Download Here we would like to find the three indicated node... method. So I’ll begin at node 1.

yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Mercury-arc valve wikipedia , lookup

Electrical ballast wikipedia , lookup

Resistive opto-isolator wikipedia , lookup

Stray voltage wikipedia , lookup

Three-phase electric power wikipedia , lookup

Mathematics of radio engineering wikipedia , lookup

Topology (electrical circuits) wikipedia , lookup

Mains electricity wikipedia , lookup

Opto-isolator wikipedia , lookup

TRIAC wikipedia , lookup

Rectiverter wikipedia , lookup

Buck converter wikipedia , lookup

Metadyne wikipedia , lookup

Current source wikipedia , lookup

Two-port network wikipedia , lookup

Ohm's law wikipedia , lookup

Alternating current wikipedia , lookup

Network analysis (electrical circuits) wikipedia , lookup

Signal-flow graph wikipedia , lookup

Here we would like to find the three indicated node voltages using the node voltage
So I’ll begin at node 1.
We want to sum the currents that are exiting the node, and set the result to zero.
The first current is associated with the 2 amp source. The arrow is pointing out of the
node, so we write the value as positive.
Next we have the value of the 15 ohm resistor, so we write V1 on the tail side of the
arrow, and zero on the head side of the arrow. We can subtract these values and divide
them by the value of the resistor.
For the next current, we can write V1-V2 divided by 10 and set the result equal to zero.
At node two, there are four branches connected to it.
I’ll start here. This current that is exiting is V2-V1 divided by 10.
The next current that is exiting is -4 since it is entering the node.
Next we have V2-V3 divided by 5 ohms.
The last current is entering the node so it is subtracted, but since it is a value of negative
three it becomes positive 3.
These currents sum to zero.
Finally at node three we have three branches connected.
This current is V3-V2 divided by 5.
The next current is V3 divided by 20.
Lastly, we have current leaving the node so we write the value -3.
We then set the result equal to zero.
We can then rewrite these equations in matrix form.
I’ve gathered my coefficients together for each of the node voltages and put the constants
on the right side.
Now I can convert these to decimal form and put them in a matrix.
We notice that we have symmetry in this matrix.
This is a good way to check your work.
Here we have the same values for the off-diagonals.
This will always be the case in circuits that contain entirely independent current sources
when we use nodal voltage method.
We can then solve using a matrix solver.
We find that V1, V2, and V3 have the values shown.