# Nodal voltages

Can someone direct me to a tutorial for solving network
nodal voltages.
Also need to learn how to convert voltage source to a current source.

Related Electrical Engineering News on Phys.org
a text book should have everything you need..your in philly they must have a kick a*s Library there..

Yea I was there today, will be back tomorrow.

SGT
celect said:
Can someone direct me to a tutorial for solving network
nodal voltages.
Also need to learn how to convert voltage source to a current source.
You cannot convert a voltage source into a current source. What you can do is to convert a voltage source in series with a resistor into a current source in parallel with the same resistor. This is Norton's theorem.

Ok I'm reading that chapter now, I attached a sample I'm working on now.

#### Attachments

• 5.4 KB Views: 421
SGT
The solution you posted is wrong. The value of the current source is V/Z and what must be in parallel with the source is the series impedance, not the parallel you draw.

Regarding the right most circuit component, the symbol represents an inductor; yet the units is given in ohms.

inductor are coils , therefore i believe they can have resistance,
anyway,Checkout OpenCourseWare at MIT

So if I need to find E/Z = 20/9.9 = 2A

Now all I do is redraw circuit with 2A current source with
impedance parallel in circuit?

If you want to do a source transformation on this circuit, you should probably look up phasors.

SGT
celect said:
So if I need to find E/Z = 20/9.9 = 2A

Now all I do is redraw circuit with 2A current source with
impedance parallel in circuit?
No, you must make
$$I=\frac{20[cos(20)+jsin(20)]}{5.6+j8.2}$$

Last edited by a moderator:
I'm confused on how to get proper result

I have now I = 20V <20 degree/ 5.6+J8.2
and I found in book the way to draw circuit.
but I don't understand how to use the equation with cos and j 20 sin.

That's just Euler's formula and phasor notation:
"A-angle-theta" is phasor notation for A*ej*theta, which by Euler's formula is A[cos(theta) + j*sin(theta)], where A is a scalar value.

SGT
You can also write 5.6 + j8.2 as 9.93 <55.67
So, $$I=\frac{20 <20}{9.93 <55.67}$$
That is an easiest form to evaluate the quocient.

Here is a problem I got wrong, with the solution.
I don't understand how they got answer.

I don understand how you get
5.6 + j8.2 as 9.93 <55.67
how does 5.6 become 9.33
and j.2 = 55.67

#### Attachments

• 2.4 KB Views: 387
SGT
celect said:
Here is a problem I got wrong, with the solution.
I don't understand how they got answer.

I don understand how you get
5.6 + j8.2 as 9.93 <55.67
how does 5.6 become 9.33
and j.2 = 55.67
I have transformed the complex number from its cartesian form to its polar form.
Z = x + jy = R<φ
Where R is the modulus of the complex:
$$R = \sqrt{x^2 + y^2}$$
and
$$\phi = tan^{-1}\frac{y}{x}$$

Thanks its coming together now.

so my current = 20/9.93 = 2A

I don't understand how to get degrees yet
$$\phi = tan^{-1}\frac{y}{x}$$

do I take 8.3/5,5 tan and then use above formula?

what is the -1 how do i use on calculator?

SGT
celect said:
Thanks its coming together now.

so my current = 20/9.93 = 2A

I don't understand how to get degrees yet
$$\phi = tan^{-1}\frac{y}{x}$$

do I take 8.3/5,5 tan and then use above formula?

what is the -1 how do i use on calculator?
$$tan^{-1}$$ means arctan. It is the arc whose tangent is y/x.
And your current is I = 2 <-35.67.
When dividing complex numbers in the polar form you divide the moduli and subtract the phases.