Simple dc circuit, strange layout

In summary, the current in this circuit appears to go in opposite directions and ultimately meet. It's hard to visualize, but maybe there is a simpler explanation behind it. I wish I could redraw the circuit in a simpler fashion, but that is often possible.
  • #1
DivGradCurl
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I'm having difficulty to understand a circuit from my textbook (figure attached). Essentially, the currents seem to go in opposite directions and ultimately meet. It looks strange (circled area), but maybe there is a simple explanation behind it. I wish I could redraw the circuit in a simpler fashion, which is often possible; I can't visualize it this time.

Here is what the problem states:

For the given circuit, find [tex]\frac{V_0}{V_s}[/tex] in terms of [tex]\alpha[/tex], [tex]R_1[/tex], [tex]R_2[/tex], [tex]R_3[/tex], and [tex]R_4[/tex]. If [tex]R_1 = R_2 = R_3 = R_4[/tex], what value of [tex]\alpha[/tex] will produce [tex]\left| \frac{V_0}{V_s} \right| = 10[/tex]?

Here is what I think (I may be wrong!):

1. The right side gives:

[tex]V_0 = \left( \frac{R_3 R_4}{R_3 + R_4} \right) \alpha I_0[/tex]

2. The left side gives

[tex]V_s = \left( R_1 + R_2 \right) I_0[/tex]

3. The first and second expressions yield

[tex]\frac{V_0}{V_s} = \frac{R_3 R_4}{\left( R_1 + R_2 \right) \left( R_3 + R_4 \right)} \alpha[/tex]

4. If [tex]R_1 = R_2 = R_3 = R_4 = R[/tex], then

[tex]\frac{V_0}{V_s} = \frac{\alpha}{4}[/tex]

5. If [tex]\left| \frac{V_0}{V_s} \right| = 10[/tex], then [tex]\alpha = \pm 40[/tex].

Any help is highly appreciated
 

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  • #2
Your math looks fine.. A good way to think about this kind of circuit representation is, the right circuit is dependent on the left.. On the left side, you have current [itex]I_o[/itex]... The right side uses this same current [itex]I_o[/itex] and is scaled by [itex] \alpha[/itex] (also known as dependent-current-source). Don't worry about the connection node in between. That is your lowest potential (typically ground). In your diagram, both circuits are sharing this common ground.

You will see lots more of this kind of circuit diagram, if you learn about http://people.deas.harvard.edu/~jones/es154/lectures/lecture_3/bjt_models/bjt_models.html [Broken] of bipolar junction transistors.
 
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  • #3
Wow, can we give this guy a prize for such a nice post?
 
  • #4
Accolade on Post Constrution

Here.. here.. Nice job thiago, on your post!
 
  • #5
Thank you, folks! It did not make sense when I posted it and your input helped a lot.
 

1. What is a simple dc circuit?

A simple dc circuit is an electrical circuit that uses direct current (DC) to power a device or system. It consists of a power source, such as a battery, and a load, such as a light bulb, connected by wires.

2. What makes the layout of a dc circuit strange?

The layout of a dc circuit may be considered strange if it deviates from the typical or standard circuit design. This could include unusual placement of components, non-linear connections, or unconventional use of materials.

3. How does a simple dc circuit work?

A simple dc circuit works by allowing electric current to flow from the positive terminal of the power source, through the circuit and its components, and back to the negative terminal. This flow of current powers the load and allows it to function.

4. What are the basic components of a simple dc circuit?

The basic components of a simple dc circuit include a power source, such as a battery or power supply, wires to connect the components, and a load, such as a light bulb or motor. Other components, such as switches and resistors, may also be included depending on the design and purpose of the circuit.

5. How can I troubleshoot a strange layout in a dc circuit?

If you encounter a strange layout in a dc circuit, the first step is to carefully examine the circuit diagram and make sure all the components are connected correctly. You can also use a multimeter to test the voltage and current at different points in the circuit to identify any potential issues. If necessary, consult a circuit diagram or seek help from an experienced electrician or engineer.

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