Hello,
Ive tried source transformation to solve this and keep getting the wrong answer. I applied superposition and get the right answer but the book asks me to solve using nodal analysis, and in the picture below, I coulda sworn you could add voltage sources in series but when I do I still get the wrong answer. Any help would be great.
Thanks.

 

the book didn't label the ground/reference point there, I did btw.

 

Show as your attempt.

 

All current leaving nodes is positive.

Node v1 I get;
v1(1/5+1/10+1/5) - v2(1/5) - (21/5)=0
v1(0.5)-v2(0.2)= 4.2
Node v2 I get:
-v1((1/5)+v2(1/5+1/10+1/5)+(10.5/5)=0
v1(0.2)+v2(0.5)=2.1

Using matix I get:
v1= 12v
v2=9v

I know this is wrong, but im guessing its how I labeled my ground since the sources are both flipped flopped. The 21V is normal(imho) and the 10.5V is flipped. So how would I determine the proper voltage as my ground? I can solve this problem using mesh and superposition since their independent sources, but im just lost with nodal. Ive tried for a couple hours now, I know its something very simple im missing. If you need a picture of the current leaving the nodes let me know, and ill post.

 

just realized instead of:
v1(0.2)+v2(0.5)=2.1
its:
-v1(0.2)+v2(0.5)=2.1

 

Hello,
Ive tried source transformation to solve this and keep getting the wrong answer. I applied superposition and get the right answer but the book asks me to solve using nodal analysis, and in the picture below, I coulda sworn you could add voltage sources in series but when I do I still get the wrong answer. Any help would be great.
Thanks.

You can add voltage sources in series, but to do that, they have to be in series. The two sources in your circuit are NOT in series. To be in series requires that whatever current flows in one of them HAS to flow in the other. In this circuit the current flowing in one has three possible paths to follow, only one of which flows through the other source.

 

All current leaving nodes is positive.

Node v1 I get;
v1(1/5+1/10+1/5) - v2(1/5) - (21/5)=0
v1(0.5)-v2(0.2)= 4.2
Node v2 I get:
-v1((1/5)+v2(1/5+1/10+1/5)+(10.5/5)=0
v1(0.2)+v2(0.5)=2.1

Using matix I get:
v1= 12v
v2=9v

I know this is wrong, but im guessing its how I labeled my ground since the sources are both flipped flopped. The 21V is normal(imho) and the 10.5V is flipped. So how would I determine the proper voltage as my ground? I can solve this problem using mesh and superposition since their independent sources, but im just lost with nodal. Ive tried for a couple hours now, I know its something very simple im missing. If you need a picture of the current leaving the nodes let me know, and ill post.

Need to watch your signs. You also need to track your units properly.

\(
\text{Node V1:} \; V_1 \( \frac{1}{5 \Omega } \; + \; \frac{1}{5 \Omega } \; + \; \frac{1}{10 \Omega } \) \; - \; V_2 \( \frac{1}{5 \Omega } \) \; = \; 21\, V \( \frac{1}{5 \Omega} \)
\text{Node V2:} \;-V_1 \( \frac{1}{5 \Omega } \) \; + \; V_2 \( \frac{1}{5 \Omega } \; + \; \frac{1}{10 \Omega } \; + \; \frac{1}{5 \Omega } \) \; = \; -10.5\, V \( \frac{1}{5 \Omega} \)
\)

Multiply both sides of each equation by 10 Ω:

\(
\text{Node V1:} \; V_1 \( 2 \; + \; 2 \; + \; 1 \) \; - \; V_2 \( 2 \) \; = \; 21\, V \( 2 \)
\text{Node V2:} \;-V_1 \( 2 \) \; + \; V_2 \( 2 \; + \; 1 \; + \; 2 \) \; = \; -10.5\, V \( 2 \)
\)

Simplify:

\(
\text{Node V1:} \; 5 \cdot V_1 \; - \; 2 \cdot V_2 \; = \; 42\, V
\text{Node V2:} \;-2 \cdot V_1 \; + \; 5 \cdot V_2 \; = \; -21\, V
\)

Notice what happens if we add these two equations together:

\(
3 \cdot V_1 \; + \; 3 \cdot V_2 \; = \; 21\, V
3 \(V_1 \; + \; V_2 \) \; = \; 21\, V
V_1 \; + \; V_2 \; = \; 7 \, V
\)

This provides a very valuable check on your final result. If the sum of V1 and V2 isn't 7V, then you know that you have a problem.

You always want to look for these little checks as we all make the kinds of mistakes you made. In fact, in solving for the actual voltages I copied the -21 V as just 21 V, with the result being that I got V1 = 9V and V2 = 12 V, but because I already had in my hip pocket that the sum had to be 7 V, I knew I had made a mistake and so was able to find it almost immediately.

 

Need to watch your signs. You also need to track your units properly.

\(
\text{Node V1:} \; V_1 \( \frac{1}{5 \Omega } \; + \; \frac{1}{5 \Omega } \; + \; \frac{1}{10 \Omega } \) \; - \; V_2 \( \frac{1}{5 \Omega } \) \; = \; 21\, V \( \frac{1}{5 \Omega} \)
\text{Node V2:} \;-V_1 \( \frac{1}{5 \Omega } \) \; + \; V_2 \( \frac{1}{5 \Omega } \; + \; \frac{1}{10 \Omega } \; + \; \frac{1}{5 \Omega } \) \; = \; -10.5\, V \( \frac{1}{5 \Omega} \)
\)

Multiply both sides of each equation by 10 Ω:

\(
\text{Node V1:} \; V_1 \( 2 \; + \; 2 \; + \; 1 \) \; - \; V_2 \( 2 \) \; = \; 21\, V \( 2 \)
\text{Node V2:} \;-V_1 \( 2 \) \; + \; V_2 \( 2 \; + \; 1 \; + \; 2 \) \; = \; -10.5\, V \( 2 \)
\)

Simplify:

\(
\text{Node V1:} \; 5 \cdot V_1 \; - \; 2 \cdot V_2 \; = \; 42\, V
\text{Node V2:} \;-2 \cdot V_1 \; + \; 5 \cdot V_2 \; = \; -21\, V
\)

Notice what happens if we add these two equations together:

\(
3 \cdot V_1 \; + \; 3 \cdot V_2 \; = \; 21\, V
3 \(V_1 \; + \; V_2 \) \; = \; 21\, V
V_1 \; + \; V_2 \; = \; 7 \, V
\)

This provides a very valuable check on your final result. If the sum of V1 and V2 isn't 7V, then you know that you have a problem.

You always want to look for these little checks as we all make the kinds of mistakes you made. In fact, in solving for the actual voltages I copied the -21 V as just 21 V, with the result being that I got V1 = 9V and V2 = 12 V, but because I already had in my hip pocket that the sum had to be 7 V, I knew I had made a mistake and so was able to find it almost immediately.

...facepalm, its one of those days. After getting the wrong answer multiple times using nodal and trying multiple methods and THEN getting the right answer, I think I was just psyching myself out. Thanks, I got it now.

 

...facepalm, its one of those days. After getting the wrong answer multiple times using nodal and trying multiple methods and THEN getting the right answer, I think I was just psyching myself out. Thanks, I got it now.

I more of a mesh person anyways, but the book asked to solve using nodal so....

 

I more of a mesh person anyways, but the book asked to solve using nodal so....

It's good to know this about yourself. The next step is to make sure that you use nodal analysis whenever you can, instead of mesh, so that you get to the point where you are equally comfortable with both and can choose which to use based on the circuit being analyzed. You want as many tools in your toolbox as you can pack in there and you want to know when and how to use each. If the only tool you know how to use is a hammer, then every problem looks suspiciously like a nail.