Circuits Problem(s)

[~Zero~]

So in this course I am taking, there is one chapter on circuits and I am really not familiar with them at all. So I am looking for you guys to explain some of these problems to me in the most simplest way possible, since its my first time ever trying to learn these things lol.

I will post more, after I at least attempt them, I attempted this, and I got somewhere (i think, not entirely sure). Here is my horribly drawn circuit



Thanks in advanced.

[LongGone]

Redrew your diagram:


Voltage/electric potential is measured at a point. Think of it like water or air flowing; fluids flow from a high pressure (i.e. high voltage) point to a lower one. The reason the voltage drops is due to resistance.

The 35V supply starts off with that much V, then it drops across the resistors. By the time it reaches back the negative terminal, the voltage is drained to 0. Same goes for the 50V supply. That means, the bottom side of the diagram as marked above is a shared 0 volts (vague explanation)

You want the Voltage drop across the resistors, V1 and V2. Since all the remaining voltage that goes across the 3ohm resistors drops to 0, the voltages before it crosses them can be marked as V1 and V2. Redrawing this diagram:



The two points V1 and V2 are the electric potentials of the junction.

[Reason: If we take say, V1. We start with V1, and as current flows across the 3ohm, there is a voltage drop of V1, so V1-V1=0V (because, we have to end with 0V on the bottom side of the circuit)]


Voltage node method states that we can take any point of the circuit, and write the sum of all the current flowing in or out of the point which should equal 0.



In this case for V1,

I1 + I2 + I3 = 0

Note that I=V/R,

(35-V1)/5 + (V2-V1)/2 + (0-V1)/3 = 0

Similarly, we can write an equation for the other node V2. Thus with two equations, you can obtain V1 and V2; the potentials at those two junctions. Or in this case, the potential drop across the two 3ohm resistors

[LongGone]

(someone correct me if I'm wrong, especially terminology; I'm somewhat rusty at this)

A simpler example


A battery supplies 1.5V, attached in series is a 3ohm resistor.

The voltage of 1.5V gets fully drained across that one single resistor. So, at point z we can assume the potential to be 0V, or ground.



What is the potential at x? It is 1.5V.
What is the potential at y? It is 1.5V still.
The potential at z? 0V, because it is all drained.

Since this is just a series circuit, the current is the same throughout the entire loop. I1=I2=I3

Where I2=V/R=1.5/3=0.5A

---------------------------------------------------------

What if we have two resistors in series?



What is the voltage at x? 1.5V
What is the voltage at y? We don't know yet, call it Y
What is the voltage at z? 0V, because by this point, all V is drained across the two resistors, where we are on the same side as the negative terminal of the battery

We can apply Voltage Node method here, at point Y. All current flowing into the point will sum to 0. Or in this case, total current flowing in, I1 equals the total current flowing out, I2

I1=I2

(1.5-Y)/3=(Y-0)/3
Y=0.75V

Since this is a symmetrical circuit, intuitively we have 0.75 volts dropped across the first 3ohm res, then another 0.75 volts dropped across the second 3ohm res. The voltage drop is balanced across both resistors.

Voltage Node method in this case only has two connections on the Y node. Your question has 3, so its a little more complex needing simultaneous equations

[~Zero~]

Okay so, lets say for V2



I5 + I4 -I2 = 0 ?

(50v - V2)/I5 + (0-V2)/I4 - (V2 - V1)/I2 = 0
50v = I5(-V2/I4) - I5((V2-V1/I2)) + V2
50v = V2 [1 + I5[-1/I4]] - [I5/I2] - V1/I2
50v = V2 [1 + 10[-1/3]] - [10/2] - V1/3
V1 = 3 * V2 [1 + 10[-1/3]] - [10/2] - V1/3
V1 = 3 * (-2.3333 V2 - 5 - 50v)
V1 = -6.99v2 - 15 - 150v

????? Did i derive that right, i think it would probably be easier for me to do it on paper than think about it on the computer i will check it again

[LongGone]

You mixed up between current and resistance, I=current

but with the equations I think you have the right idea...

I ended up with V1=12.644V and V2=12.131V

[~Zero~]

Okay I am probably just failing at algebra at the moment

ended up getting V1 = -7.82 volts and V2 = 4.088 volts

I am gonna try it once more


Edit: Quick question. Why do you make all the currents go into V1 - is there any reason or could do you do it anyway. Like making all the currents leaving V1?

[LongGone]

Its preference. You can make all currents leave V1, or all currents enter V1. If you do the algebra, you will get back the same thing. You can even have I1 and I2 flow in and I3 leaving; which means I1+I2=I3. Just be careful with the direction of the arrows and the positive/negative signs

[~Zero~]

Did it again and got

v2 = 5.72 v
v1 = 6.77 v

So I think I am doing something wrong still, do you think you could show me your work long?

For v1
I had ( I1 going in and I3 and I2 going out)
so
I1 = I2 + I3

For v2
I had (I5 going in and I2 going in and I3 going out)
I5 + I2 = I3

And after deriving
I got
V1 = 18.6667 V2 - 100
V2 = 10.3333 V1 - 70

Edit: did it again and got
V1 = 14.35
V2 = 13.05
>____<
I dont understand lol

In the process of doing it again, will type all my work into MSword then copy paste over here

[~Zero~]

Double Post

Using this


Did This






V2 = 14.28 +0.428 V1
V1 = 6.78 + 0.484 V2

V2 = 14.28 + 0.428 (6.78 + 0.484 V2)
V2 = 14.28 + 2.91 + 0.912 V2
V2 – 0.912 V2 = 17.19
0.88 V2 = 17.19
V2 = 19.53
V1 = 6.78 + 0.484 (19.53)
V1 = 16.23

Edit: replaced the deriving with pictures so its easier to read - hopefully someone can spot whats wrong with my math now xD

Okay i finally got the same answer as long gone, i just havent been checking my resistors closely enough, I have a V2 - 50 / 3 when it should be a V2 - 50 / 10

DERP DERP DERP DERP DERP that is all