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Today is full of studying studying studying! With AP and IB exams coming up, there is no time to spare! Today’s main topic - direct current circuits! Hope everyone is having a nice day!
boooooooooo is anyone really good at physics and wants to help me. its about DC circuits. im trying to figure out capacitance (C) by using my slope which is my linearized Voltage/time (sec) and my resistance(R) 4.35k and probably other things. im just super confuseddddd :)Â
When this happens, it’s called circuit loading, and it’s a phenomenon with powerful and far-reaching consequences. Real power supplies (and other devices required to power others) often have output resistances measured in milliOhms. Conversely, devices which are designed for measurement (like a scope or a mulitmeter) or designed to accept power from another device are made with very high input resistances to reduce the chance of their loading down the source.
Suppose you’re hooking up a device to a circuit and you want to know how much power you can deliver to it. More than that, you want to make sure you give your new device as much power as you possibly can.
We know power is (I^2)*R, or (V^2)/R, and we know the maximum value of a function occurs when its derivative is zero. So let’s run with that and see what we get.
This is kind of an interesting result. It looks like the maximum power we can deliver to a load occurs when the load’s resistance is equal to the circuit resistance. This is a useful thing to know when you have a circuit that you want to drive something, or a circuit that should be driving something, but isn’t.
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Suppose you have some shiny new device that you want to connect to a circuit you already have. It’s basically a black box - you don’t know the details of what’s in it, just that it’s got a bunch of voltages, currents, and resistances inside.
How do you figure out how it’s going to behave with your circuit?
The big question here, of course, is how to determine the values of the voltage source, current source, and resistance. It’s actually pretty easy. The voltage source is the open circuit voltage of the device - that is, the voltage you would see if you took a voltmeter and measured the voltage across its terminals without it connected to anything. The current source is likewise the short circuit current - the current you would measure if you put an ammeter between the terminals without them being connected to anything else.
Knowing these measurements, you have a useful way of representing your black box in a circuit diagram and you can get an accurate idea of how it will affect your original circuit.
How do I solve the voltage across a capacitor at a given time
How do I solve the voltage across a capacitor at a given time?
I am taking a circuits course currently and am trying to remember how to solve for voltages across resistors and capacitors. If someone can explain how to do problems a and b, I would appreciate it. Thanks!
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Rather than solve it for you: You have a parallel DC circuit. In one path 2 resistors, in the other a cap and a…
I absolutely hate night classes. Everything about them makes me miserable. It would be one thing if there were projects that were hands on but five hours of monotone book lessons about electromagnetic currents, resistors, and AC and DC currents is pure hell. >.< I just want to walk around, stretch my legs, and not sit down the whole time. :( Ugh.