Finding Values in a Complex Circuit

  • Thread starter Thread starter KieranRye
  • Start date Start date
  • Tags Tags
    Circuit Complex
AI Thread Summary
To find values across a complex circuit, apply Kirchhoff's First and Second Laws by breaking the circuit into three loops: AXYD, XBCY, and ABCD. Start by writing Kirchhoff's Voltage Law (KVL) for each loop, ensuring to account for current splits at junctions. Designate currents for each path, such as I1, I2, and I3, and express voltages in terms of these currents using Ohm's Law. An example for loop AXYD shows how to set up the equation with the resistances and voltage sources involved. This method simplifies the analysis and helps in solving for the unknown values in the circuit.
KieranRye
Messages
3
Reaction score
0

Homework Statement



I need to find all the values essentially across the entire circuit

Homework Equations



Kirchoffs First and Second Law

The Attempt at a Solution



Attempted to find currents and resistance multiple time; didn't know how to work with the two power sources or the capacitor

An explanation of the assignment can be found here, along with a diagram of the circuit
 

Attachments

  • KirchoffsLaws.pdf
    KirchoffsLaws.pdf
    145.3 KB · Views: 1,035
  • vjae.png
    vjae.png
    6.2 KB · Views: 670
Last edited:
Physics news on Phys.org
KieranRye said:

Homework Statement



I need to find all the values essentially across the entire circuit

Homework Equations



Kirchoffs First and Second Law

The Attempt at a Solution



Attempted to find currents and resistance multiple time; didn't know how to work with the two power sources or the capacitor

An explanation of the assignment can be found here, along with a diagram of the circuit

Assume the switches have been closed for a while. Start by writing KVL around the AXYD loop. Show us what you get.
 
As gneill put it, it helps to split them up, and get individual equations. I would split the circuit up into three different loops.

loop AXYD

loop XBCY

loop ABCD

And write out the equations in terms of voltages. V=IR. Also, don't forget to take into account that your current will split or join at a junction point.
 
How do I account for the two voltage IN though. I'm sorry to put you through this but can you show me an example of say AXYD so I can figure out how to do it? I was never taught how, just handed the formula.
 
It will probably be easiest for you to first designate your currents through the loops. Which, you can choose arbitrarily. So, to start you off, I would compel you to glance at the end of page 3 in your attached pdf, to give you something to look at, when taking into account Kirchoff's second law.

So along, imagine the current I1 traveling along the wire from Y→D→A→X . You'll notice that the wire doesn't split along this path, so the current through that length of the wire is the same. When you get to X, you'll notice the wire splits, and thus, the current must split into two different paths.

So, (again, I would recommend you to look at the figure in your notes), let's say the current splits into I2 and I3 . I2 follows the wire from X→B→C→Y , while I3 goes from X→Y

In this case, this means the current splits. I1 = I2 + I3

Now that we can identify each individual current, it makes it a LOT easier to follows Kirchoff's first law. That the total voltages over the whole loop should be equal to zero.

In conjunction with Ohm's law, you can write out these equations for each loop.

(I know this may sound super confusing, with all these laws, but it's not that bad)

So just write each loop in terms of V
 
  • Like
Likes 1 person
To help start you off to better understand, let's go around the loop from Y→D→A→X→Y

We'll hit the resistor of 820Ω. Since we already defined the current I1 traveling through this part of the wire,

V820Ω = I1(820Ω)

now following this along clockwise

V820Ω + V5V + V180Ω + V680Ω + V3.3V + V220Ω = 0

I1(820Ω) - 5V +I1(180Ω) + I3(680Ω) + 3.3V + I3(220Ω) = 0

You're just changing everything to voltages to be able to apply Kirchoff's first law, that all voltages in a loop must equal 0.

Now do that for each of the two other loops A→B→C→D→A and X→B→C→Y→X
 
  • Like
Likes 1 person
Alright, I'll try it myself on this one again (so I learn how to do it!) and the other two in the morning. Thank you for your help!
 
Back
Top