Solving for resistance in a circuit using KCL

[DODGEVIPER13]

Homework Statement

If I=3A and the 18 V source delivers 8 A of current, what is the value of R?(Hint: You need Ohms law as well as KCL)

Homework Equations

i1+i1+i1+...+in=0

The Attempt at a Solution

Work is attached

 

[NascentOxygen]

Current I is not shown. Can you mark it in?

Is that a 13A current marked from the unknown voltage source?

 

[DODGEVIPER13]

Sorry I messed up (I is the same as ix), I just labeled it incorrect. Yes it is from the unknown voltage.

 

[DODGEVIPER13]

ix=3A

 

[DODGEVIPER13]

Is my question confusing? Or is it just that it has not been answered yet?

 

[DODGEVIPER13]

I think I get it, I just need help with when I go around a circuit telling whether the circuit element should be + or -.

 

[gneill]

Your question is confusing, and it seems to get more so each time you add a "clarification"!

You've labelled your diagram with two currents. One, i_x, seems to be associated with the 6Ω resistor branch. But then you say it's associated with the "unknown voltage source", presumably Vx. But you've also labelled the current coming from Vx with a value of what appears to be 13A. Then you say that I is ix and that ix = 3A Adding to the mystery is that you've written three equations without describing what they are meant to represent, with sequences like " +18 -5 - 5" that are hard to fathom from the given diagram, since no particular current or resistor values are specified to let us know what path is being followed through the circuit.

In your problem statement you ask, "What is the value of R?". But there is no component labelled "R" on the diagram; Only one labelled R_A seems to be an unknown.

Furthermore, since the 6Ω resistor is paralleled by the 18V supply the current through it cannot be anything but 3A, so being given its value (i_x) doesn't add any information. And R_A is also paralleled by the 18V supply, so unless you know more about the currents supplied or absorbed by both voltage supplies, it doesn't look hopeful for finding a particular value for R_A.

 

[DODGEVIPER13]

Yes ix is with the 6 ohm resistor, sorry the 13A is with unknown voltage source that is what I meant when I posted but I got rushed. Also there is no current I in the problem sorry that was a typo I meant for that to be ix, which again goes with the 6 ohm resistor. The three equations I wrote are three separate attempts to get the answer to the problem, I wanted to know which was correct way to approach it. R was also a typo it should be RA.

 

[DODGEVIPER13]

Hopefully that clears it up the 18v gives 8A of current to the circuit. The unknown source gives 13 A and the current across the 6 ohm resistor labeled ix is 3 A. The three equations are not all being used to achieve the solution they are simply my three attempts sorry for the confusion I will do better next time I post.

 

[gneill]

So, now that the fog has lifted , does this represent the circuit you are analyzing, with all the given voltages, currents, and resistances? Are the current directions correct as drawn?

I take it that you are to find R_A and V_x?

 

[DODGEVIPER13]

Yes this is the correct circuit.

 

[gneill]

Yes this is the correct circuit.

Okay then, it looks like there's enough information to find both R_A and V_x. Nothing too fancy is required.

Suppose you take the bottom "rail" as the reference node. If you apply KCL to the top node (where the three resistors meet), what must be the current through the R_A branch?

 

[DODGEVIPER13]

ok well Vx is any easy one its 65 volts because V=ir so 13(5)=65. RA is a bit more difficult for me so I assume it would be this way first I find for the resistance of the voltage source 65/13=5 ohms so 0=-5(ohms from voltage source) - 5(ohms from resistor)+6(3)-8RA. solving for RA i get 1 ohm which is correct but did I do it right?

 

[DODGEVIPER13]

if I am correct why is what I did correct wht are the two resistancees negative and why the would I also not have found for the resistance in the 18 V source. Ugh or do I do it the way 0 = RA -6 ohms + 5 ohms gives RA=1 ohm

 

[gneill]

ok well Vx is any easy one its 65 volts because V=ir so 13(5)=65.

No, that's the potential drop across the 5Ω resistor. The current is 13A and the resistance is 5Ω, so the potential drop is I*R = 13*5 = 65V. If the potential drop across the resistor is 65V, and the left end of the resistor is at 18V, what must be the potential at its other end (The Vx end)?

RA is a bit more difficult for me so I assume it would be this way first I find for the resistance of the voltage source 65/13=5 ohms so 0=-5(ohms from voltage source) - 5(ohms from resistor)+6(3)-8RA. solving for RA i get 1 ohm which is correct but did I do it right?

An ideal voltage source has no resistance, and by your description you seem to be adding resistances and voltages together, which is not correct. So no, your approach is not correct.

Instead, use KCL at the node where the resistors all join to find the current through R_A. You already know the potential across it.

 

[DODGEVIPER13]

So Vx=47 volts because 65V-18V=47V right. 0=iA-13+3, so the current through RA is 10 A that doesn't semm right ugh??

 

[gneill]

So Vx=47 volts because 65V-18V=47V right.

What direction is the 13A current flowing? So what is the direction of the potential drop across the 5Ω resistor?

0=iA-13+3, so the current through RA is 10 A that doesn't semm right ugh??

Sum the known currents flowing into the node. Subtract the known currents flowing out. What's left?

 

[DODGEVIPER13]

0=-iA+13+8-3 so iA=18A then my potential is 18v= 18RA so RA=1 ohm. But why is iA negative?? as to the Vx question it is going to the left across the 5 ohm resistor right so would that be -65+18=-47 volts

 

[gneill]

0=-iA+13+8-3 so iA=18A then my potential is 18v= 18RA so RA=1 ohm. But why is iA negative??

It's negative for the same reason that you've made ix negative: Because it is flowing out of the node, and you've assumed that currents flowing into the node are positive.

as to the Vx question it is going to the left across the 5 ohm resistor right so would that be -65+18=-47 volts

Nope. Draw the potential change across the resistor on the diagram. Which end is "+" and which end is "-"? Starting at the left end of the resistor, at a potential of 18V, do you add or subtract the 65V to get to the top of Vx?

 

[DODGEVIPER13]

well since I have done all else the answer must be 18+65=83 volts but I would assume the reason is the plus is on the right and minus on the left of the resistor. I guess this would be because the Vx positive side is up so therefore the resistors pos side must be on the right.

 

[gneill]

well since I have done all else the answer must be 18+65=83 volts but I would assume the reason is the plus is on the right and minus on the left of the resistor. I guess this would be because the Vx positive side is up so therefore the resistors pos side must be on the right.

The direction of the potential change across the resistor is due to the direction of the current flow through it; Potential drops in the direction of current flow. The current direction was specified, thus the direction of the potential drop is determined accordingly. Thus the result is 83 V.

 

[DODGEVIPER13]

Cool thanks man I understand it now