Find the current and Voltage of the following pieces

[itzernie]

Homework Statement

Above is the circuit I have to deal with. We are given the V_in = 6V, V_out = 3.367 V and all the R values. We are asked to find the remaining values (The I's and V's)

Homework Equations

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Kirchhoff Voltage and current laws
V = IR
R's in series add
Voltage from a source has the same value on a resistor if they are in parallel

The Attempt at a Solution

At the moment I did not really start plugging in numbers, but i tried thinking of how to get certain values:

I₂: looks like V on it is 6 since parallel to V in so 6/20000
I₅: same procedure as stated for ^
V₃: Use a ratio between the two resistors in that node line to see how much voltage goes to that as opposed to R₄
V₁: confused because when I use the KVL i get V_in - I2R2 + V₁ = 0 but the first 2 arguments in that phrase would already = 0 unless I₂ is not as simple as stated above.
V₆ : completely unsure

** Also to add, we have not learned current or voltage division yet and I feel like that may play a role because I am not sure how to find R_eq of this nor understand what V_out does to help.

 

[TomHart]

I2: looks like V on it is 6 since parallel to V in so 6/20000

How can that be unless there is no voltage drop across R1?

Edit: Clarification: Even if there was no voltage drop across R1, it would still not be true that R2 is in parallel with Vin, if that is what you were saying.

Edit 2: Is there any current flowing through R6? If so, where does it go?

 

[itzernie]

Just reading that alone makes complete sense. I was just reading in my notes that voltage in parallel remains the same but I'm seeing that does not apply to this case. I feel like my issue is that I can't really figure out where to properly start? I can not directly find V₁ since I do not know the current coming from the 6v source and I can't figure out how to get that without finding R_eq

 

[TomHart]

Consider my "Edit 2" above where I ask about current flowing through R6.

 

[itzernie]

Edit 2: Is there any current flowing through R6? If so, where does it go?

Sorry if I seem uneducated in this topic. I know it's not an excuse but we get prelab assignments due before the actual lectures and we don't have text I can really learn from. To answer your question, I believe there is a current that flows through R₆ and out to the output?

Edit: this is the first time I have ever seen a V_out

 

[TomHart]

I believe there is a current that flows through R6 and out to the output?

Okay, that is not an unreasonable answer. But what is at the output that will allow current to flow? Is there anything?

 

[itzernie]

Is there anything?

Nope, the circuit is broken at that point, so despite there being some V_out, there is no current at the output?

 

[TomHart]

That's how I saw it too - well, unless we consider that there was a voltmeter to measure the voltage. But that voltmeter will have a very high resistance and therefore, negligible current. So if we know (at least we think we know) that no current flows through R6, then what is the voltage at the other side of R6?

 

[itzernie]

Hmmm. My intuition is telling me 3.367 V since that V_out value must come from somewhere and there is no current between the resistor and the output

 

[NascentOxygen]

My intuition is telling me 3.367 V

You have an impressive 4-significant-figure intuition? Where did this value spring from?

 

[ehild]

That's how I saw it too - well, unless we consider that there was a voltmeter to measure the voltage. But that voltmeter will have a very high resistance and therefore, negligible current. So if we know (at least we think we know) that no current flows through R6, then what is the voltage at the other side of R6?

No, it is not sure that the output current is zero. You can check easily: in this case the voltage across R2, R3+R4, and R5 is Vout, and the sum of the currents through the resistors is equal to I1. Also,
Vout + I1 R1 = Vin. Is it true?

 

[ehild]

You have an impressive 4-significant-figure intuition? Where did this value spring from?

Vout was given as 3.367 V.

 

[NascentOxygen]

Vout was given as 3.367 V.

Ah, that simplifies things!

 

[TomHart]

No, it is not sure that the output current is zero. You can check easily: in this case the voltage across R2, R3+R4, and R5 is Vout, and the sum of the currents through the resistors is equal to I1. Also,
Vout + I1 R1 = Vin. Is it true?

@ehild, you are absolutely correct. I retract everything I previously said. Thank you for correcting me.
@itzernie, please take note of ehild's post.

 

[ehild]

@ehild, you are absolutely correct. I retract everything I previously said. Thank you for correcting me.
[

Check: The current through R6 might be zero with the data given!

 

[TomHart]

Check: The current through R6 might be zero with the data given!

Well, I tried to check a while ago (when I first read your post) and it looked like the currents did not match. But now, since you brought it up, and since I have so little confidence in my abilities right at the moment . . .
I1 = (6-3.367)/2k = 1.3165 mA
I2 = 3.367/20k = 0.16835 mA
I4 = 3.367/(2.2k + 4.7k) = 0.48797 mA
I5 = 3.367/5.1k = 0.66020 mA
I2 + I4 + I5 = 0.16835 + 0.48797 + 0.66020 = 1.31652 mA

So it appears that the currents do match. However, due to my excessive fumbling through this problem, I have very little confidence that my calculations here are correct. If someone offered me a bet where they would pay me $100 if my calculations are correct, and I would have to pay them one donut if my calculations are wrong, I would probably have to consider that a bad risk for me.

I think it is bedtime. Good night Mary Ellen. Good night John boy. Good night Jim Bob.

 

[ehild]

So it appears that the currents do match. However, due to my excessive fumbling through this problem, I have very little confidence that my calculations here are correct.

Yes, they do :) Good night.

 

[itzernie]

Sorry for the late response! I it was 4am on my last post so I passed out lol. Anyway @ehild what you said in the way specific R's add up to V_out and how I₁ is the result of all currents from the other resistors makes sense to me. Thanks ! @TomHart I will work on the problem asap and let you know if my answers are close enough to where you can take the donut risk.

 

[itzernie]

OK so I sat down and really went over what was talked about here. I fully understand 2 of the 3 statements that @ehild made.

I'm just confused with how did you know the sum of the current on the resistors R2-R5 had to equal the I1 current? Is it because If you were to simplify all of those resistors together into one you would have a junction up top with 3 directions to go. Left is where I1 comes from, right has no current , and down is the combined resistor where I_sum must go. Therefore by KCL we know at a junction the current sum must = 0 and therefore the I_{sum of R} = I₁ ?

I can see why Vout + I1 R1 = Vin. - this alone can help find I₁ and V1
Also I see why R₂ through R₅ have the same Voltage as V_out.

Using that statement alone helped me find I2,I4,V3,I5.
V6 is in this case 0V.
I believe that is all of the variables (and @TomHart, I basically computed the same numbers so you can make the bet hopefully)

 

[ehild]

OK so I sat down and really went over what was talked about here. I fully understand 2 of the 3 statements that @ehild made.

I'm just confused with how did you know the sum of the current on the resistors R2-R5 had to equal the I1 current?

It would not be true if there was current through R6.
You have a two-port circuit, with some input and output voltages and currents. Usually such circuit is part of a bigger one, so the current can flow out of it. You would be able to determine the currents with the information given, and the result is Iout = 0 if Vout=3.367 V. With an other output voltage this current would not be zero.

Is it because If you were to simplify all of those resistors together into one you would have a junction up top with 3 directions to go. Left is where I1 comes from, right has no current , and down is the combined resistor where I_sum must go. Therefore by KCL we know at a junction the current sum must = 0 and therefore the I_{sum of R} = I₁ ?

Yes, assuming that the current through R6 is zero, the current through R1 is the same as the current through the combined resistors.

I can see why Vout + I1 R1 = Vin. - this alone can help find I₁ and V1
Also I see why R₂ through R₅ have the same Voltage as V_out.

Using that statement alone helped me find I2,I4,V3,I5.
V6 is in this case 0V.
I believe that is all of the variables (and @TomHart, I basically computed the same numbers so you can make the bet hopefully)

It is correct.
Can you write the equations for arbitrary V_out and I_out?

 

[TomHart]

I'm just confused with how did you know the sum of the current on the resistors R2-R5 had to equal the I1 current? Is it because If you were to simplify all of those resistors together into one you would have a junction up top with 3 directions to go.

Just to point out what may be obvious, one side of each of resistors R1, R2, R3, R5 and R6 all connect to the same node. The reason it is true that the currents through those 3 resistor paths is equal to I1 is because of Kirchhoff's current law - that the sum of the currents flowing into a node must be equal to the sum of the currents flowing out of that node. That's how you know that I1 = I2 + I4 + I5 + I6. But as we (painfully for me) found out, I6 = 0. Now you could have picked the direction of each of those currents however you wanted. And if, when you worked the math, it turned out the result of one of them was negative, you would know that the direction you guessed turned out to be the the opposite direction of the actual current flow for that path.

But yes, you could have reduced R2 through R5 to a single equivalent resistor.

 

[itzernie]

Can you write the equations for arbitrary Vout and Iout

for any V_out similar to this one I'm thinking:
V_out = V_{in + any other V sources} - V_n n standing for summed Resistors in series/parallel with the V_{in + any other V sources} hmmm I never actually thought about an I_out in this case. I know you can have a Voltage without a current though. Wouldn't it always be 0 in cases just like this one?

but in general I would like to say :

I_out = I_{remaining amount from I1 after distributing it through the circuit} ?

 

[itzernie]

That's how you know that I1 = I2 + I4 + I5 + I6. But as we (painfully for me) found out, I6 = 0.

ahh. that is the line that helps clarify everything. and basically since you found I2,4,5 from ohms law where the V was V out you could see that combo already = I1. Therefore I6 would have to be 0

 

[ehild]

If something is connected to the output terminals, I6 can be different from zero, and there is a voltage drop across R6.

Then I₁=I_eq+I₆ and Vin = I₁R₁+I_eqR_eq, I_eqR_eq=I₆R₆+Vout.
Solve for I₆.
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