#dc circuits

[quietimtryingtostudy]

Physics questions series: Current Electricity (part 3)

Anyone solving these questions, you can reblog and add your solutions to these questions, or if you want you can ask/submit questions too.

Q.3) Find the current drawn from the battery:

1.

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2.

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Masterlist

[study-snails]

Saturday, 04/29/2017 - 12:42pm

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!

[srrybabe]

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  :) 

[engineeringtldr]

More Than You Ever Wanted to Know About Electrical Engineering: Circuit Loading

Here’s another neat trick you can do with Thévenin equivalents. Suppose we have a generic circuit, represented by its Thévenin equivalent, and we’re attaching a load to it.

What voltage does our load see here? This circuit is just a basic voltage divider, so we can find the answer pretty easily.

Take a look at that for a minute, and think about what happens if the circuit’s Thévenin resistance is large compared to the resistance of the load. You’ll get a much smaller voltage across the load. In other words, a circuit with a large output impedance makes a lousy power supply, since the voltage it can provide drops when you put a load on it.

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.

[engineeringtldr]

More Than You Ever Wanted to Know About Electrical Engineering: Maximum Power Transfer

Thévenin’s theorem can be a powerful tool for analysis. Because it allows you to simplify any resistive circuit to a single voltage source and resistance, it lets you make general, broad-reaching predictions about the behavior of these circuits.

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.

Let’s represent the existing circuit with a Thévenin equivalent circuit to make things easy. We’ll call the new load R_L.

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.

[engineeringtldr]

More Than You Ever Wanted to Know About Electrical Engineering, Part 9: Thévenin and Norton Equivalent Circuits

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?

Fortunately, you don’t need to know the details of what’s inside your black box. As long as the unknown device is a purely resistive load (no capacitance or inductance, which we’ll talk about pretty soon), you can represent it with a single voltage source and a resistance in series (a Thévenin equivalent circuit), or a single current source and a resistance in parallel (a Norton equivalent 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.

You can get the resistance the same way - it’ll be whatever resistance you’d see across the terminals with an ohmmeter. Or, Ohm’s Law still applies here - you have a voltage and a current, and the resistance (usually called the Thévenin resistance) is the ratio of the two.

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.

[jasonyeaman]

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!

        Rather than solve it for you: You have a parallel DC circuit. In one path 2 resistors, in the other a cap and a…

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[kaitiikat]

5-10 :((

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.