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 \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
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