DC Circuit Analysis: Kirchhoff's Laws

[gkamal]

1. Homework Statement

A DC circuit is shown below. Note the ground (V=0) in the circuit

1- calculate the current through the 5Ω resistance. Use "+" sign for the current directed from right to left, and "-" sign if the current flows from left to right.

2- The sign of the potential at a is:

3- The magnitude of the potential at a is:

Homework Equations

I1=I2+I3
-11I2+7-5I1=0
19-8I3-6I3+7-5I1=0

I1 being the current flowing clockwise in the loop from 7V going down [a] and back to 7V
I2 being the current going down [a] through the 11ohm resistor
I3 is the loop going anti-clockwise from 19V up the 11ohm resistor and back to 19 V

3. Attempt at solution :

I found the answer for number 1 to be 1.12 A but I am not sure if its right and not sure what is the sign for the direction.Also, i found the answer for number 2 to be positive but I do not understand it fully so please explain , for number 3 I have no clue what to do

 

[gneill]

Can you show the details of your calculation for part a? We can't comment on or help you with what we don't see.

 

[gkamal]

Those are my calculations, please ignore the + and - on the resistors that was just me playing around

 

[gneill]

It would be very helpful if you could type in your equations and explain your steps as you go. Otherwise helpers have to puzzle out where you started on the image and what you're intention was. Further, if they come across what appears to be an error, how are they to quote it and point it out? Many will just pass on making the effort and move on to another post.
That being said, I have spotted what appears to be a sign error in your expression for I3 derived from the outer loop KVL equation. Check your algebra there.

 

[gkamal]

I don't see what is wrong with it , the equation would be [-26-5I1]/-14 since it's a -14 i could just flip the signs of the numerator to get rid of the minus tell me if I'm wrong.Also please help me with number 3 , I have no idea where to start. I also added my equations above so please check them if that would make this clearer to u

 

[gneill]

I don't see what is wrong with it , the equation would be [-26-5I1]/-14 since it's a -14 i could just flip the signs of the numerator to get rid of the minus tell me if I'm wrong.Also please help me with number 3 , I have no idea where to start. I also added my equations above so please check them if that would make this clearer to u

When you moved the $-5I_1$ from the left side to the right side, why didn't it change sign?

 

[gkamal]

okay so i get 1.38 for I1 after fixing that equation thank you I don't know how I missed that one and also for the direction part of the question this would flowing from left to right so the answer should be -1.38 right?.Please give me a hint for number 3 because i really have no idea where to start...

 

[gneill]

All you need to be able to do is sum up the potential changes on a "KVL walk" from the reference node to node a. What path can you follow where you know all the potential changes?

 

[gkamal]

I find I2 then calculate the voltage drop across the 11 ohm resistor and since i have I1 i can calculate the voltage drop across the 5ohm resistor.So I did that and i get -6.89 and i add both voltage drops across the resistors and after i add 7 i get 0.1075V does that look reasonable to you ?

 

[gneill]

You only need one path from the reference node to node a. Any path that takes you there is fine. You can follow either your yellow path or your green path. You don't need both. This is not a closed loop KVL walk; it's a walk from a starting point (the reference node) to another point where you want to know the potential.

But if you have already solved for $I_2$ then you should know that the potential at node a is the same as the potential across the 11 Ω resistor, since it's the only component in that path from the reference node to node a.

 

[gkamal]

thank you very much I got the right answers you were very helpful

 

[gneill]

thank you very much I got the right answers you were very helpful

You're welcome