Solve this Circuit using both Nodal Analysis and Mesh Analysis

In summary, the OP attempted to solve a homework equation using mesh analysis and nodal analysis, but his current assumption was incorrect. He should use a single node nodal equation to solve for V20 and V30.
  • #1
11
1

Homework Statement


[/B]
Determine, Using the values given in TABLE A, the current I in the circuit of figure 2 by
(a) Mesh analysis
(b) Nodal analysis

Circuit diagram and given variable attached.
V1 = 120
V2 = j120
V3 = 14.142+j14.142
Z1= 2
Z2=-J5
Z3=4
Z4=-J5
Z5=J4

Homework Equations


[/B]
KVL
KCL
Ohms law
Supernode

The Attempt at a Solution


[/B]
(a) complete, answer (-9.2+j17.2)

(b)

Ref node 0 on bottom line of diagram.
Node V10 between V1 and Z1 = 120V
Node V40 between V2 and Z3 = j120
Node V20 between V3, Z4 and Z1
Node V30 between V3, Z5 and Z3.

Replace V30 with super node. V20-V3)

(V1-V20/Z1)+(0-V20/Z4)+(-(V20-V3/Z5)+(V2-(V20-V3/Z3)=0

(V1-V20/Z1) - (V20/Z4) + (V3-V20/Z5) + (V2-V20+V3/Z3) = 0

-V20(1/Z1+1/Z4+1/Z5+1/Z3) + (V1*1/Z1) + (V2*1/Z3) + (V3*1/Z3) + (V3*1/Z5) = 0

-V20((1/2)+(1/-j5)+(1/j4)+(1/4) + (120*(1/2)+(V2*1/4)+(14.142+j14.142*1/4)+(14.142+j14.142*1/j4)=0

-V20(0.75+j0.05) + (60+j37.071)=0

(60+j37.071)=V20(0.75+j0.05)

V20 = (60+j37.071)/(0.75+j0.05)

V20 = (82.9266 +j43.8996)

V20/Z4 = I

i =-8.77992 +j16.5853. Which is off from my mesh answer. working backwards i expected due to the current from the mesh answer, times Z4 to be looking for a V20 voltage of (86+j46) can anyone help. Thanks in advance.

 

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  • #2
In your first equation, did you mean:

V1-V20/Z1

or
(V1-V20)/Z1
 
  • #3
V1-V20/Z1
 
  • #4
If you are using nodal analysis, the current is given by the quantity - difference in node voltages - divided by the impedance between them. You need to revisit this.
 
  • #5
Thanks Magoo it's really appreciated as I've been struggling with this for weeks though I'm still not following. As my V20 is connected via the reference node the difference in quantity would be 0-V20 no? or am I missing the point that node V20 would be V1-V20. If you could point me in the direction of any additional reading material it would be a massive help as the college notes for the whole course I'm on are, being polite, less than useless.
 
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  • #6
I redrew your circuit below.

upload_2018-9-13_15-40-5.png


With the reference node 0, there is no need to introduce V20 and V40. V20 is simply V1 and V40 is V2.

Let's say you have two nodes, Vhere and Vthere. Start at Vhere.

The term, (Vhere - Vthere)/Z, will represent a current in the direction of Vthere.

In your actual case, there will be a few "Vthere" terms, but the procedure is still the same.

See if any of this helps.

I don't have any references to suggest. Why not do a search on nodal equations for circuit analysis or something like that.
 

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  • #8
magoo said:
I redrew your circuit below.

View attachment 230649

The OP described the designators he is using in post #1:

"Ref node 0 on bottom line of diagram.
Node V10 between V1 and Z1 = 120V
Node V40 between V2 and Z3 = j120
Node V20 between V3, Z4 and Z1
Node V30 between V3, Z5 and Z3."

Your diagram is different.
 
  • #9
Yes, you are correct. The point is you still don't need to introduce V10 and V40. They are V1 and V2, respectively.
 
  • #10
magoo said:
Yes, you are correct. The point is you still don't need to introduce V10 and V40. They are V1 and V2, respectively.

True enough, and the OP didn't use V10 and V40 in his equations, but you showed two different nodes both labeled V20, which he does use.
 
  • #11
Ok, here is a corrected drawing.

upload_2018-9-14_0-36-38.png


My suggestion to the OP would be to write the nodal equations for node V20 and for node V30 based on the methodology that I outlined.
 

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  • #12
Thanks guys.

So reference node is defined at the bottom.

Trivial Nodes would be V10 & V40 as they are actually just V1 and V2 respective

So my super node is defined as V20-V30=V3.

My currents assumption will be all leaving the supernode so I will get the equation.

V20-V1/Z1+V20/Z4+V30/Z5+V30-V2/Z3=0

-V1/Z1+V20(1/Z1+1/Z4)+V30(1/Z5+1/Z3+-V2/Z3=0

Correct method so far?
 
  • #13
Rather than work with a supernode involving V20, V30 and V3, I would suggest sticking with nodal equations for a single node like V20 or V30.

The nodal equations must take into account Vhere and Vthere to do it correctly.

Your V20 - V1/Z1 should be (V20 - V1)/Z1. This procedure follows for the other currents as well.
 
  • #14
pgetts said:
Thanks guys.

So reference node is defined at the bottom.

Trivial Nodes would be V10 & V40 as they are actually just V1 and V2 respective

So my super node is defined as V20-V30=V3.

My currents assumption will be all leaving the supernode so I will get the equation.

V20-V1/Z1+V20/Z4+V30/Z5+V30-V2/Z3=0

-V1/Z1+V20(1/Z1+1/Z4)+V30(1/Z5+1/Z3+-V2/Z3=0

Correct method so far?

Substitute (V20 - V3) for V30, plug in numerical values for everything, and you're good to go.
 
  • #15
Ah thanks fellas. Finally got the same result as my Mesh analysis. Thanks again.
 
  • #16
Glad to hear that.
 

What is Nodal Analysis?

Nodal Analysis is a method used in circuit analysis to determine the voltage at each node (connection point) in a circuit. It involves creating equations based on Kirchhoff's Current Law and Ohm's Law to solve for the unknown voltages.

What is Mesh Analysis?

Mesh Analysis is a method used in circuit analysis to determine the current flowing through each branch of a circuit. It involves creating equations based on Kirchhoff's Voltage Law and Ohm's Law to solve for the unknown currents.

When should I use Nodal Analysis?

Nodal Analysis is typically used for circuits with multiple parallel branches and a smaller number of nodes. It is also useful for circuits with voltage sources.

When should I use Mesh Analysis?

Mesh Analysis is typically used for circuits with multiple series branches and a smaller number of meshes. It is also useful for circuits with current sources.

Can I use both Nodal Analysis and Mesh Analysis together?

Yes, it is possible to use both methods together to solve a circuit. This can be helpful in more complex circuits where one method may be easier to apply in certain parts of the circuit.

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