# Simple dc circuit, strange layout!

1. Mar 19, 2006

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

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2. Mar 19, 2006

### 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 small signal models of bipolar junction transistors.

Last edited: Mar 20, 2006
3. Mar 19, 2006

### Cyrus

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

4. Mar 20, 2006