Homework Help: Circuit Quesiton

1. The problem statement, all variables and given/known data

See Attachment

2. Relevant equations



3. The attempt at a solution

I know how to do A this is an easy question. I'm not sure how to do B.

For A there are two essential nodes. There are seven nodes.

For B I'm not sure. I always drew my current so that way it went from positive to negative (this is after all the direction of current I believe). This creates a problem because when I try to solve the problem using the node voltage method I get about 1.714 for the top essential node which I'm not sure if it's right or anything but it seems a little bit off.

I don't understand why the right most branch with the current source... why is it that the whole branch is regulated to be 1 A even before the current source and before the 1 ohm resistor but after the node? This never made much sense to me.

I am also not sure how to solve using the node voltage method because I called the top essential node node voltage 1

(V1 - 6)/2

is a correct way to the difference in voltage between V1 and the bottom essential node for the reference node which I put a ground. At least I think it is but I'm not sure how to deal with the 1 ohm resistor before the 6V voltage source in my calculations.

Check the second attachment for my drawings of current etc.

As you can see all the currents go into the top essential node which is a problem.

Thanks for any help at all.

 

Done Editing

Ya

I understand why the red line is 1 A as it comes after the source. I however don't understand why before the current source (the blue line) is also one amp, as it comes before the current source.

see attachment

 

hm I've never thought about it like that lol

um can you help me with part B? I think I got the direction of the currents messed up as well lol

 

Can I combine the resistors in the left most branch (the one with the 6V power source and two resistors 2 ohm, 1 ohm) into an equialent one ohm resistor of 3 ohm? I'm not sure if I can or not because of the power source inbetween them.

I have identified the current in the second branch (the one with the 4V source and the 2 ohm source) as 2 A.

 

Start from the left and number the 4 branches from 1 to 4

The second branch
[itex]\frac{V_{1}-4}{2}[/itex]
The third branch
[itex]\frac{V_{1}}{4}[/itex]
The first branch
[itex]\frac{V_{1}-6}{3}[/itex]
The fourth branch
[itex]\frac{V_{1}}{4}[/itex]

[itex]\frac{V_{1}-4}{2} + \frac{V_{1}}{4} + \frac{V_{1}-6}{3} + \frac{V_{1}}{4} = 0[/itex]
[itex]6V_{1}-24+3V_{1}+4V_{1}-24+3V_{1}=0[/itex]
[itex]16V_{1}=48[/itex]
[itex]V_{1}=3 V[/itex]

I feel as if this answer might be wrong for [itex]V_{1}[/itex] as my friend said he got different answers to the questions in part B. Is it wrong? I don't see what I'm doing wrong here. I thought about just counting 1 amp as the current in the fourth branch due to the current source. But then...

[itex]\frac{V_{1}}{4}=1, V_{1}=4 V[/itex]

I'm not sure which answer is right if any of them. Thanks for your help. Also thanks for helping me before in that other circuit problem. I ended up getting the right answer and checked that I got the same answers as several other people in my class. You were big help.