What Is the Correct Node Analysis for Calculating Voltages at Nodes A and B?

AI Thread Summary
The discussion focuses on using mesh analysis to determine the currents in three meshes and subsequently calculating the voltages at nodes A and B. The initial mesh analysis yielded current values I1 = 0.046, I2 = 0.398, and I3 = -0.5908. However, the node analysis presented equations that incorrectly mixed voltages and currents, leading to confusion. A correction was suggested, emphasizing that the terms should reflect proper voltage differences, specifically changing the input to (V1 - V2 - 30)/40. The final equations must adhere to Kirchhoff's Current Law (KCL), ensuring that the sum of the currents at the node equals zero for accurate voltage calculations.
Cricket.sign
Messages
2
Reaction score
0

Homework Statement


Using mesh analysis work out the currents flow through each of the three meshes. From these current values calculate the voltage of node A and node B with respect to the ground node. XX V has a value of 13.

http://www.flickr.com/photos/29385059@N07/2744576925/


Homework Equations


KCL
KVL


The Attempt at a Solution


Well I figured out the mesh analysis easy enough. Obtaining, I1=.046, I2=.398 and I3=-.5908. Got to the node analysis and haven't got a clue. I came up with

1. V1/15 = 3 + [V1 -(V2+30)/40] + [V1-V2]/10

2. [V1-V2]/10 + [V1 -(V2+300)/40] = V2/20 + [(V2+17)/25]

Which gave me V1 = 187.5
V2 = 105.5
 
Physics news on Phys.org
Cricket.sign said:

Homework Statement


Using mesh analysis work out the currents flow through each of the three meshes. From these current values calculate the voltage of node A and node B with respect to the ground node. XX V has a value of 13.

http://www.flickr.com/photos/29385059@N07/2744576925/


Homework Equations


KCL
KVL


The Attempt at a Solution


Well I figured out the mesh analysis easy enough. Obtaining, I1=.046, I2=.398 and I3=-.5908. Got to the node analysis and haven't got a clue. I came up with

1. V1/15 = 3 + [V1 -(V2+30)/40] + [V1-V2]/10

2. [V1-V2]/10 + [V1 -(V2+300)/40] = V2/20 + [(V2+17)/25]

Which gave me V1 = 187.5
V2 = 105.5

In your two equations there is a term where you mix voltages and currents:
1. [V1 -(V2+30)/40] where V1 is a voltage and (V2+30)/40 is a current
2. [V1 -(V2+300)/40] the same

Redo your equations
 
[V1 -(V2+30)/40] is actually incorrectly input. It should be (V1-V2-30)/40. Not sur if that changes anything, but I think it should
 
Cricket.sign said:
[V1 -(V2+30)/40] is actually incorrectly input. It should be (V1-V2-30)/40. Not sur if that changes anything, but I think it should

It changes, but it is not correct yet. According to KCL, the sum of the currents out of a node must be zero. Using your reference it should be:
\frac{V_1}{15}+3+\frac{V_1-V_2-30}{40}+\frac{V_1-V_2}{10}=0
 
Back
Top