# Solving a Nodal Analysis: Find Va(t)

• eehelp150
In summary, the homework statement is:Find Va(t) using Nodal Analysis.The Attempt at a Solution is:W = 250Z36mH = j9 ohmZ80mH = j20 ohmZ0.25mF = -j16 ohm (is capacitor impedance always negative? It says so in my instructor's notes but I forgot why).The Nodal Equations are:Node1:\frac{Node1-20<0}{8+9j} + \frac{Node1}{-j16}=0Node2:-1.2<45° + \frac
eehelp150

## Homework Statement

Find Va(t) using Nodal Analysis:

Zc = 1/(jwC)
Zl = jw*L

## The Attempt at a Solution

W = 250
Z36mH = j9 ohm
Z80mH = j20 ohm
Z0.25mF = -j16 ohm (is capacitor impedance always negative? It says so in my instructor's notes but I forgot why)

20cos(250t) -> 20 <0° V
1.2cos(250t+45°) -> 1.2<45° A

Are the following Nodal equations correct? Is Node1 simply Va?
Node1:
$$\frac{Node1-20<0}{8+9j} + \frac{Node1}{-j16}=0$$

Node2:
$$-1.2<45° + \frac{Node2-4Va}{j20}=0$$

Node1 and Node2 are just one node. Nodes are not defined by black dots. Those two dots are connected by a wire of perfect conductivity and the voltages at the two dots are the same, so they are really only one node.

The Electrician said:
Node1 and Node2 are just one node. Nodes are not defined by black dots. Those two dots are connected by a wire of perfect conductivity and the voltages at the two dots are the same, so they are really only one node.
I assume Va is the voltage across the Capacitor?

Last edited:
That looks OK; now all you have to do is solve for Va.

By the way, the reason that capacitor impedance is negative is because 1/j = -j

The Electrician said:
That looks OK; now all you have to do is solve for Va.

By the way, the reason that capacitor impedance is negative is because 1/j = -j
Of course. I feel like an idiot now.

The Electrician said:
That looks OK; now all you have to do is solve for Va.

By the way, the reason that capacitor impedance is negative is because 1/j = -j
I ended up getting
Va = 4528.6 - 911.4i
Is that correct?

No, I don't get that. As a sanity check, for Va to be 4000 volts plus seems a little high to me.

What happened to your equation in post #3?

The Electrician said:
No, I don't get that. As a sanity check, for Va to be 4000 volts plus seems a little high to me.

What happened to your equation in post #3?

Don't know what happened to it

I plugged the equation into MyAlgebra.com and it spit out the answer. Here is the equation again:
$$\frac{Va-20}{8+j9}-\frac{Va}{j16}-\frac{3Va}{j20} = \frac{1.2\sqrt{2}}{2}+ \frac{1.2\sqrt{2}}{2}i$$

eehelp150 said:
I ended up getting
Va = 4528.6 - 911.4i
Is that correct?
It's not what I get, if that helps

gneill said:
It's not what I get, if that helps
I tried another calculator and got:
1.89293-12.2816 i
is that correct?

eehelp150 said:
I tried another calculator and got:
0.848528+0.848528 i
is that correct?
Nope.

Sometimes plugging in all the complex values into a node equation and lugging them around in the reduction is more tedious and error prone than using symbols. Why not assign variables to the branch impedances and the current and solve? That way we can follow your steps.

gneill said:
Nope.

Sometimes plugging in all the complex values into a node equation and lugging them around in the reduction is more tedious and error prone than using symbols. Why not assign variables to the branch impedances and the current and solve? That way we can follow your steps.
I read the output wrong, it was 1.89293-12.2816 i.

eehelp150 said:
Don't know what happened to it

I plugged the equation into MyAlgebra.com and it spit out the answer. Here is the equation again:
$$\frac{Va-20}{8+j9}-\frac{Va}{j16}-\frac{3Va}{j20} = \frac{1.2\sqrt{2}}{2}+ \frac{1.2\sqrt{2}}{2}i$$

I think you may be having a problem with your complex constant j. In Mathematica there is a difference between j9 and j 9.

Without a space between the j and the 9 (in other words, like this j9), Mathematica treats that as a variable name. With a space, like j 9, that is treated like j*9, the product of j and 9. Also at the very end of your equation you have i instead of j.

What does MyAlgebra use for the complex constant, j or i? Whichever it is, put the right one in your equation and instead of j9 and j20, etc., put j*9 and j*20 (or possibly i*9 and i*20 if i is the symbol for the complex constant) and run the equation again.

The Electrician said:
I think you may be having a problem with your complex constant j. In Mathematica there is a difference between j9 and j 9.

Without a space between the j and the 9 (in other words, like this j9), Mathematica treats that as a variable name. With a space, like j 9, that is treated like j*9, the product of j and 9. Also at the very end of your equation you have i instead of j.

What does MyAlgebra use for the complex constant, j or i? Whichever it is, put the right one in your equation and instead of j9 and j20, etc., put j*9 and j*20 (or possibly i*9 and i*20 if i is the symbol for the complex constant) and run the equation again.

eehelp150 said:
I read the output wrong, it was 1.89293-12.2816 i.
This is correct.

## 1. What is nodal analysis?

Nodal analysis is a method used in circuit analysis to determine the voltage at each node in a circuit. It is based on Kirchhoff's Current Law (KCL) which states that the sum of currents entering a node must equal the sum of currents leaving the node.

## 2. Why is nodal analysis used?

Nodal analysis is used because it is a systematic and efficient method for solving complex circuits. It allows for the simplification of a circuit into smaller sections, making it easier to analyze and solve. Additionally, it can be used to solve circuits with multiple independent sources.

## 3. What are the steps for solving a nodal analysis?

The steps for solving a nodal analysis are:

• Identify and label all nodes in the circuit
• Apply KCL at each node to create equations
• Solve the resulting system of equations for the unknown node voltages

## 4. How do you find Va(t) using nodal analysis?

To find Va(t) using nodal analysis, you must first solve for the voltages at each node in the circuit. Then, use Ohm's Law to calculate the voltage across any resistors connected to the node where Va(t) is located. The sum of these voltages will equal Va(t).

## 5. Can nodal analysis be used for circuits with capacitors and inductors?

Yes, nodal analysis can be used for circuits with capacitors and inductors. However, additional equations must be included to account for the transient response of these components. This is known as modified nodal analysis and involves solving a system of differential equations.

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