Simple dc circuit, strange layout

[DivGradCurl]

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 \frac{V_0}{V_s} in terms of \alpha, R_1, R_2, R_3, and R_4. If R_1 = R_2 = R_3 = R_4, what value of \alpha will produce \left| \frac{V_0}{V_s} \right| = 10?

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

1. The right side gives:

V_0 = \left( \frac{R_3 R_4}{R_3 + R_4} \right) \alpha I_0

2. The left side gives

V_s = \left( R_1 + R_2 \right) I_0

3. The first and second expressions yield

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

4. If R_1 = R_2 = R_3 = R_4 = R, then

\frac{V_0}{V_s} = \frac{\alpha}{4}

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

Any help is highly appreciated

 

[Ouabache]

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 I_o... The right side uses this same current I_o and is scaled by \alpha (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 of bipolar junction transistors.

 

[Cyrus]

Wow, can we give this guy a prize for such a nice post?

 

[Ouabache]

Accolade on Post Constrution

Here.. here.. Nice job thiago, on your post!

 

[DivGradCurl]

Thank you, folks! It did not make sense when I posted it and your input helped a lot.