Nodal Analysis Problem: Where Did I Go Wrong in Using the Superposition Method?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
13 replies · 2K views
terryds
Messages
392
Reaction score
13

Homework Statement



1214gf6.png


The Attempt at a Solution


[/B]
At node B,

##i_s + 3 = i_2##
## 3 + 3 = i_2##
##i_2 = 6A##
##V_2 = i_2/R_2 = 3V##

But, it seems that the answer is wrong. Using superposition method, the voltage is 4V
Please tell me where I got wrong
 
Physics news on Phys.org
I can't understand your node equation. Can you explain the terms in detail? The current source term is clear, it's a fixed 3 A source, but I don't see any influence from the resistor values.
 
gneill said:
I can't understand your node equation. Can you explain the terms in detail? The current source term is clear, it's a fixed 3 A fGsource, but I don't see any influence from the resistor values.

The resistor 2 kilo ohms is at the right. Of course, it doesn't influence the independent current source.

I label the three points at the top as A , B , and C. Forget about what Va means. Let's call it Vb since it is on the point B.

At the point B, the current which goes in are the current generated by Vs and ##R_1## and the independent current source. The current goes out from point B is ##I_2##.
So,
in=out
##Vs/R_1 + 3 A = I_2##
##3 V / 1 k ohms + 3 A = I_2##
##I_2 = 3 + 3 * 10^-3 A = 3.003 A##
##V_2 = 3.003 A * 2 * 10 ^ 3 = 6006 A##

Which seems a wrong answer... In the book which use the superposition method, the ##I_2## is 1 mA and the voltage ##V_2## is 4 V... Please help me where I got wrong
 
terryds said:
At the point B, the current which goes in are the current generated by Vs and ##R_1## and the independent current source. The current goes out from point B is ##I_2##.
So,
in=out
##Vs/R_1 + 3 A = I_2##
The source current is not VsR1. What is the potential difference across R1? Does it not depend on the potential of the node B?
 
ehild said:
The source current is not VsR1. What is the potential difference across R1? Does it not depend on the potential of the node B?

Hmm.. You're right..

##\frac{V_B-V_A}{R_1}+3=i_2##
##\frac{V_B-V_s}{R_1}+3=i_2##
##\frac{V_B-3}{1000}+3=i_2##
##V_B - 3 + 3000 = i_2##
##V_B - i_2 = -2997##
##V_B -\frac{V_2}{2000} = -2997##

I'm stuck here

How to relate ##V_2## and ##V_B## ?

If I'm not wrong, ##V_A## = ##V_B## = ##V_C## because they are parallel, right ?
And, ##V_C## is ##V_2##, so ##V_B## = ##V_2## ?

Am I wrong about this voltage equality ??
 
If I'm not wrong, ##V_A## = ##V_B## = ##V_C## because they are parallel, right ?
And, ##V_C## is ##V_2##, so ##V_B## = ##V_2## ?

Am I wrong about this voltage equality ??[/QUOTE]
No, that is right, VB=V2.
 
ehild said:
If I'm not wrong, ##V_A## = ##V_B## = ##V_C## because they are parallel, right ?
And, ##V_C## is ##V_2##, so ##V_B## = ##V_2## ?

Am I wrong about this voltage equality ??
No, that is right, VB=V2.

If it's right, then ##V_2 = V_B =V_A = V_S = 3 V ##?? But, the answer is 4V :cry:

However, by substituting ##V_2## = ##V_B##, I get
##V_B - \frac{V_B}{2000} = -2997##
##2000V_B-V_B = -5994000##
##1999V_B = -5994000##
##V_B = \frac{-5994000}{1999} = -2998.49##

which seems to be an impossible answer..
Please help me..
 
terryds said:
##\frac{V_B-V_A}{R_1}+3=i_2##
You have a sign error in the first line. You assumed the current flowing from left to right, that means it is (VA-VB)/R1. The current flows in the direction of decreasng potential.
I think the current is given in mA instead of A, it is 3 mA instead of 3 A.
 
  • Like
Likes   Reactions: terryds
You sure the ammeter is measuring 3 Amperes and not 3 milliamperes
 
ehild said:
You have a sign error in the first line. You assumed the current flowing from left to right, that means it is (VA-VB)/R1. The current flows in the direction of decreasng potential.
I think the current is given in mA instead of A, it is 3 mA instead of 3 A.

##\frac{3-V_B}{1000} + 3 * 10^-3 = \frac{V_B}{2000}##
##3-V_B+3=\frac{V_B}{2}##
##V_B = 4 Volt##

Thanks for your help...

Anyway, you haven't answered my question :
If it's right, then V2=VB=VA=VS=3V?? But, the answer is 4V :cry:
Why ##V_B## is not equal to ##V_S## since it is parallel ??
 
mpresic said:
You sure the ammeter is measuring 3 Amperes and not 3 milliamperes

Maybe it is an error in the book.. It should be 3 miliamps not 3 amps
 
ehild said:
V2=VB, but not equal to VA. The source and the R2 resistors are not parallel. When are two elements connected in parallel?

I think ##V_S## and ##I_S## are in parallel. So, I think VA = VB..
I think two elements connected in parallel means that they are opposite direction..
But, Hmm...
So, the better definition is parallel-connection means that the currents are divided to sections, right ?
So, the voltage source is neither parallel to the current source (vertical line B) nor the R2 (vertical line C), right ?
 
Parallel connection of two element means that both terminals of one of them is common with the other one.The voltage is the same across both elements.
Series connection means one common terminal, and nothing else is connected to that. The same current flows through both elements.
 
  • Like
Likes   Reactions: terryds