How Do You Solve a Norton Equivalent Circuit with a Dependent Source?

In summary, a Norton Equivalent Circuit is a simplified representation of a complex network of electrical components that behaves in the same way as the original network in terms of current and voltage at a specific load. It differs from a Thevenin Equivalent Circuit in that it represents a voltage source in parallel with a resistor rather than in series. The purpose of finding a Norton Equivalent Circuit is to simplify analysis of the original network at a specific load. To calculate the Norton Equivalent Current and Resistance, the original network's voltage sources must be short-circuited and current sources must be open-circuited, respectively. However, Norton Equivalent Circuits are limited in their applicability to only linear, time-invariant networks and may not accurately represent the original network
  • #1
rafehi
49
1

Homework Statement



See picture:

http://img88.imageshack.us/img88/8319/87671967.jpg

The answer is given as phasors
I = 0.3536[tex]\angle[/tex]45
Z = 2.828[tex]\angle[/tex]-45

Homework Equations


The Attempt at a Solution



ZL = jwL = 2j Ohms
ZC = -j/(wC) = -4j Ohms

Can't do it by zeroing the sources as there's a dependent source (if it's possible, we haven't been taught it).

So then we'll have to find both the open source voltage VOC and the short circuit current ISC, correct?

Starting with VOC with a node b/w the inductor and resistor (which is equal to VOC because there is no voltage drop through the resistor due to no current going through it):
(groud at bottom node)
VOC: [tex]\frac{Voc}{-4j}[/tex] + [tex]\frac{Voc - 2}{2}[/tex] = 1.5IL

IL = [tex]\frac{Voc - 2}{2}[/tex]

Subbing in and arranging gives:

[tex]\frac{Voc}{-4j}[/tex] = [tex]\frac{Voc - 2}{4}[/tex]

Solving gives:
Voc = -1.414[tex]\angle[/tex]45,

however I'm fairly sure it's wrong given the answers ( IZ != V).I'm not sure how to find ISC because of the dependent current source. Have tried and can't get the correct answer - not sure if I'm going about it the right way.
i1 = top loop current
i2 = left loop
i3 = right loop

Taking positive to be CW,
i1 = 1.5iL
iL = i1 - i2

Therfore,
iL = 2i2

Loop two (left):
-2i2 - 4j (i2 - i3) = 2[tex]\angle[/tex]0

Loop three (right):
2i3 - 4j (i3 - i2) = 0

Solving the two linear equations gives
i3 = iN = 2j

which isn't the correct answer.Any help would be greatly appreciated...
 
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  • #2


your approach to solving this problem should be systematic and based on fundamental principles and equations. Here are some steps you can follow to solve this problem:

1. Identify the unknown quantities in the circuit: In this case, the unknown quantities are the current I and the impedance Z.

2. Write down the equations that relate these unknown quantities to the given information: In this case, the equations are I = 0.3536∠45 and Z = 2.828∠-45.

3. Use the given information to solve for the unknown quantities: In this case, you can use the given equations to solve for I and Z. Note that the given values for I and Z are in phasor form, so you need to convert them to Cartesian form before solving for the unknown quantities.

4. Check your solution: Once you have solved for I and Z, you can check your solution by substituting the values into the given equations. Your solution should satisfy the given equations.

5. If your solution does not satisfy the given equations, check your calculations and make sure you have not made any mistakes. If you are unable to find the mistake, you may need to approach the problem from a different angle or seek help from a classmate, tutor, or instructor.

6. If your solution satisfies the given equations, then you have successfully solved the problem! Congratulations!
 

Related to How Do You Solve a Norton Equivalent Circuit with a Dependent Source?

What is a Norton Equivalent Circuit?

A Norton Equivalent Circuit is a simplified representation of a complex network of electrical components, which behaves in the same way as the original network in terms of current and voltage at a specific load.

How is a Norton Equivalent Circuit different from a Thevenin Equivalent Circuit?

The main difference between a Norton Equivalent Circuit and a Thevenin Equivalent Circuit is that the former represents a voltage source in parallel with a resistor, while the latter represents a voltage source in series with a resistor.

What is the purpose of finding a Norton Equivalent Circuit?

The purpose of finding a Norton Equivalent Circuit is to simplify a complex network of electrical components into a single equivalent circuit, which can be used to analyze the behavior of the original network at a specific load without having to consider the internal details of the network.

How do you calculate the Norton Equivalent Current and Resistance?

To calculate the Norton Equivalent Current, you need to short-circuit all voltage sources in the original network and calculate the total current flowing through the short circuit. To calculate the Norton Equivalent Resistance, you need to open-circuit all current sources in the original network and calculate the equivalent resistance between the load terminals.

What are the limitations of Norton Equivalent Circuits?

Norton Equivalent Circuits are only valid for linear, time-invariant networks and cannot be used for non-linear or time-varying networks. Additionally, they are only accurate at a specific load and may not accurately represent the behavior of the original network at other loads.

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