Welcome to my educational engineering channel. I am a professor of engineering and this channel is dedicated to teaching all about the basic concepts of engineering and design. Opinions expressed are solely my own and do not express the views or opinions of my employer.
Do you recommend reading the textbook along with these videos? I felt like the mention of P_load formula at the very end was kinda of confusing. I understood the part about having a higher resistance makes current lower and all that...but I dont really get the part of the max power being when R_AB = Rthev.
Can you give a quick example of what Thevenin theorem is useful for application wise? Im a bit confused because we can find Vthev and Rthev, but then how do you know how much current will be draw? I guess if you had a motor with some resistance as the load, finding VThev might help? Thanks!
Batteries are a good example. We commonly model batteries as a simple voltage source. If we model them this way, we make the assumption it can supply infinite current. Modeling a battery as a voltage source and an internal resistance (Rth) becomes a better approximation. We can actually calculate the maximum current that a battery can supply to a circuit. You are also correct with the motor example, if we are analyzing multiple alternatives for motor design it would be much easier to just solve the circuit with a source and a resistor instead of trying to solve the complex circuit for every motor alternative.
I had one more question, for source transformations, I'm confused because I would assume that when making a voltage source a current source in parallel with a resistor, the current will now be split, and a lower current will flow through the resistor, creating a different voltage than the original voltage source.
You are correct that the current in the parallel resistor will “split” between the resistor and the load in the current-source version, but the total current delivered to the external load remains consistent with what the voltage source would provide. My advice would be to draw a simple circuit with a voltage source and 2 resistors in series. Solve for the voltage across the second resistor using traditional voltage divider. Next, analyze the same circuit by first doing a source transformation with the voltage source and first resistor. You will find the voltage and current through the second resistor to be the same as the first scenario. You can study these two examples and I think you will more deeply understand what is going on with source transformation, rather than just following the set rules I laid out in the video.
I am a bit confused why we can just draw a reference GND anywhere in a circuit and find voltage... I feel like that almost changes the circuit.... Can I always just keep a gnd in any circuit problem and use it to help me assume other voltages and currents?
You can put a reference anywhere you want, its commonly placed on the - side of the voltage supply. This ensures all the node voltages will be positive when solving. If you randomly place a GND reference on a diagram, you will find that you will have negative values in voltage when solving.
You guys should really look up how to derive the voltage divider formula. He should have covered that. Now im kind of worried that this series wont go in depth at all...
sure there is someone on youtube who explains it a hundred times better than others, especially my lecturer. what would i do without you thank you soooo much
Good question, we model op amps as black boxes so we don’t really know what circuitry exists on the inside. We just use this rule to analyze ideal op amps. In the real world, they most likely would not be equal without feedback. In this analysis, we also don’t model the power source for powering the amp. Depending on the power source, it also places more constraints on the op amp. In the real world, there are op amps that excel in some applications but will perform poorly in others. It is important to study the specifications of op amps to select the best performer for your real world application.
Voltage is always in reference point (commonly ground - 0 V). In this example a voltage source is connected between two nodes, in this case we call it V2 and V3. If V3 was ground, then V2 would be 5 V. However in this case, V3 is not at 0V, it is located in the circuit where the - side of the power source is not connected to ground. So we must write the equation as V2 = V3 + Vs, where Vs is the voltage of the power source (in this case 5V). When solving the system of equation we find that V2 = 9.2 V and V3 = 4.2 V. This checks with our equation 9.2 = 4.2 + 5.
@@ProfessorJohns I think it also helps them to understand it in terms of "higher and lower" potential. The V2 is on the higher potential side (+) of the 5V, and the V2 is on the lower end. So thats why its V2 = V3 +5V (at least I think so lol)
Mine is showing this error: Traceback (most recent call last): File "main.py", line 1, in <module> ImportError: no module named 'RPi' MicroPython v1.22.0 on 2023-12-27; Raspberry Pi Pico with RP2040 Type "help()" for more information.
import RPi.GPIO as GPIO import time import hx711 # Configuração de GPIOs dt_pin = 17 clk_pin = 18 # Configuração do HX711 hx = hx711.HX711(dt_pin, clk_pin) # Calibração (opcional) # Se você já possui os valores de offset e scale da calibração anterior, # utilize-os aqui para maior precisão. hx.set_scale(318.64) # Fator de escala hx.set_offset(117800) # Offset # Função para ler o peso def read_weight(): try: # Leia o valor bruto do HX711 weight_raw = hx.get_raw_data() # Converta o valor bruto em peso weight = hx.get_weight(scale=318.64, offset=117800) # Imprima o valor do peso print(f"Peso: {weight} kg") except Exception as e: print(f"Erro ao ler peso: {e}") # Loop principal while True: read_weight() time.sleep(1) # Ajuste o tempo de intervalo entre as leituras
Você pode digitar comandos `pip` no terminal do seu Raspberry Pi. Aqui estão os passos para fazer isso: 1. **Abra o Terminal**: Você pode encontrar o aplicativo de terminal no menu do Raspberry Pi OS, ou pode usar o atalho `Ctrl + Alt + T`. 2. **Digite o comando pip**: Uma vez que o terminal esteja aberto, você pode digitar o comando `pip` para instalar a biblioteca `RPi.GPIO` ou qualquer outro pacote Python. Por exemplo: ```sh pip install RPi.GPIO ``` Se você estiver usando Python 3, pode ser necessário usar `pip3` em vez disso: ```sh pip3 install RPi.GPIO ``` 3. **Execute o comando**: Pressione `Enter` para executar o comando. Se você encontrar problemas de permissão, pode adicionar `sudo` antes do comando para executá-lo com permissões de superusuário: ```sh sudo pip install RPi.GPIO ``` ou ```sh sudo pip3 install RPi.GPIO ``` Seguindo esses passos, você pode instalar os pacotes Python necessários no seu Raspberry Pi.
Interesting video. What about the torque reaction from the torque going through the rear wheel to the road. This would also be trying to lift the front of the bike so shouldn't it be considered too?
The torque at the rear wheel is taken into account in this problem. Since we are tasked to solve for acceleration, we don’t frame the problem in terms of rear wheel torque. The acceleration we solve for in this problem is the effect of torque on the rear wheel. At the end of the video we solve for the frictional force between the road and the tire. If we knew the diameter of the wheel, we could solve for the resultant rear wheel torque needed to lift the front tire off the ground. Also, note that this is a static problem, we are not considering dynamic effects (the rider shifting weight, effects of “popping the clutch, etc.) It is a simplified analysis, in reality, by shifting your weight and using the clutch advantageously it would require much less acceleration to lift the front wheel off the ground. Thanks for taking the time to watch the video.
For the special case of nodal analysis. You used the formula V2=V3+5. Why is it not V3=V2+5 ? Should the voltage not be higher at the node after the voltage source?
Notice the + and - on the 5V power source. V2 must be 5V higher than V3. "After" the voltage source is the node on the + side of the voltage source. If the power source was arranged in the opposite direction your suggested formula would be correct. Hope this helps.
