====== Block 07 — Power-relevant figures ====== ===== Learning objectives ===== * Define and compute **input/output power**, **losses**, **efficiency** $\eta$ and **utilization rate** $\varepsilon$ for DC sources and loads. * Use the **real source model** with internal resistance $R_{\rm i}$ to compute operating point $(U_{\rm L}, I_{\rm L})$, $P_{\rm L}$ and $P_{\rm loss}$. * Understand the different design goals: - **High efficiency** (power engineering): $R_{\rm L} \gg R_{\rm i}$. - **Maximum power transfer** (communications): $R_{\rm L} = R_{\rm i}$. * Combine efficiencies along a **power-flow chain**. * Relate these figures to **Thevenin/Norton** equivalents and the **loaded voltage divider**. ~~PAGEBREAK~~ ~~CLEARFIX~~ ===== Preparation at Home ===== And again: * Please read through the following chapter. * Also here, there are some clips for more clarification under 'Embedded resources'. For checking your understanding please do the following exercises: * 7.1 * E3.3.3 ~~PAGEBREAK~~ ~~CLEARFIX~~ ===== 90-minute plan ===== - Warm-up (8 min): recall passive/active sign convention; quick unit check for $P=U\cdot I$. - Core concepts (35 min): real source model; definitions of $\eta$ and $\varepsilon$; design goals; chain efficiency. - Worked example (10 min): battery + internal resistance + load. - Two-port view & loaded divider (12 min): quick Thevenin/Norton recap; loaded divider formulas. - Practice (20 min): 3 short exercises (see panels below). - Wrap-up (5 min): summary + pitfalls. ===== Conceptual overview ===== - Real sources are modeled by an **ideal source** plus **internal resistance** $R_{\rm i}$; the terminal voltage **drops under load**. - **Efficiency** $\eta$ compares *delivered* to *drawn* power. In the simple DC source–load case, $\displaystyle \eta=\frac{R_{\rm L}}{R_{\rm L}+R_{\rm i}}$ (dimensionless). High-efficiency design wants $R_{\rm L}\gg R_{\rm i}$. - **Utilization rate** $\varepsilon$ compares delivered power to the **maximum** available from the ideal source: $\displaystyle \varepsilon=\frac{R_{\rm L}R_{\rm i}}{(R_{\rm L}+R_{\rm i})^2}$. It peaks at $R_{\rm L}=R_{\rm i}$ with $\varepsilon_{\max}=25~\%$. This is the **maximum power transfer** condition. - Different goals → different $R_{\rm L}$: * **Power engineering**: maximize $\eta$ → $R_{\rm L}\gg R_{\rm i}$. * **Communications** (matching, antennas, RF): maximize $P_{\rm L}$ → $R_{\rm L}=R_{\rm i}$, $\eta=50~\%$. ~~PAGEBREAK~~ ~~CLEARFIX~~ ===== Core content ===== ==== Power Measurement ==== First, it is necessary to consider how to determine the power. The power meter (or wattmeter) consists of a combined ammeter and voltmeter. In the wattmeter with the circuit symbol can be seen as a round network with crossed measuring inputs. The circuit also shows one wattmeter each for the (not externally measurable) output power of the ideal source $P_\rm S$ and the input power of the load $P_\rm L$. {{drawio>LeistungsmessungLineareSpannungsquelle.svg}} ~~PAGEBREAK~~ ~~CLEARFIX~~ ==== Power and Characteristics in Diagrams ==== The simulation in shows the following: * The circuit with linear voltage source ($U_0$ and $R_\rm i$), and a resistive load $R_\rm L$. * A simulated wattmeter, where the ammeter is implemented by a measuring resistor $R_\rm S$ (English: shunt) and a voltage measurement for $U_\rm S$. The power is then: $P_\rm L = {{1}\over{R_\rm S}}\cdot U_\rm S \cdot U_\rm L$. * in the oscilloscope section (below). * On the left is the power $P_\rm L$ plotted against time in a graph. * On the right is the already-known current-voltage diagram of the current values. * The slider load resistance $R_\rm L$, with which the value of the load resistance $R_\rm L$ can be changed. Now try to vary the value of the load resistance $R_\rm L$ (slider) in the simulation so that the maximum power is achieved. Which resistance value is set? \\ {{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0mQrFaB2AzADgJwIGyU1qpDvBPOiIgCyqUCmAtGGAFABuIOVIATFZJ25gq3AQJ6JEYpPBYB3QeBHgwPJaJYAnFWuFDkOdVHBxI83pAp7e-I2e38B1xzemOzCl3wGpUh71DmAQFcroGecEZUodYeIF620dwBZgBGIKhCYMjgmDk8yBBmAB4ZVpC68IhgCPHgFGwA9gA2AC4AhgDmdAA6AM59jQCumgDGdCylEmroiKg8uvh1YBQADo1ydJr9PT0AdgC2dK1bAJZ7nUGWRtmGsSyr8THKSWFFgajIPuhCL5HW7xKZRA6EM6BmmGWFGa5zo7U0kxBah4xBAmGSSyEFAAqoiCLxVCZkmBqOA1ECwERwMhdFRdMhSSsQAAlRHCO7IKxUKzoCBCNSnRF+fzMDKcgncLEsoXocRgQwYEWQ-kgAAy5lQ8Fp3Awvw0pXg1zA2EowlytBV6tKVEwtBWtEN1VBUJAuNKqBNqjUHruEmWaitZTt8rKVmNLoAkiw6eJbNZfHdlAJMDgNX4jJrtYF3cgcs6qKKsC7mo12gATFiNZbGAs4O3QCSMKwN+CMGrGNQ6yu8HIQHL8TCQnjQKSwP0QIqwSDc8G8iQsDB1JyRdxq0tl-rMuh9U59Dp7cabgD66qAA noborder}} ~~PAGEBREAK~~ ~~CLEARFIX~~ shows three diagrams: * Diagram top: current-voltage diagram of a linear voltage source. * Diagram in the middle: source power $P_\rm S$ and consumer power $P_\rm L$ versus delivered voltage $U_\rm L$. * Diagram below: Reference quantities over delivered voltage $U_\rm L$. The two powers are defined as follows: * source power: $\, \, \large{ P_\rm S = U_0 \cdot I_\rm L} $ * consumer power: $\large{ P_\rm L = U_\rm L \cdot I_\rm L} $ - Both power $P_\rm S$ and $P_\rm L$ are equal to 0 without current flow. The source power becomes maximum, at maximum current flow, that is when the load resistance $R_\rm L=0$. In this case, all the power flows out through the internal resistor. The efficiency drops to $0~\%$. This is the case, for example, with a battery shorted by a wire. - If the load resistance becomes just as large as the internal resistance $R_\rm L=R_\rm i$, the result is a voltage divider where the load voltage becomes just half the open circuit voltage: $U_\rm L = {{1}\over{2}}\cdot U_{\rm OC}$. On the other hand, the current is also half the short-circuit current $I_\rm L=I_{\rm SC}$, since the resistance at the ideal voltage source is twice that in the short-circuit case. - If the load resistance becomes high impedance $R_{\rm L}\rightarrow\infty$, less and less current flows, but more and more voltage drops across the load. Thus, the efficiency increases and approaches $100~\%$ for $R_{\rm L}\rightarrow\infty$. {{drawio>StromSpannungsDiagrammMitLeistung.svg}} The whole context can be investigated in this [[https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgzCAMB0mQrFaB2MAOAnAgbJDmxJt4BGEeDckAFgngFMBaEkgKADcRGNsuAmeL25Cw8PlAnVIVabOjxWAJxDZqXUeJJ806sRJJxIrAO4gSyEXr6QdjDVCXhr652GeMBvaQaOm30jz41a1s+IIdTElx+cKiYtV8uHnikoU8HZTC1OysbXXE5QzY+LRBkxlUzUuZtCQhrQzgAKgAKAENGACMASlbOxgBjbpMnAPt0bPtE-xSJlKMAczMMcTtNFa442VZrZGXVypINxjRpNQAFAH0AGVZOzb4RDEoSakpmNDIje+wyYRAwtIwF5WAAPB6IRhSSjUIRSXhqEg6AA2AEsAHb0NqKAA6AGcAGoAe2RABc2gt6PiAMpEgCuigG9DBmzAe2YeTsAQMiEROjapPxpIAFlS8bSGUyRuZLJpajkCiMQikogEsg4AA7gNCTPRzQKIiTTdzjE3OIxs6TaWymsZ6L4RUb8QROg0OcGMZB-bTsij8HU0Mw6ACqLMY8HZ2lWpH4yBhZkV4JYk1ijxcZD5IAASiySIR1EizG4C7yEyBUSztGk0Ig+Ph-eJMznwYQ1adwERY3tEeJbuCpJH69Q8-wMKXSn3wBgIFyIFJIa3A4XQy2DOpkGowHE7F6l72WYQZ+ggZy2d2gyAAJKsS0ufIqXWaI1hvPsvNA2xac+F85E4z0XE8QAWSxPEGXoABbeh0VJPE7lZdkx21cA9iMXYJAqBIuHbC5LmpVgljGApNi8BwSlWcpKhOblaiBAFGmadoul6JihkcNZ7yhG09EKOA2FMf4FS4ahuMVSI0nCQT1TQ0o+GQVYRIBNBP1KK0QE6AB6docQ1GBIBIYYlmBT9eGMsxSLQyA9jMriaDgYSdDUehyXgoI-mSYdTJBcE3M2KyaC3Py1EzABhNpkQGOlkQFVEiXRfEiQAM3xABRRLEtRAZUWggYAE8CNSFIpLybYln+AJki2Mj-LkhSdFqrhKlw-ClmSaRKosnZ-PKYt5LvNRADHgPhGC6vZKLUPrMMDIadKVZT73VITLL2BrmF4Vbi1wycqIMHQSHgbIeSXflBTxEUxWuIk2gAE1GgFJPsIIxmBabhru5UhKRSZqGCkAZo1O7BJ29wmr+khZvQwTixqIjppIEbIcUoSprsF7BvBgHIeh+w1vUNGwZGj0I3UNA0iPbBHLLX9-1xHF0RAto8WgSt7LsDAyD4MBJlQPcQGpgCcTphmmdYIkAQfSQokbaBnlluX5YwJg4hIJXUPF8QIEvFpg26fFzmRIlSVFmhDWkUFPn03gSGgAQkDIDmkFWCwTfAJAEAkNKMqynLcvxZy2mNl6HXNgw4kQGAHafT0reIv5nYlRkxX5wCrmpfEWgWRR6Gg3W8TadFrvxS6br1v8BbxK5rnTrPruGdBXYAMQdHw7NZshrkZ0ljFRa6ALO-PC7xLMblYIA|Simulation with a resistor]] or [[https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgzCAMB0lw7ADgIwCYCsVrOWMyAWRA9MSANnMVXInUREwLoFMBaHAKADcQ2BOcnwxCBosOlRRpBSI2lyY6TgCcQ5AnwlS0DNtunI4kTgHcQyeOMkhUkPQZNqNctqkRyXw5ELlGT5mB2wqiadnqooVBmFuSukZo+IZoBfILJaaIi0WoJWjbh+VKKsHDInKhoIOlsGhZVHO7SEHbGcABUABQAhmwARgCUXX1sAMYDMUGuBmDERdGBwW5Rs5rLKZwA5hb8Uvo6u3xJihWQ8Dt7dciHbB4gmgAKAPoAMpx9RzRa-PwWBL8cFDRD7kZCZWzBMC+TgAD0+mDYsl+BFEsiEiQYABsAJYAOxY3RUAB0AM4ANQA9piAC7dTYsUkAZQpAFcVKMWLCjmBzhx7FpXEYmBYGN1qaTqQALBkk5lsjkxSzWHRNfQ2VKFdaxeJREwAB3AczVUlWGQgfkmSxmVsgGx5cncDhsU3mYIdlvi8BNSzyJjhbHgYLciF56AB7k0GJAAFUuWx0Lz3Ht0EHUPBkRZilycGsgokvvswVGAErZshaZAMXB7fDCqrYrl4PbuTBgAhrJP3EUgUtwttZO4GNzwc75kBvOGyRP8BgEXDCfh1qQT8D8CD6SAQWQIshyKOxvtGLTwTT4cSBrtVFdkdezORkBwjy8MACSnHtWhrNjqxos0lQca4LyuD3noaCjt2DwUqYLDEiSACyBIkmyLAALYsLi1Iku83K8ouhrgOcJh2ERfB1K4dyPE8jJbNI+x0cc0SVHsNQ-ncjQMPethtB0PT9EMfHjKon7zIiTrFFgxjlOYYiiUQ8ypEkWqyVqxFVGmezyY6RxVA6IB9AA9D0RJ6jAkDIBM2xQmBQjWbECinOcdlifccB8PJmgsLSOGhEG6RzrZ0Jwr5RxnPc+ChZG3YAMLdJiowspiYrYhSuKkhSABmpIAKIZRl2KjNiGGjAAnrRKl5BV-InNssmuOkjHEWFGnuQwLW1FFzw0ds6RyA1vhMWFNRBCAXoiZogBjwKgbCOdUoh1GNHVdlNJkxNpv6VY4s3tUk7UjVRK4-kYVboGsQrPiAYoStKpIvBS3QACbbVEG2yFoULLdN238r+la5u2n2rSR4JLY0rh1JNyBA2FKkjWD40gBNyAzcDLm-gGNYA5D0O8nDQ5KXg6KI8jcYJloiBZLeVCXlIUEwcSRK4oh3QktA2b4foIYQgCUIZlUdOwUSjPM6znAZbYiARJL4A+tLI3INAmBmQBxYS1LbVvcGnF-sEMAtAo2Bhsgp38MwkC7JwFK2OoKQgDCKCQDYBDQD8rtu+7-DsEkyBe6RxSK-+4CSZAEkvp00YDKSDyYhS1KW-ciTSPbRgPUICuhB7meexwae+zrWBuowfvSLl+WFcVJWkl53TxzoScO6n8gwGCUj+0rmYG+3erQYLJKpddMox4974MBAABihf+K5bkcOOLPUqY2IPbBJK0riD2ksWrycEAA|this one with a variable load]]. ==== The Efficiency ==== To understand the lower diagram in , the definition equations of the two reference quantities shall be described here again: The **efficiency** $\eta$ describes the delivered power (consumer power) concerning the supplied power (power of the ideal source): \begin{align*} \eta = {{P_{\rm out}}\over{P_{\rm in}}} = {{R_{\rm L}\cdot I_{\rm L}^2}\over{(R_{\rm L}+R_{\rm i}) \cdot I_{\rm L}^2}} \quad \rightarrow \quad \boxed{ \eta = {{R_{\rm L}} \over {R_{\rm L}+R_{\rm i}}} } \end{align*} Once we want to get the **relative maximum power** out of a system (so maximum power related to the input power) the efficiency should go towards $\eta \rightarrow 100\%$. This situation close to (1.) in . Application: - In __power engineering__ $\eta \rightarrow 100\%$ is often desired: We want the maximum power output with the lowest losses at the internal resistance of the source. Thus, the internal resistance of the source should be low compared to the load $R_{\rm L} \gg R_{\rm i} $. The **utilization rate** $\varepsilon$ describes the delivered power $P_{\rm out}$ concerning the maximum possible power $P_{\rm in, max}$ of the ideal source. Here, the currently supplied power is not assumed (as in the case of efficiency), but the best possible power of the ideal source, i.e. in the short-circuit case: \begin{align*} \varepsilon = {{P_{\rm out}}\over{P_{\rm in, max}}} = {{R_{\rm L} \cdot I_{\rm L}^2}\over{{U_0^2}\over{R_i}}} = {{R_{\rm L}\cdot R_{\rm i} \cdot I_{\rm L}^2}\over {U_0^2}} = {{R_{\rm L}\cdot R_{\rm i} \cdot \left({{U_0}\over {R_{\rm L}+R_{\rm i}}}\right)^2} \over {U_0^2}} \quad \rightarrow \quad \boxed{\varepsilon = {{R_{\rm L}\cdot R_{\rm i}} \over {(R_{\rm L}+R_{\rm i})^2}} = {{R_{\rm L}} \over {(R_{\rm L}+R_{\rm i})}}\cdot {{R_{\rm i}} \over {(R_{\rm L}+R_{\rm i})}}} \end{align*} ==== The Utilization Rate ==== In other applications, the **absolute maximum power** has to be taken from the source, without consideration of the losses via the internal resistance. This corresponds to the situation (2.) in . For this purpose, the internal resistance of the source and the load are matched. This case is called **{{https://en.wikipedia.org/wiki/Impedance_matching|impedance matching}} ** (the impedance is up to for DC circuits equal to the resistance). The utilization rate here becomes maximum: $\varepsilon = 25~\%$. Application: - In __communications engineering__ the impedance matching of the source (the antenna) and the load (the signal-acquiring microcontroller) uses resistors, capacitors, and inductors. There, we want to get the maximum power out of an antenna. For this purpose, the internal resistance of the source (e.g., a receiver) and the load (e.g., the downstream evaluation) are matched. An example can be seen in this {{electrical_engineering_1:anp084a_en_-_impedance_matching_for_near_field_com.pdf#page=4|application note for near field communication}}. - Furthermore, also for __photovoltaic cells__ one wants to get the maximum power out. In this case, the concept is often called **{{https://en.wikipedia.org/wiki/Maximum_power_point_tracking|Maximum Power Point Tracking (MPPT)}} ** The impedance matching/power matching is also [[https://www.youtube.com/watch?v=BJLlXUD6CsM|here]] explained in a German video. ~~PAGEBREAK~~ ~~CLEARFIX~~ **Given:** $U_0=12.0~\rm V$, $R_{\rm i}=0.50~\Omega$, $R_{\rm L}=5.0~\Omega$. **Find:** $U_{\rm L}$, $I_{\rm L}$, $P_{\rm L}$, $\eta$, $\varepsilon$. \begin{align*} I_{\rm L} &= \frac{12.0~{\rm V}}{0.50~\Omega+5.0~\Omega} = 2.182~{\rm A} \\ U_{\rm L} &= I_{\rm L} R_{\rm L}= 2.182~{\rm A}\cdot 5.0~\Omega = 10.91~{\rm V} \\ P_{\rm L} &= U_{\rm L}I_{\rm L} = 10.91~{\rm V}\cdot 2.182~{\rm A}= 23.8~{\rm W} \\ P_{\rm in,max}&= \frac{U_0^2}{R_{\rm i}}=\frac{(12.0~{\rm V})^2}{0.50~\Omega}=288~{\rm W} \\ \eta &= \frac{R_{\rm L}}{R_{\rm i}+R_{\rm L}}=\frac{5.0~\Omega}{5.50~\Omega}=0.909=90.9~\% \\ \varepsilon &= \frac{R_{\rm L}R_{\rm i}}{(R_{\rm L}+R_{\rm i})^2}=\frac{5.0\cdot 0.50}{(5.50)^2}=0.0826=8.26~\% \end{align*} Interpretation: very **efficient** (small $R_{\rm i}$) but using only **8.26 %** of the source’s ideal maximum capability $U_0^2/R_{\rm i}$—which is fine for power engineering aims. ~~PAGEBREAK~~ ~~CLEARFIX~~ ==== Power-flow chains (series stages) ==== The usable (= outgoing) $P_{\rm O}$ power of a real system is always smaller than the supplied (incoming) power $P_{\rm I}$. This is due to the fact, that there are additional losses in reality. \\ The difference is called power loss $P_{\rm loss}$. It is thus valid: $P_{\rm I} = P_{\rm O} + P_{\rm loss}$ Instead of the power loss $P_{\rm loss}$, the efficiency $\eta$ is often given: $\boxed{\eta = {{P_{\rm O}}\over{P_{\rm I}}}\overset{!}{<} 1}$ For cascaded conversions (cf. ), the **overall efficiency is the product** of stage efficiencies: $\boxed{\eta = {{P_{\rm O}}\over{P_{\rm I}}} = {\not{P_{1}}\over{P_{\rm I}}}\cdot {\not{P_{2}}\over \not{P_{1}}}\cdot {{P_{\rm O}}\over \not{P_{2}}} = \eta_1 \cdot \eta_2 \cdot \eta_3}$ {{drawio>Leistungsfluss.svg}} ~~PAGEBREAK~~ ~~CLEARFIX~~ ===== Exercises ===== {{fa>pencil?32}} A source has $U_0=9.0~\rm V$, $R_{\rm i}=1.0~\Omega$. - (a) Choose $R_{\rm L}=9.0~\Omega$. Compute $I_{\rm L}$, $U_{\rm L}$, $P_{\rm L}$, $\eta$, $\varepsilon$. - (b) Choose $R_{\rm L}=1.0~\Omega$. Repeat. Which choice maximizes $P_{\rm L}$? Which yields higher $\eta$? **Strategy:** use the boxed formulas in this block; for (b) note $R_{\rm L}=R_{\rm i} \Rightarrow \eta=50~\%$. {{fa>pencil?32}} A battery (stage 1) feeds a DC/DC converter (stage 2) which feeds a sensor (stage 3). Their efficiencies are $\eta_1=0.93$, $\eta_2=0.90$, $\eta_3=0.80$. - Compute $\eta_{\rm total}$. - If the battery provides $5.0~\rm W$, what power reaches the sensor? {{fa>pencil?32}} For the company „HHN Mechatronics & Robotics“ you shall analyze a competitor product: a simple drilling machine. This contains a battery pack, some electronics, and a motor. For this consideration, the battery pack can be treated as a linear voltage source with $U_{\rm s} = ~11 V$ and internal resistance of $R_{\rm i} = 0.1 ~\Omega$. The used motor shall be considered as an ohmic resistance $R_{\rm m} = 1 ~\Omega$. The drill has two speed-modes: - max power: here, the motor is directly connected to the battery. - reduced power: in this case, a shunt resistor $R_{\rm s} = 1 ~\Omega$ is connected in series to the motor. {{drawio>sketchDrillingMachine.svg}} Tasks: - Calculate the input and output power for both modes. - What are the efficiencies for both modes? - Which value should the shunt resistor $R_s$ have, when the reduced power should be exactly half of the maximum power? - Your company uses the reduced power mode instead of the shunt resistor $R_{\rm s}$ multiple diodes in series $D$, which generates a constant voltage drop of $U_D = 2.8 ~V$. \\ What are the input and output power, such as the efficiency in this case? You can check your results for the currents, voltages, and powers with the following simulation: {{url>http://www.falstad.com/circuit/circuitjs.html?hideSidebar=true&ctz=CQAgjCAMB0l5AWAnC1b0DYqwExgZDgUgOxgkkIKVYDMkIArCArUwKYC0YYAUAG4gkWHCRxCsYHAA4ocguAgNl0RrwBOE8DPBJxU2Sr6acjEWJCnJeucZC0cDcnUfbD4ASzjgSWAk51lL0todxU1AGVdfR0rNzlxADMAQwAbAGd2OVpeAHdLM3AqAutxSDyS+IcAwwq4nnF-RTKKpoavGqgK6p8XJxty-J6wG3qB1u9nDuiu-LabYfH84UsLONEWgA9LCBGSe1lHBgRLE4BVXm3aQ8h9xgY8Y9OQdMv4m3xmJGYTohAAJTe+CwKCKSHAhBY4FkAEs3jJxEQIDhQUQTr8ToDtijDghZLQpAUICcwLIALZvVhODDHajgDBsDEgC7bKiyMAYfYIGk+J6kkAAETetAQ+gZ9kY4I5fmeWIlUrI9kV5HRz1e2PwQmR9CEqr+EV4QA noborder}} #@TaskTitle_HTML@#3.3.3 Power of two pole components #@TaskText_HTML@# Two heater resistors (both with $R_\rm L = 0.5 ~\Omega$) shall be supplied with two lithium-ion-batteries (both with $U_{\rm S} = 3.3 ~\rm V$, $R_{\rm i} = 0.1 ~\Omega$). 1. What are the possible ways to connect these components? #@HiddenBegin_HTML~Solution333_1,Solution~@# {{drawio>electrical_engineering_1:diagram333_1.svg}} #@HiddenEnd_HTML~Solution333_1,Solution ~@# 2. Which circuit can provide the maximum power $P_{\rm L ~max}$ at the loads? #@HiddenBegin_HTML~Solution333_2,Solution~@# At the maximum utilization rate $\varepsilon = 0.25$ the maximum power $P_{\rm L ~max}$ can be achieved. \\ The utilization rate is given as: \begin{align*} \varepsilon &= {{P_{\rm out}}\over{P_{\rm in, max}}} \\ &= {{R_{\rm L}\cdot R_{\rm i}} \over {(R_{\rm L}+R_{\rm i})^2}} \\ \end{align*} As near the resulting equivalent internal resistance approaches the resulting equivalent load resistance, as higher the utilization rate $\varepsilon$ will be.\\ Therefore, a series configuration of the batteries ($2 R_{\rm i} = 0.2~\Omega$) and a parallel configuration of the load (${{1}\over{2}} R_{\rm L}= 0.25~\Omega$) will have the highest output. #@HiddenEnd_HTML~Solution333_2,Solution ~@# #@HiddenBegin_HTML~Result333_2,Result~@# The following configuration has the maximum output power. {{drawio>electrical_engineering_1:diagram333_3.svg}} #@HiddenEnd_HTML~Result333_2,Result~@# 3. What is the value of the maximum power $P_{\rm L ~max}$? #@HiddenBegin_HTML~Solution333_3,Solution~@# The maximum utilization rate is: \begin{align*} \varepsilon &= {{{{1}\over{2}} R_{\rm L} \cdot 2 R_{\rm i} } \over { ({{1}\over{2}} R_{\rm L} + 2 R_{\rm i} )^2}} \\ &= { {0.25 ~\Omega \cdot 0.2 ~\Omega } \over { ( 0.25 ~\Omega + 0.2 ~\Omega )^2}} \\ &= 24.6~\% \end{align*} Therefore, the maximum power is: \begin{align*} \varepsilon &= {{P_{\rm out}}\over{P_{\rm in, max}}} \\ \rightarrow P_{\rm out} &= \varepsilon \cdot P_{\rm in, max} \\ &= \varepsilon \cdot {{U_s^2}\over{R_{\rm i}}} \\ &= 24.6~\% \cdot {{(3.3~\rm V)^2}\over{0.1~\Omega}} \\ \end{align*} #@HiddenEnd_HTML~Solution333_3,Solution~@# #@HiddenBegin_HTML~Result333_3,Result~@# \begin{align*} P_{\rm out} = 26.8 W \end{align*} #@HiddenEnd_HTML~Result333_3,Result~@# 4. Which circuit has the highest efficiency? #@HiddenBegin_HTML~Solution333_4,Solution~@# The highest efficiency $\eta$ is given when the output power compared to the input power is minimal. \\ A parallel configuration of the batteries (${{1}\over{2}} R_{\rm i} = 0.05~\Omega$) and a series configuration of the load ($2 R_{\rm L}= 1.0~\Omega$) will have the highest efficiency. #@HiddenEnd_HTML~Solution333_4,Solution~@# #@HiddenBegin_HTML~Result333_4,Result~@# {{drawio>electrical_engineering_1:diagram333_4.svg}} #@HiddenEnd_HTML~Result333_4,Result~@# 5. What is the value of the highest efficiency? #@HiddenBegin_HTML~Solution333_5,Solution~@# The efficiency $\eta$ is given as: \begin{align*} \eta &= { {2 R_{\rm L} }\over{ 2 R_{\rm L}+ {{1}\over{2}} R_{\rm i} }} \\ &= { { 1.0~\Omega }\over{ 1.0~\Omega + 0.05~\Omega }} \end{align*} #@HiddenEnd_HTML~Solution333_5,Solution~@# #@HiddenBegin_HTML~Result333_5,Result~@# \begin{align*} \eta = 95.2~\% \end{align*} #@HiddenEnd_HTML~Result333_5,Result~@# \\ \\ #@HiddenBegin_HTML~Details333,Detailed Comparison~@# {{drawio>electrical_engineering_1:diagram333_2.svg}} #@HiddenEnd_HTML~Details333,Detailed Comparison~@# #@TaskEnd_HTML@# ~~PAGEBREAK~~ ~~CLEARFIX~~ ===== Summary ===== - Real sources: $U_{\rm L}=U_0 \dfrac{R_{\rm L}}{R_{\rm i}+R_{\rm L}}$, $I_{\rm L}=\dfrac{U_0}{R_{\rm i}+R_{\rm L}}$; $P_{\rm L}=\dfrac{U_0^2 R_{\rm L}}{(R_{\rm i}+R_{\rm L})^2}$. - **Efficiency**: $\displaystyle \eta=\dfrac{R_{\rm L}}{R_{\rm i}+R_{\rm L}}$; maximize by $R_{\rm L}\gg R_{\rm i}$ (power engineering). - **Utilization rate**: $\displaystyle \varepsilon=\dfrac{R_{\rm L}R_{\rm i}}{(R_{\rm L}+R_{\rm i})^2}$; peak $\varepsilon_{\max}=25~\%$ at $R_{\rm L}=R_{\rm i}$ (maximum power transfer; $\eta=50~\%$). - **Chain efficiencies** multiply: $\eta_{\rm total}=\prod \eta_i$. - Thevenin/Norton help to **separate** source figures ($U_0$, $R_{\rm i}$) from the load and to reuse the same formulas. - **Max efficiency $\eta$**: $R_{\rm L}\rightarrow\infty$ (relative to $R_{\rm i}$) → small current, small loss. - **Max delivered power $P_{\rm L}$**: $R_{\rm L}=R_{\rm i}$ (impedance matching). See also {{https://en.wikipedia.org/wiki/Impedance_matching|impedance matching}} and {{https://en.wikipedia.org/wiki/Maximum_power_point_tracking|Maximum Power Point Tracking (MPPT)}} for PV systems. ===== Common pitfalls checklist ===== - Forgetting **units** in intermediate results (always write $x = \text{number} \times \text{unit}$). - Mixing up **goals**: high $\eta$ vs. high $P_{\rm L}$ lead to **different** $R_{\rm L}$. - Using **ideal source** formulas for a **real** source (always include $R_{\rm i}$). - Ignoring the **sign convention** when interpreting $P=U\cdot I$ (source vs. load). ~~PAGEBREAK~~ ~~CLEARFIX~~