- #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
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