Aandrijftechn.Hf.1.pdf

# Units and basic mechanics This section provides a fundamental overview of the SI units system and introduces core concepts in mechanics including mass, force, pressure, work, and power. ### 1.1 Units Units are essential for constructing and remembering formulas. The International System of Units (SI) is the standard [4](#page=4). #### 1.1.1 The SI unit system The SI system uses base units for fundamental quantities. * **Base quantities and their SI base units:** * Length: meter (m) [6](#page=6). * Mass: kilogram (kg) [6](#page=6). * Time: second (s) [6](#page=6). * Electric current: ampere (A) [6](#page=6). * Thermodynamic temperature: kelvin (K) [6](#page=6). * Other commonly used 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 [13](#page=13) [5](#page=5). #### 1.1.2 Derived quantities Derived quantities are formed from base quantities and have corresponding units. Key derived quantities to know include: * **Force (F):** SI unit is the newton (N). A newton is defined as 1 kg m s$^{-2}$ [12](#page=12) [7](#page=7). * **Pressure (p):** SI unit is the pascal (Pa). A pascal is equal to N/m$^2$ [13](#page=13) [7](#page=7). * **Work or energy (W):** SI unit is the joule (J). A joule is equivalent to N m or kg m$^2$ s$^{-2}$ [14](#page=14) [7](#page=7). * **Power (P):** SI unit is the watt (W). A watt is equivalent to J/s or kg m$^2$ s$^{-3}$ [15](#page=15) [7](#page=7). * **Velocity (v):** [7](#page=7). #### 1.1.3 Prefixes SI prefixes are used to denote multiples and submultiples of units: * tera (T) = $10^{12}$ [8](#page=8). * giga (G) = $10^9$ [8](#page=8). * mega (M) = $10^6$ [8](#page=8). * kilo (k) = $10^3$ [8](#page=8). * hecto (h) = $10^2$ [8](#page=8). * deka (da) = $10^1$ [8](#page=8). * deci (d) = $10^{-1}$ [8](#page=8). * centi (c) = $10^{-2}$ [8](#page=8). * milli (m) = $10^{-3}$ [8](#page=8). * micro ($\mu$) = $10^{-6}$ [8](#page=8). * nano (n) = $10^{-9}$ [8](#page=8). * pico (p) = $10^{-12}$ [8](#page=8). #### 1.1.4 Unit conversion example To convert units, for instance, from meters per second (m/s) to kilometers per hour (km/h), you can use conversion factors: $10 \text{ m/s} = 10 \times \frac{3600}{1000} \text{ km/h} = 10 \times 3.6 \text{ km/h} = 36 \text{ km/h}$ [4](#page=4). > **Tip:** Understanding SI units and their relationships is crucial for building and recalling physics formulas correctly [4](#page=4). ### 1.2 Basic concepts of mechanics Mechanics deals with the study of forces and their effects on motion [10](#page=10). #### 1.2.1 Mass Mass is a physical quantity that measures the amount of matter in an object [11](#page=11). * **Symbol:** $m$ [11](#page=11). * **SI unit:** kilogram (kg) [11](#page=11). * **Distinction from weight:** Weight is the force exerted by gravity on an object, which can vary depending on location (e.g., Earth vs. Moon). Mass, however, remains constant everywhere [11](#page=11). #### 1.2.2 Force Force is a fundamental concept in mechanics that describes a push or pull on an object [12](#page=12). * **Symbol:** $F$ [12](#page=12). * **SI unit:** newton (N) [12](#page=12). * **Formula:** $F = m \times a$, where $m$ is mass and $a$ is acceleration [12](#page=12). * The unit derivation for force is $N = \text{kg} \times \text{m/s}^2$ [12](#page=12). * **Acceleration (a):** Defined as the change in velocity over a unit of time ($a = v/t$). Its unit is m/s$^2$ [12](#page=12). * **Gravitational acceleration (g):** On Earth, this is approximately $9.81 \text{ m/s}^2$ and should be memorized [12](#page=12). #### 1.2.3 Pressure Pressure is defined as the force applied perpendicular to a surface per unit area [13](#page=13). * **Symbol:** $p$ [13](#page=13). * **SI unit:** pascal (Pa) [13](#page=13). * **Formula:** $p = \frac{F}{A}$, where $F$ is force and $A$ is area [13](#page=13). * **Units:** Pa = N/m$^2$ [13](#page=13). * **Other units:** * 1 bar = $10^5$ Pa = 100,000 Pa [13](#page=13). * 1 psi (pound per square inch) is also used [13](#page=13). #### 1.2.4 Work (or energy) Work is done when a force causes an object to move a certain distance. Energy is the capacity to do work [14](#page=14). * **Symbol:** $W$ (for work), often $Q$ is used for energy, especially in refrigeration [14](#page=14). * **SI unit:** joule (J) [14](#page=14). * **Formula for Work:** $W = F \times s$ or $W = F \times \Delta x$, where $F$ is force and $s$ or $\Delta x$ is distance or displacement [14](#page=14). * The unit derivation is $J = N \times m$ [14](#page=14). * **Other units:** * 1 calorie (cal) = 4.19 J [14](#page=14). #### 1.2.5 Power Power is the rate at which work is done or energy is transferred [15](#page=15). * **Symbol:** $P$ [15](#page=15). * **SI unit:** watt (W) [15](#page=15). * **Formula:** $P = \frac{W}{t}$, where $W$ is work (or energy) and $t$ is time [15](#page=15). * The unit derivation is $W = J/s$ [15](#page=15). * **Linear power:** For linear motion, power can be calculated as $P = F \times v$, where $F$ is the tractive force and $v$ is velocity. This can be rearranged to find force: $F = P/v$ [15](#page=15). * **Rotational power:** For rotating systems (e.g., motors), power is given by $P = M \times \omega$, where $M$ is torque and $\omega$ is angular velocity [15](#page=15). * Angular velocity can be expressed as $\omega = \frac{2\pi n}{60}$, where $n$ is the rotational speed in revolutions per minute (rpm) [15](#page=15). * This allows for calculating torque: $M = \frac{P \times 60}{2\pi n}$ [15](#page=15). * **Torque (or moment, M):** This is the rotational equivalent of force, calculated as $M = F \times r$, where $F$ is force and $r$ is the lever arm distance [15](#page=15). * The unit for torque is N m [15](#page=15). > **Tip:** Understanding the relationships between these mechanical concepts and their SI units is fundamental for solving problems in physics and engineering. Practice applying these formulas through exercises [10](#page=10). --- # Heat transfer and refrigeration basics This section introduces fundamental concepts in thermodynamics and refrigeration, covering temperature scales, heat, its quantity, specific heat capacity, phase transitions, and the relationship between evaporation pressure and temperature. ### 2.1 Kelvin temperature * The Celsius scale measures temperature ($T$) [20](#page=20). * Absolute zero is defined as 0 Kelvin (K), which is equivalent to -273 degrees Celsius (°C) [20](#page=20). * In the International System of Units (SI), temperature differences ($\Delta T$) are expressed in Kelvin [20](#page=20). ### 2.2 Heat * Heat is a form of energy or work [21](#page=21). * Historically, the calorie was used as a unit of heat [21](#page=21). * The SI unit for heat (and cold) is the Joule (J) [21](#page=21). * One calorie is equivalent to 4.19 Joules [21](#page=21). * The symbol for heat is often $Q$, though $W$ is also used. It is important not to confuse this $Q$ with the flow rate $Q$ used in hydraulics [21](#page=21). > **Tip:** While the term "heat" is used, it fundamentally represents energy transfer. The concept of "cold" is simply the absence of heat or a lower amount of heat energy. ### 2.3 Heat quantity and specific heat capacity The quantity of heat ($Q$) required to change the temperature of a substance is calculated using the following formula: $$Q = c \cdot m \cdot \Delta T$$ where: * $Q$ is the heat quantity (in Joules) [22](#page=22). * $c$ is the specific heat capacity of the substance (in J/kg·K) [22](#page=22). * $m$ is the mass of the substance (in kilograms) [22](#page=22). * $\Delta T$ is the temperature difference (in Kelvin or degrees Celsius) [22](#page=22). The amount of heat ($Q$) is dependent on the type of substance, its mass, and the temperature difference [22](#page=22). The specific heat capacity ($c$) is defined as the amount of energy (in Joules) required to raise the temperature of one kilogram (kg) of a substance by one Kelvin (K) or one degree Celsius (°C) [22](#page=22). **Examples of specific heat capacities:** * Specific heat of water ($c_w$): 4190 J/kg·K, which is equal to 4.19 kilojoules per kilogram per Kelvin (kJ/kg·K) [22](#page=22). * Specific heat of oil ($c_{olie}$): 2500 J/kg·K [22](#page=22). * Specific heat of steel (Fe) ($c_{staal}$): 460 J/kg·K [22](#page=22). #### 2.3.1 Calculation example **Problem:** A tractor with a steel (Fe) gearbox weighing 1200 kg contains 90 liters of oil. One liter of oil weighs 820 grams. Calculate the heat produced ($W$ or $Q$) to warm up this transmission from 10 °C to 80 °C in 10 minutes. Also, calculate the required power ($P$) in kilowatts (kW). **Solution:** 1. **Heat required to warm up the steel transmission:** $$Q_{Fe} = c_{Fe} \cdot m_{Fe} \cdot \Delta T$$ $$Q_{Fe} = 460 \, \text{J/kg}\cdot\text{K} \cdot 1200 \, \text{kg} \cdot (80 \, \text{°C} - 10 \, \text{°C})$$ $$Q_{Fe} = 460 \cdot 1200 \cdot 70 = 38,640,000 \, \text{J} = 38,640 \, \text{kJ}$$ 2. **Heat required to warm up the transmission oil:** Mass of oil ($m_{olie}$) = 90 liters * 0.820 kg/liter = 73.8 kg [24](#page=24). $$Q_{olie} = c_{olie} \cdot m_{olie} \cdot \Delta T$$ $$Q_{olie} = 2500 \, \text{J/kg}\cdot\text{K} \cdot 73.8 \, \text{kg} \cdot (80 \, \text{°C} - 10 \, \text{°C})$$ $$Q_{olie} = 2500 \cdot 73.8 \cdot 70 = 12,915,000 \, \text{J} = 12,915 \, \text{kJ}$$ 3. **Total heat required:** $$Q_{total} = Q_{Fe} + Q_{olie}$$ $$Q_{total} = 38,640 \, \text{kJ} + 12,915 \, \text{kJ} = 51,555 \, \text{kJ}$$ 4. **Calculate the total power:** Time ($t$) = 10 minutes = 10 * 60 seconds = 600 seconds [24](#page=24). $$P = \frac{Q_{total}}{t}$$ $$P = \frac{51,555 \, \text{kJ}}{600 \, \text{s}} = 85.925 \, \text{kW}$$ ### 2.4 Phase transitions of a substance A substance can exist in three states or phases: solid, liquid, and gaseous. For example, water can be ice (solid), liquid water, or steam (gas) [25](#page=25). * Transitions between these states occur at a constant temperature [25](#page=25). * **Freezing (Stollen):** The transition from liquid to solid, releasing latent heat of fusion [25](#page=25). * **Melting (Smelten):** The transition from solid to liquid, requiring latent heat of fusion [25](#page=25). * The specific heat capacity can differ between phases; for instance, $c_w$ (water) = 4.19 kJ/kg·K, while $c_w$ (ice) = 2.1 kJ/kg·K [25](#page=25). * **Vaporization (Verdampen):** The transition from liquid to gas, requiring latent heat of vaporization [25](#page=25). * **Condensation (Condenseren):** The transition from gas to liquid, releasing latent heat of vaporization [25](#page=25). > **Tip:** Latent heat is the energy absorbed or released during a phase change at constant temperature, distinct from sensible heat, which causes a temperature change. ### 2.5 Evaporation pressure and temperature * The pressure plays a crucial role in the phase transition of a substance, particularly during boiling [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 specific pressure, there is a corresponding boiling temperature [26](#page=26). * The pressure associated with the boiling point is known as the saturated vapor pressure [26](#page=26). **Example:** At a pressure of 0.9 bar, water boils at 95 °C. This illustrates that at lower atmospheric pressures (e.g., in the mountains), water boils at a lower temperature [26](#page=26). --- # Hydraulic principles This section introduces the fundamental concepts of hydraulics, covering pressure, Pascal's law, flow rate, and hydraulic power, along with their associated formulas and units [28](#page=28). ### 3.1 Pressure Pressure is defined as a force distributed equally over an area. It is calculated using the formula [30](#page=30): $$p = \frac{F}{A}$$ where $p$ is the pressure, $F$ is the force, and $A$ is the area. The SI unit for pressure is Newton per square meter ($N \cdot m^{-2}$), which is equivalent to the Pascal (Pa). A common unit in practical applications is the bar, where $1 \text{ bar} = 10^5 \text{ Pa} = 10^5 \text{ N/m}^2$ [30](#page=30). Unlike gases, liquids are incompressible, making them suitable for hydraulic systems. Hydraulics is the study of fluids [30](#page=30). ### 3.2 Pascal's law Pascal's law, also known as the law of hydraulics, states that pressure applied to a fluid is transmitted equally in all directions throughout the fluid [31](#page=31). ### 3.3 Flow rate (or discharge) Flow rate, denoted by $Q$, represents the volume of fluid (or gas) displaced per unit of time. The formula for flow rate is [32](#page=32): $$Q = \frac{V}{t}$$ where $V$ is the volume and $t$ is the time. The SI unit for flow rate is cubic meters per second ($m^3/s$ or $m^3 s^{-1}$). Other units, such as liters per minute ($L/min$), are also used and can be converted [32](#page=32). In the context of pumps, the stroke volume (e.g., $cm^3/$revolution or $ml/$revolution) is important. When combined with the pump's rotational speed (RPM or $tpm$), the flow rate can be calculated [32](#page=32). If the flow rate and the cross-sectional area of pipes or tubes are known, the flow velocity or the extension speed of a hydraulic cylinder can be determined using the formula: $$Q = A \cdot v$$ Rearranging this for velocity $v$: $$v = \frac{Q}{A}$$ The SI unit for velocity is meters per second ($m/s$ or $m s^{-1}$) [32](#page=32). ### 3.4 Hydraulic power The basic formula for power is $P = W/t$, where $W$ is work and $t$ is time. The SI unit of power is the Watt (W), which is equivalent to Joules per second ($J/s$). Work ($W$) can be expressed as Force ($F$) times distance ($d$), so $W = F \cdot d$. Substituting this into the power formula gives $P = (F \cdot d)/t$. In SI units, force is $kg \cdot m/s^2$, so $P = (kg \cdot m/s^2) \cdot m / s = kg \cdot m^2 / s^3$ [33](#page=33). The unit for hydraulic power is also Watts (W). The formula for hydraulic power is [33](#page=33): $$P = p \cdot Q$$ The units for this formula can be derived as follows: $W = \text{Pa} \cdot m^3/s$ $W = (N/m^2) \cdot m^3/s$ $W = [(kg \cdot m/s^2) / m^2 \cdot m^3/s$ $W = (kg \cdot m/s^2 \cdot m^3) / (m^2 \cdot s) = kg \cdot m^4 / (m^2 \cdot s^3) = kg \cdot m^2 / s^3$ [33](#page=33). ### 3.5 Exercises #### 3.5.1 Exercise 1 **Given:** Hydraulic pump with a flow rate of 40 liters per minute ($L/min$). The inner diameter of the cylinder is 5 cm. The load to be moved is 3000 kg. **Calculate:** The (hydraulic) power in kilowatts ($kW$) of the oil pump. **Tips for solving:** * First, convert the flow rate $Q$ to the SI unit $m^3/s$. * Use the formula $P = p \cdot Q$. * To calculate the pressure $p$, you will need the force $F (= m \cdot a)$ and the area $A$. The area $A$ is the inner surface area of the cylinder piston: $A = \pi \cdot r^2$ or $A = \pi \cdot D^2 / 4$, where $r$ is the radius and $D$ is the inner diameter of the cylinder piston. * Ensure you are using the correct units in the formulas [34](#page=34). #### 3.5.2 Exercise 2 **Given:** Hydraulic pump with a stroke volume of 90 $cm^3/$revolution. The tractor engine rotates at 2000 revolutions per minute ($RPM$). There is a gear reduction of 24/48 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. **Required:** * Calculate the flow rate, pressure, and (hydraulic) power of the oil pump. * Calculate the extension speed of the piston. **Note:** The stroke length of 1 meter is not directly needed for these calculations, but it would be useful if you wanted to determine how long the extension takes [35](#page=35). --- # Electrical concepts and circuit analysis This section outlines the fundamental concepts of electricity, drawing parallels with hydraulic principles, and details the analysis of series, parallel, and mixed resistor circuits. ### 4.1 Fundamentals of electricity Electricity is defined as the movement of electrons from one atom to another, constituting an electron flow [39](#page=39). #### 4.1.1 Chemical background of electricity Matter is composed of molecules, which are in turn made up of atoms. Atoms consist of a positively charged nucleus and negatively charged electrons orbiting it. For example, a copper atom (Cu) has 29 protons in its nucleus and 29 electrons, with one electron in its outermost shell. Electrons can be either bound to their atoms or free to move. The ease with which a free electron can jump from one atom to another is what makes a substance a good conductor. This movement of free electrons from atom to atom through a substance is known as electric current [39](#page=39) [40](#page=40) [41](#page=41). * **Good conductors** (e.g., copper, silver) have fewer than four electrons in their outermost shell. Copper, with only one electron in its outer shell, is an excellent conductor [41](#page=41). * **Insulators** have more than four electrons in their outermost shell [41](#page=41). #### 4.1.2 Electric charge In an electrically neutral atom, the number of positive charges (protons) equals the number of negative charges (electrons). An excess or deficiency of electrons can arise, for instance, through friction, leading to static electricity. When static electricity flows through a conductor, it becomes dynamic electricity. This movement of electrons flows from the negative (-) to the positive (+) terminal of a conductor, sustained by an electrical source [42](#page=42). #### 4.1.3 Potential difference and voltage (U) For an electron flow to occur in a conductor, there must be a difference in charge at both ends of the conductor. This difference in charge is analogous to the height difference ('h') between water containers in hydraulics, representing an electrical potential difference. The charge difference between the ends of a conductor is called voltage, measured in volts (V). In hydraulics, pressure difference ($\Delta p$) is comparable to voltage (U) in electricity [43](#page=43). Voltage can be generated chemically and/or mechanically. The relationship between voltage (U), current (I), and resistance (R) can be understood through a hydraulic analogy: Flowing water = pressure / pipe resistance. Correspondingly, in electricity, current ($I$) equals voltage ($U$) divided by resistance ($R$), expressed by Ohm's Law [44](#page=44): $$I = \frac{U}{R}$$ [44](#page=44). #### 4.1.4 Current intensity (I) The strength of an electron flow is comparable to the flow rate of water in a pipe between two reservoirs. Electric current is known as 'le courant' or 'current' [45](#page=45). * **AC (Alternating Current)** is a type of current [45](#page=45). * **DC (Direct Current)** is another type of current [45](#page=45). The SI unit for current intensity is the ampere, symbolized by 'A'. The hydraulic equivalent of current intensity is the flow rate (Q) [45](#page=45). #### 4.1.5 Resistance (R) Electrical resistance is akin to the resistance a water current encounters in a pipe due to factors like roughness, diameter, and length. The SI unit for resistance is the ohm, symbolized by '$\Omega$' [46](#page=46). The resistance ($R$) of an electrical cable depends on: * The type of material (resistivity, $\rho$) [46](#page=46). * The cross-sectional area ($A$) [46](#page=46). * The length ($l$) [46](#page=46). The resistivity ($\rho$) of a material is its specific resistance. Its unit is $\Omega \cdot \text{mm}^2/\text{m}$ [46](#page=46). Two key formulas related to resistance are: 1. **Pouillet's Law:** $$R = \rho \frac{l}{A}$$ [47](#page=47). For copper at 15°C, the resistivity is $\rho_{15°\text{C}}(\text{Cu}) = 0.0175 \, \Omega \cdot \text{mm}^2/\text{m}$ [47](#page=47). 2. **Ohm's Law:** $$U = I \cdot R$$ [47](#page=47). #### 4.1.6 Examples of resistance calculations > **Example 1:** Calculate the resistance of a 20 m long copper wire with a 1.5 mm$^2$ cross-sectional area connected to a lamp. The current flows both ways, so the total length of copper wire is 40 m. > > Using Pouillet's Law: > $R = \rho \frac{l}{A} = 0.0175 \, \Omega \cdot \text{mm}^2/\text{m} \times \frac{2 \times 20 \, \text{m}}{1.5 \, \text{mm}^2} = 0.4667 \, \Omega$ [48](#page=48). > **Example 2:** A copper wire with a 2.5 mm$^2$ cross-sectional area has a resistance of 7 $\Omega$. Calculate its length. > > Rearranging Pouillet's Law to solve for length ($l$): > $l = \frac{R \cdot A}{\rho} = \frac{7 \, \Omega \times 2.5 \, \text{mm}^2}{0.0175 \, \Omega \cdot \text{mm}^2/\text{m}} = 1000 \, \text{m}$ [48](#page=48). > **Example 3:** A 50 m long copper wire has a resistance of 1.75 $\Omega$. Calculate its cross-sectional area in mm$^2$ and its diameter in mm. > > First, calculate the cross-sectional area ($A$) using Pouillet's Law: > $A = \frac{\rho \cdot l}{R} = \frac{0.0175 \, \Omega \cdot \text{mm}^2/\text{m} \times 50 \, \text{m}}{1.75 \, \Omega} = 0.5 \, \text{mm}^2$ [49](#page=49). > > To find the diameter ($D$), use the area formula for a circle, $A = \pi r^2$ or $A = \frac{\pi D^2}{4}$: > $r^2 = \frac{A}{\pi} = \frac{0.5 \, \text{mm}^2}{\pi}$ > $r = \sqrt{\frac{0.5}{\pi}} \approx 0.4 \, \text{mm}$ > $D = 2r = 2 \times 0.4 \, \text{mm} = 0.8 \, \text{mm}$ [49](#page=49). ### 4.2 Switching of resistors Resistors can be connected in series, parallel, or a combination of both (mixed circuits) [51](#page=51). #### 4.2.1 Series connection In a series circuit, the current ($I$) is the same through all components. The total resistance ($R_{tot}$) is the sum of the individual resistances [52](#page=52): $$R_{tot} = R_1 + R_2 + R_3 + \dots$$ [52](#page=52). The total voltage ($U_{tot}$) is the sum of the voltage drops across each resistor: $$U_{tot} = U_1 + U_2 + U_3 + \dots$$ [52](#page=52). > **Example (Series Circuit):** Four resistors with values 5 $\Omega$, 8 $\Omega$, 20 $\Omega$, and 3 $\Omega$ are connected in series across a 12 V source. > > 1. **Total resistance:** > $R_{tot} = 5 \, \Omega + 8 \, \Omega + 20 \, \Omega + 3 \, \Omega = 36 \, \Omega$ [52](#page=52). > > 2. **Total current:** Using Ohm's Law ($I = U/R$): > $I_{tot} = \frac{12 \, \text{V}}{36 \, \Omega} = \frac{1}{3} \, \text{A} \approx 0.33 \, \text{A}$ [53](#page=53). > This current is the same for all resistors ($I_1 = I_2 = I_3 = I_4 \approx 0.33 \, \text{A}$). > > 3. **Voltage drops across each resistor:** Using Ohm's Law ($U = I \cdot R$): > $U_1 = 0.33 \, \text{A} \times 5 \, \Omega = 1.65 \, \text{V}$ [53](#page=53). > $U_2 = 0.33 \, \text{A} \times 8 \, \Omega = 2.64 \, \text{V}$ [53](#page=53). > $U_3 = 0.33 \, \text{A} \times 20 \, \Omega = 6.60 \, \text{V}$ [53](#page=53). > $U_4 = 0.33 \, \text{A} \times 3 \, \Omega = 0.99 \, \text{V}$ [53](#page=53). > The sum of these voltage drops is $1.65 + 2.64 + 6.60 + 0.99 = 11.88 \, \text{V}$, which is approximately equal to the source voltage (12 V) due to rounding. #### 4.2.2 Parallel connection In a parallel circuit, the voltage ($U$) is the same across all components. The total current ($I_{tot}$) is the sum of the currents through each branch [54](#page=54): $$I_{tot} = I_1 + I_2 + I_3 + \dots$$ [55](#page=55). The reciprocal of the total resistance ($R_{tot}$) is the sum of the reciprocals of the individual resistances: $$\frac{1}{R_{tot}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots$$ [54](#page=54). > **Example (Parallel Circuit):** Four resistors with values 5 $\Omega$, 8 $\Omega$, 20 $\Omega$, and 3 $\Omega$ are connected in parallel across a 12 V source. > > 1. **Total resistance:** > $\frac{1}{R_{tot}} = \frac{1}{5 \, \Omega} + \frac{1}{8 \, \Omega} + \frac{1}{20 \, \Omega} + \frac{1}{3 \, \Omega} = \frac{24 + 15 + 6 + 40}{120} = \frac{85}{120}$ [54](#page=54). > $R_{tot} = \frac{120}{85} \, \Omega \approx 1.41 \, \Omega$ [54](#page=54). > > 2. **Total current:** Using Ohm's Law ($I = U/R$): > $I_{tot} = \frac{12 \, \text{V}}{1.41 \, \Omega} \approx 8.51 \, \text{A}$ (The example calculates $I_{tot}$ by summing individual currents, which is $2.4 + 1.5 + 0.60 + 4 = 8.5 \, \text{A}$) [55](#page=55). > > 3. **Currents through each resistor:** Using Ohm's Law ($I = U/R$): > $I_1 = \frac{12 \, \text{V}}{5 \, \Omega} = 2.4 \, \text{A}$ [55](#page=55). > $I_2 = \frac{12 \, \text{V}}{8 \, \Omega} = 1.5 \, \text{A}$ [55](#page=55). > $I_3 = \frac{12 \, \text{V}}{20 \, \Omega} = 0.60 \, \text{A}$ [55](#page=55). > $I_4 = \frac{12 \, \text{V}}{3 \, \Omega} = 4 \, \text{A}$ [55](#page=55). > The sum of these currents is $2.4 + 1.5 + 0.60 + 4 = 8.5 \, \text{A}$. #### 4.2.3 Mixed circuits Mixed circuits combine both series and parallel connections of resistors. In practice, most circuits encountered are mixed [56](#page=56). The general strategy for analyzing mixed circuits is: 1. Decompose the circuit into simpler series and parallel sub-circuits [56](#page=56). 2. Calculate the equivalent resistance of parallel sections to simplify them into single resistors [56](#page=56). 3. Repeat this process until the entire circuit is reduced to a single equivalent resistance [56](#page=56). 4. Then, work backward in reverse order to calculate unknown voltages and currents in each component [56](#page=56). 5. Results for resistance ($R$), current ($I$), and voltage ($U$) should be presented in a clear table [56](#page=56). > **Example (Mixed Circuit 1):** A circuit with resistors $R_1=2 \, \Omega$, $R_2=4 \, \Omega$, and a parallel combination of $R_3=6 \, \Omega$, $R_4=8 \, \Omega$, and $R_5=4 \, \Omega$, followed by a series resistor $R_6=3 \, \Omega$, connected to a 220 V source. > > 1. **Calculate equivalent resistance of the parallel section ($R_{3,4,5}$):** > $\frac{1}{R_{3,4,5}} = \frac{1}{6 \, \Omega} + \frac{1}{8 \, \Omega} + \frac{1}{4 \, \Omega} = \frac{4 + 3 + 6}{24} = \frac{13}{24}$ [57](#page=57). > $R_{3,4,5} = \frac{24}{13} \, \Omega \approx 1.846 \, \Omega$ [57](#page=57). > > 2. **Calculate total resistance ($R_{tot}$):** This is a series combination of $R_1$, $R_2$, $R_{3,4,5}$, and $R_6$. > $R_{tot} = 2 \, \Omega + 4 \, \Omega + 1.846 \, \Omega + 3 \, \Omega = 10.846 \, \Omega$ (The example has a typo and uses 1.85 $\Omega$, resulting in 10.85 $\Omega$) [57](#page=57). > > 3. **Calculate total current ($I_{tot}$):** > $I_{tot} = \frac{U_{source}}{R_{tot}} = \frac{220 \, \text{V}}{10.85 \, \Omega} \approx 20.28 \, \text{A}$ [58](#page=58). > > 4. **Calculate voltage drops and currents for each component:** > | Component | $R (\Omega)$ | $I (A)$ | $U (V)$ | > | :--------------- | :---------- | :------ | :------ | > | $R_1$ | 2 | 20.28 | 40.56 | > | $R_2$ | 4 | 20.28 | 81.12 | > | $R_{3,4,5}$ (parallel) | 1.85 | 20.28 | 37.52 | > | $R_6$ | 3 | 20.28 | 60.84 | > | Total | 10.85 | 20.28 | 220 | > > **Individual currents within the parallel section ($R_3, R_4, R_5$) are calculated using the voltage across the parallel section ($U_{3,4,5} \approx 37.52 \, V$):** > $I_3 = \frac{37.52 \, \text{V}}{6 \, \Omega} \approx 6.25 \, \text{A}$ [58](#page=58). > $I_4 = \frac{37.52 \, \text{V}}{8 \, \Omega} \approx 4.69 \, \text{A}$ [58](#page=58). > $I_5 = \frac{37.52 \, \text{V}}{4 \, \Omega} \approx 9.38 \, \text{A}$ [58](#page=58). > (Note: The sum of $I_3, I_4, I_5$ is $6.25 + 4.69 + 9.38 = 20.32 \, A$, which is close to the total current $I_{tot} \approx 20.28 \, A$) > **Example (Mixed Circuit 2):** A circuit with resistors $R_1=10 \, \Omega$, a parallel combination of $R_2=10 \, \Omega$ and $R_3=10 \, \Omega$, a series resistor $R_4=10 \, \Omega$, all connected to a 24 V source. > > 1. **Equivalent resistance of the parallel section ($R_{2,3}$):** > $\frac{1}{R_{2,3}} = \frac{1}{10 \, \Omega} + \frac{1}{10 \, \Omega} = \frac{2}{10}$ [60](#page=60). > $R_{2,3} = \frac{10}{2} \, \Omega = 5 \, \Omega$ [60](#page=60). > > 2. **Total resistance ($R_{tot}$):** This is a series combination of $R_1$, $R_{2,3}$, and $R_4$. > $R_{tot} = 10 \, \Omega + 5 \, \Omega + 10 \, \Omega = 25 \, \Omega$ [60](#page=60). > > 3. **Total current ($I_{tot}$):** > $I_{tot} = \frac{24 \, \text{V}}{25 \, \Omega} = 0.96 \, \text{A}$ [60](#page=60). > > 4. **Calculate voltages and currents:** > | Component | $R (\Omega)$ | $I (A)$ | $U (V)$ | > | :-------- | :---------- | :------ | :------ | > | $R_1$ | 10 | 0.96 | 9.6 | > | $R_{2,3}$ (parallel) | 5 | 0.96 | 4.8 | > | $R_4$ | 10 | 0.96 | 9.6 | > | Total | 25 | 0.96 | 24 | > > **Individual currents within the parallel section ($R_2, R_3$):** Since $R_2=R_3=10\,\Omega$ and the voltage across them is $4.8\,V$: > $I_2 = I_3 = \frac{4.8 \, \text{V}}{10 \, \Omega} = 0.48 \, \text{A}$ [60](#page=60). > **Example (Mixed Circuit 3):** A circuit with resistors $R_1=10 \, \Omega$, a parallel combination of $R_2=10 \, \Omega$, $R_3=10 \, \Omega$, $R_4=10 \, \Omega$, $R_5=10 \, \Omega$, and $R_6=10 \, \Omega$, all connected to a 24 V source. Note that $R_2$ and $R_5$ are in series, forming a $20 \, \Omega$ resistance. This combination is in parallel with $R_3$ and $R_4$ (each $10 \, \Omega$). > > 1. **Series combination of $R_2$ and $R_5$:** > $R_{2,5} = R_2 + R_5 = 10 \, \Omega + 10 \, \Omega = 20 \, \Omega$ [61](#page=61). > > 2. **Equivalent resistance of the parallel section ($R_{2,3,4,5}$):** This section consists of $R_{2,5}$ (20 $\Omega$) in parallel with $R_3$ (10 $\Omega$) and $R_4$ (10 $\Omega$). > $\frac{1}{R_{2,3,4,5}} = \frac{1}{R_{2,5}} + \frac{1}{R_3} + \frac{1}{R_4} = \frac{1}{20 \, \Omega} + \frac{1}{10 \, \Omega} + \frac{1}{10 \, \Omega} = \frac{1 + 2 + 2}{20} = \frac{5}{20}$ [61](#page=61). > $R_{2,3,4,5} = \frac{20}{5} \, \Omega = 4 \, \Omega$ [61](#page=61). > > 3. **Total resistance ($R_{tot}$):** This is a series combination of $R_1$, $R_{2,3,4,5}$, and $R_6$. > $R_{tot} = R_1 + R_{2,3,4,5} + R_6 = 10 \, \Omega + 4 \, \Omega + 10 \, \Omega = 24 \, \Omega$ [61](#page=61). > > 4. **Total current ($I_{tot}$):** > $I_{tot} = \frac{24 \, \text{V}}{24 \, \Omega} = 1 \, \text{A}$ [61](#page=61). > > 5. **Calculate voltages and currents:** > | Component | $R (\Omega)$ | $I (A)$ | $U (V)$ | > | :--------------- | :---------- | :------ | :------ | > | $R_1$ | 10 | 1 | 10 | > | $R_{2,3,4,5}$ (parallel) | 4 | 1 | 4 | > | $R_6$ | 10 | 1 | 10 | > | Total | 24 | 1 | 24 | > > **Individual currents within the parallel section ($R_2, R_3, R_4, R_5$):** The voltage across this section is 4 V. > $I_2 = \frac{4 \, \text{V}}{10 \, \Omega} = 0.4 \, \text{A}$ (This is incorrect in the document, it should be $0.2 \, A$ for $R_2$ and $R_5$ which are in series, and $0.4 \, A$ for $R_3, R_4$) > $I_5 = \frac{4 \, \text{V}}{10 \, \Omega} = 0.4 \, \text{A}$ (This is incorrect in the document, it should be $0.2 \, A$) > $I_3 = \frac{4 \, \text{V}}{10 \, \Omega} = 0.4 \, \text{A}$ [61](#page=61). > $I_4 = \frac{4 \, \text{V}}{10 \, \Omega} = 0.4 \, \text{A}$ [61](#page=61). > > The total current through the parallel section is $I_{tot} = I_1 = I_{2,3,4,5} = I_6 = 1 \, A$. > The voltage across $R_{2,3,4,5}$ is $U_{2,3,4,5} = I_{tot} \times R_{2,3,4,5} = 1 \, A \times 4 \, \Omega = 4 \, V$. > > Now, let's correctly calculate the currents within the parallel branches: > The 20 $\Omega$ branch ($R_2$ and $R_5$ in series) has current $I_{2,5} = \frac{4 \, V}{20 \, \Omega} = 0.2 \, A$. > The $R_3$ branch has current $I_3 = \frac{4 \, V}{10 \, \Omega} = 0.4 \, A$. > The $R_4$ branch has current $I_4 = \frac{4 \, V}{10 \, \Omega} = 0.4 \, A$. > > Therefore, the currents in the table should be: > | Component | $R (\Omega)$ | $I (A)$ | $U (V)$ | > | :-------- | :---------- | :------ | :------ | > | $R_1$ | 10 | 1 | 10 | > | $R_2$ | 10 | 0.2 | 2 | > | $R_3$ | 10 | 0.4 | 4 | > | $R_4$ | 10 | 0.4 | 4 | > | $R_5$ | 10 | 0.2 | 2 | > | $R_6$ | 10 | 1 | 10 | > | Total | 24 | 1 | 24 | > > The document's table presents: > | n | R ($\Omega$) | I (A) | U (V) | > | --- | :---------- | :---- | :---- | > | 1 | 10 | 1 | 10 | > | 2 | 10 | 0.2 | 2 | > | 3 | 10 | 0.4 | 4 | > | 4 | 10 | 0.4 | 4 | > | 5 | 10 | 0.2 | 2 | > | 6 | 10 | 1 | 10 | > | 2,5 | 20 | 0.2 | 4 | (This row correctly shows the series $R_{2,5}$ component) > | 2,3,4,5 | 4 | 1 | 4 | (This row correctly shows the parallel combination $R_{2,3,4,5}$) > | Tot. | 24 | 1 | 24 | --- ## 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 units | The International System of Units (SI) is the modern form of the metric system and the world's most widely used system of measurement. It is based on seven base units: the meter, kilogram, second, ampere, kelvin, mole, and candela. | | Base quantities | These are the fundamental physical quantities that are defined independently of each other. In the SI system, there are seven base quantities: length, mass, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity. | | Derived quantities | These are quantities derived from the base quantities by combination, using multiplication and division. Examples include force, pressure, energy, and power. | | Prefixes | Prefixes are used in the metric system to denote multiples or submultiples of units. Examples include kilo (10^3), mega (10^6), milli (10^-3), and micro (10^-6). | | Force | A force is an influence that can change the motion of an object. It is a vector quantity, meaning it has both magnitude and direction. The SI unit of force is the Newton (N). | | Pressure | Pressure is defined as force per unit area. It is the measure of how concentrated a force is over a surface. The SI unit of pressure is the Pascal (Pa), which is equal to one Newton per square meter. | | Work | Work is done when a force causes a displacement. In physics, work is calculated as the product of the force and the distance moved in the direction of the force. The SI unit of work is the Joule (J). | | Energy | Energy is the capacity to do work. It exists in many forms, such as kinetic, potential, thermal, electrical, and chemical energy. The SI unit of energy is the Joule (J). | | Power | Power is the rate at which work is done or energy is transferred. It is calculated as work divided by time. The SI unit of power is the Watt (W). | | Kelvin temperature | Kelvin is the base unit of thermodynamic temperature in the International System of Units (SI). It is an absolute scale, meaning 0 K represents absolute zero, the theoretical point at which all molecular motion ceases. | | Specific heat capacity | Specific heat capacity is the amount of heat energy required to raise the temperature of one unit of mass of a substance by one degree Celsius or one Kelvin. It is an intrinsic property of a material. | | Pascal's law | Pascal's law states that a pressure change at any point in a confined incompressible fluid is transmitted equally throughout the fluid and to the walls of the containing vessel. This is fundamental to hydraulic systems. | | Flow rate (Debiet) | Flow rate, also known as discharge, is the volume of fluid that passes through a given surface per unit time. The SI unit for flow rate is cubic meters per second (m³/s). | | Hydraulic power | Hydraulic power is the power transmitted by a fluid in a hydraulic system. It is calculated as the product of pressure and flow rate. The SI unit is the Watt (W). | | Electric charge | Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electromagnetic field. It can be positive or negative. The SI unit of electric charge is the Coulomb (C). | | Voltage (Potential difference) | Voltage, also known as electric potential difference, is the difference in electric potential between two points. It is the driving force that causes electric charge to flow. The SI unit of voltage is the Volt (V). | | Electric current (Current intensity) | Electric current is the rate of flow of electric charge. It is analogous to the flow of water in a pipe. The SI unit of electric current is the Ampere (A). | | Electrical resistance | Electrical resistance is a measure of the opposition to the flow of electric current in a conductor. It is analogous to friction or resistance to fluid flow. The SI unit of resistance is the Ohm (Ω). | | Ohm's Law | Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance between them. Mathematically, $I = U/R$. | | Series circuit | In a series circuit, components are connected end-to-end, so that the same current flows through each component. The total resistance is the sum of individual resistances. | | Parallel circuit | In a parallel circuit, components are connected across the same two points, so that the voltage across each component is the same. The total current is the sum of the currents through each branch. | | Mixed circuit | A mixed circuit, also known as a series-parallel circuit, combines elements of both series and parallel connections. Analyzing such circuits involves breaking them down into simpler series and parallel components. |