# Three-part Problem: Thevenin + Dependent current source

• s3a
In summary, the conversation discussed a problem involving an electric circuit and its components, such as dependent current sources, nodal analysis, short circuit current, Thevenin voltage, and Thevenin resistance. The file included in the conversation contained the problem statement and an attempt at a solution, along with a request for help in identifying any flaws in the work. The main issue identified was in part b), where the voltage value of V_1 was incorrectly calculated. The correct value was determined to be V_1=0, leading to a different calculation for I_0. Further discussion also clarified some minor errors in notation and sign conventions. Overall, the conversation served to provide a thorough understanding of the problem and its solution.
s3a

## Homework Statement

The diagram of the electric circuit, the problem statement (and an attempt) are included in the file below.:
https://www.docdroid.net/RD0ZEgA/question-3.pdf

## Homework Equations

Dependent Current Source

Nodal analysis

Short Circuit Current

Thevenin Voltage

Thevenin Resistance

## The Attempt at a Solution

See the (same) included file.:
https://www.docdroid.net/RD0ZEgA/question-3.pdf

Could someone please help point out the flaws in the work? If you show me another way to do the problem's parts that would also be nice, but I'm largely interested in knowing what's wrong with the work in the attachment.

Any input would be greatly appreciated!

P.S.
I didn't attach the file because, when I tried to, the website said it's too large.

The problem i immediately saw was with the part b). You get ##V_1=0.10##. Shouldn't ##V_1## be ##V_1=0## since it's short-circuited to the ground?
Using that i get: ##\frac{0.5V}{0.25}+\frac{V_x}{3}=I_0##, where ##V_x = 0.5##. Then ##I_0=2.16666##.

Last edited:
s3a
diredragon said:
The problem i immediately saw was with the part b). You get ##V_1=0.10##. Shouldn't ##V_1## be ##V_1=0## since it's short-circuited to the ground?
Using that i get: ##\frac{0.5V}{0.25}+\frac{V_x}{3}=I_0##, where ##V_x = 0.5##. Then ##I_0=2.16666##.
Oh, yes!

Shouldn't it be like this, though (instead of what you typed)?:
(V_1 - 0.5)/0.25 + V_x/3 - I_0 = 0

(V_1 - 0.5)/0.25 + (V_1 - 0.5)/3 - I_0 = 0

(0 - 0.5)/0.25 + (0 - 0.5)/3 - I_0 = 0

(-0.5)/0.25 + (-0.5)/3 - I_0 = 0

I_0 = (-0.5)/0.25 + (-0.5)/3

I_0 = -2.16666666666666666667 A (meaning that the current is pointing upwards, instead of downwards)

s3a said:
Oh, yes!

Shouldn't it be like this, though (instead of what you typed)?:
(V_1 - 0.5)/0.25 + V_x/3 - I_0 = 0

(V_1 - 0.5)/0.25 + (V_1 - 0.5)/3 - I_0 = 0

(0 - 0.5)/0.25 + (0 - 0.5)/3 - I_0 = 0

(-0.5)/0.25 + (-0.5)/3 - I_0 = 0

I_0 = (-0.5)/0.25 + (-0.5)/3

I_0 = -2.16666666666666666667 A (meaning that the current is pointing upwards, instead of downwards)

Using these equations the ##-I_0## suggest that (following your sign rule) the ##I_0## enters the node. Since you get a negative value it means that the current flow in the opposite direction than you speculated it would. Meaning it goes downward since you draw it going upward.
See that in my equations i immediately concluded that it would go downward and i get a positive result proving that it does.

s3a
Ah, yes. I get it / part b now. Thanks. :)

About the rest, do you have an idea as to why grades are deducted? Is anything wrong there?

Edit:
I still get it, but I think you meant upwards instead of downwards and downwards instead of upwards.

s3a said:
Ah, yes. I get it / part b now. Thanks. :)

About the rest, do you have an idea as to why grades are deducted? Is anything wrong there?

Edit:
I still get it, but I think you meant upwards instead of downwards and downwards instead of upwards.

The a) part seems correct, and the c) uses the previous results so it could be because of the incorrect b) part.

s3a
For what it's worth, in response to the what you seem to have edited away about the "+" for the last part of the first equation in part a, it does seem to have a "+" omitted, but the work continues as if it were there and as if the -0.5 + V_1 (including the "-" in front of the 0.5) were all on the numerator of the fraction. (You probably noticed that, but I just wanted to state it, just in case.)

Having said that, I get it all now, and thanks again! :)

diredragon

## 1. What is the Thevenin theorem and how does it relate to dependent current sources?

The Thevenin theorem states that any linear electrical network can be replaced by an equivalent circuit consisting of a single voltage source and a single resistance connected in series. This equivalent circuit is known as the Thevenin equivalent circuit. Dependent current sources are often found in Thevenin equivalent circuits as they allow for the inclusion of dependent factors, such as temperature or light, in the circuit's behavior.

## 2. How do you find the Thevenin equivalent circuit in a circuit with a dependent current source?

To find the Thevenin equivalent circuit in a circuit with a dependent current source, you must first replace the dependent current source with an equivalent independent current source. This can be done by using the appropriate conversion formula for the dependent source. Once the dependent source has been replaced, the Thevenin equivalent circuit can be found using standard techniques.

## 3. Can a circuit with a dependent current source have multiple Thevenin equivalent circuits?

No, a circuit with a dependent current source can only have one Thevenin equivalent circuit. This is because the Thevenin theorem states that the equivalent circuit must accurately model the behavior of the original circuit, and having multiple equivalent circuits would result in different behaviors.

## 4. How does the presence of a dependent current source affect the calculation of the Thevenin resistance?

The presence of a dependent current source does not affect the calculation of the Thevenin resistance. The Thevenin resistance is still calculated using the same methods as in a circuit without a dependent source, such as using the open-circuit voltage and short-circuit current.

## 5. What are some practical applications of the Thevenin theorem and dependent current sources?

The Thevenin theorem and dependent current sources are commonly used in circuit analysis and design. They allow engineers to simplify complex circuits and account for variables such as temperature and light in their designs. They are also used in the design of electronic devices, such as sensors and amplifiers.

• Engineering and Comp Sci Homework Help
Replies
2
Views
30K
• Engineering and Comp Sci Homework Help
Replies
3
Views
1K
• Engineering and Comp Sci Homework Help
Replies
4
Views
6K
• Engineering and Comp Sci Homework Help
Replies
2
Views
10K
• Engineering and Comp Sci Homework Help
Replies
3
Views
8K
• Engineering and Comp Sci Homework Help
Replies
3
Views
15K
• Engineering and Comp Sci Homework Help
Replies
11
Views
15K
• Engineering and Comp Sci Homework Help
Replies
11
Views
8K
• Engineering and Comp Sci Homework Help
Replies
3
Views
8K
• Engineering and Comp Sci Homework Help
Replies
2
Views
3K