- #1
darwinharianto
- 42
- 0
Homework Statement
find the voltage by using norton and thevenin equation
is this method correct?
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Is Using Norton and Thevenin Equations to Find Voltage Correct?
In summary, the conversation is about finding the voltage using Norton and Thevenin equations and whether or not the method being used is correct. The experts give advice on finding the Thevenin voltage and resistance and using source transformations. The conversation also includes clarifying component values and converting a current source to its Norton equivalent.
find the voltage by using norton and thevenin equation
is this method correct?
- #2
darwinharianto
- 42
- 0
if I change the pararelled resistor and send the most right resistor to the left, the voltage changed
is it right if I am doing it like this?
thx for reply in advance
sory for bad english
is it right if I am doing it like this?
thx for reply in advance
sory for bad english
- #3
Rellek
- 107
- 13
Hey there!
You might notice that the voltage over your last resistor (on the right) will be the Voltage across your two nodes. So, if you were to find the voltage over that resistor, that will be your Thevenin voltage.
For your Thevenin resistance, try to deactivate all of your independent devices and starting at point A on your node, calculate the equivalent resistance from that point. That will be your Thevenin resistance.
From there you know the Norton circuit is a source transformation. Good luck!
You might notice that the voltage over your last resistor (on the right) will be the Voltage across your two nodes. So, if you were to find the voltage over that resistor, that will be your Thevenin voltage.
For your Thevenin resistance, try to deactivate all of your independent devices and starting at point A on your node, calculate the equivalent resistance from that point. That will be your Thevenin resistance.
From there you know the Norton circuit is a source transformation. Good luck!
- #4
darwinharianto
- 42
- 0
thx for the advice
after I deactivate, I found the resistor from paralleled 10k, 6k and 8k
then the ampere source have the resistor (4k) below it paralleled with volt source and the resistor ( from paralleled 10k, 6k and 8k).
after that I'm stuck.
any idea how to go on?
after I deactivate, I found the resistor from paralleled 10k, 6k and 8k
then the ampere source have the resistor (4k) below it paralleled with volt source and the resistor ( from paralleled 10k, 6k and 8k).
after that I'm stuck.
any idea how to go on?
- #5
gneill
Mentor
- 20,989
- 2,934
Hi darwinharianto, Welcome to Physics Forums.
Some of the component values in your attached figure are hard to make out. Can you clarify their values?
Some of the component values in your attached figure are hard to make out. Can you clarify their values?
- #6
Rellek
- 107
- 13
It looks like you might benefit from using source transformations to make both of your sources the same, if you want to use this method instead. I see a voltage source in series with a resistor right off the bat. Also, if you have a resistor in series with a current source, what can you say about the current in that section of wire?
From there, it seems like it would only be current division and Ohm's law to find the voltage over your resistor on the right.
From there, it seems like it would only be current division and Ohm's law to find the voltage over your resistor on the right.
- #7
NascentOxygen
Staff Emeritus
Science Advisor
- 9,242
- 1,075
Hi darwinharianto! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif
You converted the voltage source to its Norton equivalent. Why not also convert the left-most source and resistors to its Norton equivalent? You'll then have two current sources in parallel ...
... the rest is easy!
You converted the voltage source to its Norton equivalent. Why not also convert the left-most source and resistors to its Norton equivalent? You'll then have two current sources in parallel ...
... the rest is easy!
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- #8
darwinharianto
- 42
- 0
NascentOxygen said:Hi darwinharianto! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif
You converted the voltage source to its Norton equivalent. Why not also convert the left-most source and resistors to its Norton equivalent? You'll then have two current sources in parallel ...
... the rest is easy!
actually the left one is not a volt source, its an ampere source (1mA)
which is trouble me
hope this pic will be giving a better view
Last edited by a moderator:
- #9
NascentOxygen
Staff Emeritus
Science Advisor
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I know it's a current source. You can still represent it by an ideal current source in parallel with a resistance.darwinharianto said:
- #10
darwinharianto
- 42
- 0
how is that?
so the current source is multiplied? I got confused
so the current source is multiplied? I got confused
- #11
NascentOxygen
Staff Emeritus
Science Advisor
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darwinharianto said:
What is the procedure for representing any source by a Norton (or Thévenin) equivalent? Follow that.
- #12
darwinharianto
- 42
- 0
okiee
ill figure something out
ill figure something out
FAQ: Is Using Norton and Thevenin Equations to Find Voltage Correct?
What is the difference between Voltage Thevenin and Voltage Norton?
Both Voltage Thevenin and Voltage Norton are equivalent circuit models used to simplify complex circuits. The main difference between the two is that Voltage Thevenin uses a voltage source in series with a resistor, while Voltage Norton uses a current source in parallel with a resistor.
How is a Voltage Thevenin circuit calculated?
To calculate a Voltage Thevenin circuit, the circuit is first simplified by removing the load resistor. Then, the open circuit voltage is measured across the load resistor terminals. This voltage is the Thevenin voltage. The Thevenin resistance is then calculated by shorting all voltage sources and opening all current sources in the original circuit, and measuring the resistance across the load resistor terminals.
How is a Voltage Norton circuit calculated?
To calculate a Voltage Norton circuit, the circuit is first simplified by removing the load resistor. Then, the short circuit current is measured across the load resistor terminals. This current is the Norton current. The Norton resistance is then calculated by opening all voltage sources and shorting all current sources in the original circuit, and measuring the resistance across the load resistor terminals.
What is the purpose of using Voltage Thevenin and Norton equivalents?
The purpose of using Voltage Thevenin and Norton equivalents is to simplify complex circuits into equivalent circuits that are easier to analyze. These equivalents allow for easier calculation of voltage and current in a circuit, without having to analyze the entire complex circuit.
Are Voltage Thevenin and Norton equivalents always accurate?
No, Voltage Thevenin and Norton equivalents are not always accurate. They are only accurate for linear circuits, which means that the relationship between voltage and current in the circuit is linear. Non-linear elements, such as diodes and transistors, cannot be represented accurately by these equivalents.