{V}}{3 , \\Omega} = 4 , \\text{A}$ [55](#page=55). $I\_{\\text{tot}} = 2.4 + 1.5 + 0.60 + 4 = 8.5 , \\text{A}$ [55](#page=55). #### 4.3.3 Mixed circuits Mixed circuits combine both series and parallel connections of resistors. The practical application of circuits predominantly involves mixed configurations [56](#page=56). ##### 4.3.3.1 Method for analyzing mixed circuits The strategy for analyzing mixed circuits involves breaking them down into simpler series and parallel sub-circuits [56](#page=56). 1. **Simplify parallel sections**: Calculate the equivalent resistance of any parallel sub-circuits first. This reduces them to a single equivalent resistor [56](#page=56). 2. **Convert to a series circuit**: Continue simplifying until the entire circuit is represented by a single equivalent resistance [56](#page=56). 3. **Work backwards**: Once the total equivalent resistance is found, calculate the total current and voltage. Then, work back through the simplified stages to determine the unknown currents and voltages in each section and individual component [56](#page=56). 4. **Tabulate results**: Organize the calculated values for resistance (R), current (I), and voltage (U) for each component in a clear table [56](#page=56). > **Example (Mixed Circuit 1):** A circuit with a 220 V source has resistors in series and parallel configurations. The parallel section consists of R3, R4, and R5. First, calculate the equivalent parallel resistance ($R\_{3,4,5}$): $\\frac{1}{R\_{3,4,5}} = \\frac{1}{6 , \\Omega} + \\frac{1}{8 , \\Omega} + \\frac{1}{4 , \\Omega} = \\frac{4 + 3 + 6}{24} = \\frac{13}{24}$$R\_{3,4,5} = \\frac{24}{13} , \\Omega \\approx 1.846 , \\Omega$ (#page=57, 58) [57](#page=57) [58](#page=58). Assuming resistors R1 = 2 $\\Omega$, R2 = 4 $\\Omega$, and R6 = 3 $\\Omega$ are in series with this parallel combination, the total resistance would be: $R\_{\\text{tot}} = R\_1 + R\_2 + R\_{3,4,5} + R\_6 = 2 , \\Omega + 4 , \\Omega + 1.846 , \\Omega + 3 , \\Omega = 10.846 , \\Omega$ (#page=57, 58) [57](#page=57) [58](#page=58). The total current is $I\_{\\text{tot}} = \\frac{220 , \\text{V}}{10.846 , \\Omega} \\approx 20.28 , \\text{A}$. This current flows through R1, R2, and R6 [58](#page=58). The table would then detail the R, I, and U for each component. For instance, the voltage across R1 is $U\_1 = I\_{\\text{tot}} \\cdot R\_1 = 20.28 , \\text{A} \\cdot 2 , \\Omega = 40.56 , \\text{V}$. The voltage across the parallel combination (R3, R4, R5) is $U\_{3,4,5} = I\_{\\text{tot}} \\cdot R\_{3,4,5} = 20.28 , \\text{A} \\cdot 1.846 , \\Omega \\approx 37.52 , \\text{V}$ [58](#page=58). > **Example (Mixed Circuit 2):** A circuit with a 24 V source where R2 and R3 are in parallel, and this combination is in series with R1 and R4, and all are 10 $\\Omega$. > > 1. Calculate the equivalent resistance of the parallel pair R2 and R3 ($R\_{2,3}$): $\\frac{1}{R\_{2,3}} = \\frac{1}{10 , \\Omega} + \\frac{1}{10 , \\Omega} = \\frac{2}{10 , \\Omega}$$R\_{2,3} = \\frac{10}{2} , \\Omega = 5 , \\Omega$ [60](#page=60). > > 2. The circuit is now simplified to a series combination of R1, $R\_{2,3}$, and R4. $R\_{\\text{tot}} = R\_1 + R\_{2,3} + R\_4 = 10 , \\Omega + 5 , \\Omega + 10 , \\Omega = 25 , \\Omega$ [60](#page=60). > > 3. Calculate the total current: $I\_{\\text{tot}} = \\frac{24 , \\text{V}}{25 , \\Omega} = 0.96 , \\text{A}$ [60](#page=60). This total current flows through R1 and R4. The voltage across the parallel combination $R\_{2,3}$ is $U\_{2,3} = I\_{\\text{tot}} \\cdot R\_{2,3} = 0.96 , \\text{A} \\cdot 5 , \\Omega = 4.8 , \\text{V}$ [60](#page=60). > > 4. Since R2 and R3 are in parallel and share this voltage, the current through each is $I\_2 = I\_3 = \\frac{U\_{2,3}}{10 , \\Omega} = \\frac{4.8 , \\text{V}}{10 , \\Omega} = 0.48 , \\text{A}$ [60](#page=60). > > **Example (Mixed Circuit 3):** A circuit with a 24 V source. R2 and R5 are in series, R3 and R4 are in series. These two series pairs are then connected in parallel with each other. This entire parallel combination is in series with R1 and R6, all resistors being 10 $\\Omega$. > > 1. Calculate the series resistance of R2 and R5: $R\_{2,5} = R\_2 + R\_5 = 10 , \\Omega + 10 , \\Omega = 20 , \\Omega$ [61](#page=61). > > 2. Calculate the series resistance of R3 and R4: $R\_{3,4} = R\_3 + R\_4 = 10 , \\Omega + 10 , \\Omega = 20 , \\Omega$. (This step is implicitly done in the next calculation as they are treated as individual parallel components, but conceptually helpful.) > > 3. Calculate the equivalent resistance of the parallel combination of $R\_{2,5}$ and $R\_{3,4}$ (which are then grouped as $R\_{2,3,4,5}$ where $R\_{3,4}$ is treated as a single effective resistance): $\\frac{1}{R\_{2,3,4,5}} = \\frac{1}{R\_{2,5}} + \\frac{1}{R\_3} + \\frac{1}{R\_4}$ (This approach is incorrect as per the diagram, R3 and R4 are in series, and R2 and R5 are in series. The diagram implies R2, R3, R4, R5 are involved in a more complex parallel arrangement not explicitly stated as pairs). Let's re-evaluate based on the provided calculation: R2,5 = 20 $\\Omega$ is one path, and R3=10 $\\Omega$, R4=10 $\\Omega$ are other parallel paths. The provided formula seems to interpret R2, R3, R4, R5 in parallel, not as pairs in series. Re-interpreting based on the solution: R2 and R5 are treated as a single branch (20 $\\Omega$), R3 as another branch (10 $\\Omega$), and R4 as another branch (10 $\\Omega$). These three branches are in parallel. $\\frac{1}{R\_{2,3,4,5}} = \\frac{1}{20 , \\Omega} + \\frac{1}{10 , \\Omega} + \\frac{1}{10 , \\Omega} = \\frac{1 + 2 + 2}{20} = \\frac{5}{20}$$R\_{2,3,4,5} = \\frac{20}{5} , \\Omega = 4 , \\Omega$ [61](#page=61). > > 4. The circuit is now simplified to a series combination of R1, $R\_{2,3,4,5}$, and R6. $R\_{\\text{tot}} = R\_1 + R\_{2,3,4,5} + R\_6 = 10 , \\Omega + 4 , \\Omega + 10 , \\Omega = 24 , \\Omega$ [61](#page=61). > > 5. Calculate the total current: $I\_{\\text{tot}} = \\frac{24 , \\text{V}}{24 , \\Omega} = 1 , \\text{A}$. This current flows through R1 and R6 [61](#page=61). The voltage across the parallel combination $R\_{2,3,4,5}$ is $U\_{2,3,4,5} = I\_{\\text{tot}} \\cdot R\_{2,3,4,5} = 1 , \\text{A} \\cdot 4 , \\Omega = 4 , \\text{V}$ [61](#page=61). > > 6. Now, determine currents and voltages in the parallel branches: For the branch with R2 and R5 (total 20 $\\Omega$): $I\_{2,5} = \\frac{U\_{2,3,4,5}}{R\_{2,5}} = \\frac{4 , \\text{V}}{20 , \\Omega} = 0.2 , \\text{A}$. This current flows through both R2 and R5 [61](#page=61). For the branch with R3 (10 $\\Omega$): $I\_3 = \\frac{U\_{2,3,4,5}}{R\_3} = \\frac{4 , \\text{V}}{10 , \\Omega} = 0.4 , \\text{A}$ [61](#page=61). For the branch with R4 (10 $\\Omega$): $I\_4 = \\frac{U\_{2,3,4,5}}{R\_4} = \\frac{4 , \\text{V}}{10 , \\Omega} = 0.4 , \\text{A}$ [61](#page=61). The total current for the parallel section is $0.2 , \\text{A} + 0.4 , \\text{A} + 0.4 , \\text{A} = 1.0 , \\text{A}$, which matches $I\_{\\text{tot}}$ [61](#page=61). > * * * ## Common mistakes to avoid * Review all topics thoroughly before exams * Pay attention to formulas and key definitions * Practice with examples provided in each section * Don't memorize without understanding the underlying concepts

| Term | Definition | |------|------------| | SI-eenhedenstelsel (SI unit system) | A globally accepted system of measurement units that forms the basis for scientific and technical communication. It includes base units and derived units. | | Basisgrootheden (Base quantities) | Fundamental physical quantities that are defined independently and used to derive other quantities. The SI system has seven base quantities, including length, mass, and time. | | Grondeenheden (Base units) | The units corresponding to the base quantities in the SI system. Examples include meter (m) for length, kilogram (kg) for mass, and second (s) for time. | | Afgeleide grootheden (Derived quantities) | Physical quantities that are derived from base quantities through mathematical relationships. Examples include force, pressure, and energy. | | Voorvoegsels (Prefixes) | Symbols used with SI units to indicate multiples or submultiples of the unit. Examples include kilo (k) for 10^3 and milli (m) for 10^-3. | | Massa (Mass) | A fundamental property of matter that measures the amount of substance in an object. It is a scalar quantity and is constant regardless of location. | | Kracht (Force) | An interaction that, when unopposed, will change the motion of an object. It is a vector quantity and is measured in Newtons (N). | | Druk (Pressure) | The force exerted per unit area. It is a scalar quantity and is measured in Pascals (Pa). | | Arbeid (Work) | The energy transferred when a force moves an object over a distance. It is measured in Joules (J). | | Energie (Energy) | The capacity to do work. It exists in various forms, such as mechanical, thermal, and electrical energy, and is measured in Joules (J). | | Vermogen (Power) | The rate at which work is done or energy is transferred. It is measured in Watts (W). | | Wet van Pascal (Pascal's Law) | States that a pressure change at any point in a confined incompressible fluid is transmitted equally throughout the fluid. | | Debiet (Flow rate) | The volume of fluid that passes through a given cross-sectional area per unit of time. It is typically measured in cubic meters per second (m³/s) or liters per minute (l/min). | | Hydraulisch vermogen (Hydraulic power) | The power transmitted by a hydraulic fluid, calculated as the product of pressure and flow rate ($P = p \cdot Q$). | | Potentiaalverschil (Potential difference) | The difference in electric potential between two points. It is the driving force for electric current and is synonymous with voltage. | | Spanning U (Voltage U) | The electric potential difference between two points, measured in Volts (V). It is the driving force that causes electric charge to flow. | | Stroomsterkte I (Current I) | The rate of flow of electric charge, measured in Amperes (A). It represents the amount of charge passing a point per unit of time. | | Weerstand R (Resistance R) | The opposition to the flow of electric current in a conductor, measured in Ohms (Ω). It is determined by the material, length, and cross-sectional area of the conductor. | | Wet van Ohm (Ohm's Law) | A fundamental law in electricity that relates voltage, current, and resistance ($U = I \cdot R$). | | Serieschakeling (Series connection) | A circuit configuration where components are connected end-to-end, so the same current flows through each component. | | Parallelschakeling (Parallel connection) | A circuit configuration where components are connected across the same two points, so the same voltage is applied to each component. | | Gemengde schakeling (Mixed connection) | A circuit that combines both series and parallel connections of components. |