Nodal-analysis to find voltage

  • Thread starter grekin
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    Voltage
In summary, G found that the voltage at node 1 was 6.2 volts, while at node 2 the voltage was -1.5 volts. Additionally, G found that if the resistor between node 1 and node 2 was omitted, then the voltage at node 2 would be -48.8 volts.
  • #1
grekin
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Homework Statement



Use the node-voltage method to find v1 in the circuit(Figure 1) if i1 = 6.2A and i2 = 1.5A .
Use the node-voltage method to find v2 in the circuit.

Figure_P04.8.jpg


Homework Equations



G = 1/R

QPFE3Hs.png


The Attempt at a Solution



Top side going from left to right, I assigned each node as 1, 3, then 2 (because of the way v_2 is assigned).

For Node 1, I had:
(v_1)/40 - (v_3)/8 = 6.2

Node 3:
-(v_1)/8 + (v_3)/80 = 0

Node 2:
(v_2)/120 = -1.5

Solving, I got v_1 = -5.06 V and v_2 = -180 V. These are wrong (obviously). Any assistance would be appreciated.

Moving image to bottom for visibility:

Figure_P04.8.jpg
 
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  • #2
Hello G and welcome to PF. Clear story. Do you have a reason to assign node 3, or could you make do with just nodes 1 and 2 ?
Reason I ask is because V2 = V3 seems a reasonable assumption to me. (I will look up the thread where I made a complete fool of myself by completely missing such a thing -- fortunately others put me right...)

Found it -- don't laugh!
 
  • #3
BvU said:
Hello G and welcome to PF. Clear story. Do you have a reason to assign node 3, or could you make do with just nodes 1 and 2 ?
Reason I ask is because V2 = V3 seems a reasonable assumption to me. (I will look up the thread where I made a complete fool of myself by completely missing such a thing -- fortunately others put me right...)

Found it -- don't laugh!

In that case where would I include the 80 Ohm resistor? Should it only be connected to v_2 on the equation?

If so,
Node 1:
6.2 = (v_1)/40 - (v_2)/8
-1.5 = (v_2)(1/80 + 1/120) - (v_1)/8

Solving, v_1 = 3.86 V and v_2 = -48.8 V. Which is still incorrect.

I'm out of attempts for solving for v_1, which I now know is 115 V.
 
  • #4
Ah, I interpret the node a equation a little different: after all, the 8 ohm is also connected to node a !
 
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  • #5
BvU said:
Ah, I interpret the node a equation a little different: after all, the 8 ohm is also connected to node a !

Ah, I see now. I didn't realize I had to include the resistors between each node as also connected to it individually.

1:
6.2 = v_1 * (1/40+1/8) - v_2 / 8

2:
-1.5 = v_2 * (1/80 + 1/120 + 1/8) - v_1 / 8

Ended with v_1 = 114.7 and v_2 = 88 (which is correct). Thanks.
 
  • #6
Always nice to have the answer at hand... I was still plodding :smile:
And: you're welcome!
 

Related to Nodal-analysis to find voltage

1. How is nodal analysis used to find voltage?

Nodal analysis is a method used to determine the voltage at different points or nodes in an electrical circuit. It involves creating a system of equations based on Kirchhoff's Current Law (KCL) and Ohm's Law, and solving them to find the unknown node voltages.

2. What is Kirchhoff's Current Law and how is it used in nodal analysis?

Kirchhoff's Current Law states that the sum of currents entering a node must equal the sum of currents leaving that node. In nodal analysis, this law is used to create equations by setting the sum of currents at each node to zero, which helps determine the unknown node voltages.

3. Can nodal analysis be used in both DC and AC circuits?

Yes, nodal analysis can be used in both DC and AC circuits as long as the circuit is linear and the components have known resistance/impedance values. However, the equations may be more complex in AC circuits due to the inclusion of complex numbers.

4. Are there any limitations to using nodal analysis?

One limitation of nodal analysis is that it can only be used in circuits with a single reference node. Additionally, if the circuit contains dependent sources, the equations may become more difficult to solve.

5. How does nodal analysis compare to other methods of circuit analysis?

Nodal analysis is often preferred over other methods, such as mesh analysis, because it allows for fewer equations to be written and solved. It is also more flexible and can be used in a wider range of circuit configurations.

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