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Hello everyone, In the circuit shown above (solving for the open circuit voltage in a RC circuit after switching), I found that the voltage at the essential node on the top was equal to 15 volts.... and that the dependent source generated 30 volts.

Why is it that the open circuit voltage here is 45 volts? In my mind, I would assume that the 15 volts at the essential node gets subtracted (due to the polarity of the dependent source) by 30 volts and that the open circuit voltage would be -15. I got the right answer of 45 by guessing but I don't think I'll sleep well tonight not knowing why.

Thank you!

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  • \$\begingroup\$ Your \$I_\Delta\$ is pointing opposite to its actual direction. So it will be negative. That implies that the dependent source will be polarized opposite to what's shown, so it will add to, and not subtract from, the divider node voltage (which is a fifth of 75 V, or 15 V.) Therefore the voltage source adds another 30 V, making it 45 V. \$\endgroup\$
    – jonk
    Jan 20, 2020 at 11:13
  • \$\begingroup\$ Looks like I need to revisit passive sign convention, that makes total sense thank you so much for the explanation! \$\endgroup\$ Jan 20, 2020 at 16:51

1 Answer 1

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Well, we are trying to analyze the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$\text{I}_2=\text{I}_1+\text{I}_3\tag1$$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1}{\text{R}_2}\\ \\ \text{I}_3=\frac{0-\text{V}_2}{\text{R}_3} \end{cases}\tag2 $$

We also know that \$\text{V}_1-\text{V}_2=\text{V}_\text{n}\$.

Now, it is not hard to solve for \$\text{V}_2\$ and \$\text{I}_1\$:

  • $$\text{V}_2=\frac{\text{R}_2\text{R}_3\text{V}_\text{i}-\text{R}_3\left(\text{R}_1+\text{R}_2\right)\text{V}_\text{n}}{\text{R}_2\text{R}_3+\text{R}_1\left(\text{R}_2+\text{R}_3\right)}\tag4$$
  • $$\text{I}_1=\frac{\left(\text{R}_2+\text{R}_3\right)\text{V}_\text{i}-\text{R}_2\text{V}_\text{n}}{\text{R}_2\text{R}_3+\text{R}_1\left(\text{R}_2+\text{R}_3\right)}\tag5$$

Where I used Mathematica-code to solve for that:

In[1]:=FullSimplify[
 Solve[{I2 == I1 + I3, I1 == (Vi - V1)/R1, I2 == V1/R2, 
   I3 == (0 - V2)/R3, V1 - V2 == Vn}, {I1, I2, I3, V1, V2}]]

Out[1]={{I1 -> ((R2 + R3) Vi - R2 Vn)/(R2 R3 + R1 (R2 + R3)), 
  I2 -> (R3 Vi + R1 Vn)/(R2 R3 + R1 (R2 + R3)), 
  I3 -> (-R2 Vi + (R1 + R2) Vn)/(R2 R3 + R1 (R2 + R3)), 
  V1 -> (R2 (R3 Vi + R1 Vn))/(R2 R3 + R1 (R2 + R3)), 
  V2 -> (R2 R3 Vi - (R1 + R2) R3 Vn)/(R2 R3 + R1 (R2 + R3))}}

Now, we can apply the theory from above on your circuit. We know that \$\text{V}_\text{n}:=\text{k}\cdot\left(-\text{I}_1\right)\$, so we can rewrite equation \$(4)\$ as follows:

$$\text{V}_2=\frac{\text{R}_3\left(\text{k}+\text{R}_2\right)\text{V}_\text{i}}{\text{R}_2\text{R}_3+\text{R}_1\left(\text{R}_2+\text{R}_3\right)-\text{k}\text{R}_2}\tag6$$

Where I used the following code:

In[2]:=FullSimplify[
 Solve[{I2 == I1 + I3, I1 == (Vi - V1)/R1, I2 == V1/R2, 
   I3 == (0 - V2)/R3, V1 - V2 == -k*I1}, {I1, I2, I3, V1, V2}]]

Out[2]={{I1 -> ((R2 + R3) Vi)/(-k R2 + R2 R3 + R1 (R2 + R3)), 
  I2 -> ((-k + R3) Vi)/(-k R2 + R2 R3 + R1 (R2 + R3)), 
  I3 -> ((k + R2) Vi)/(k R2 - R2 R3 - R1 (R2 + R3)), 
  V1 -> (R2 (k - R3) Vi)/(k R2 - R2 R3 - R1 (R2 + R3)), 
  V2 -> ((k + R2) R3 Vi)/(-k R2 + R2 R3 + R1 (R2 + R3))}}

For your circuit we must compute \$\text{V}_2\$ when \$\text{R}_3\to\infty\$, so we get:

$$\text{V}_\text{out}:=\lim_{\text{R}_3\to\infty}\text{V}_2=\frac{\left(\text{k}+\text{R}_2\right)\text{V}_\text{i}}{\text{R}_1+\text{R}_2}\tag7$$

Where I used the following code:

In[3]:=FullSimplify[
 Limit[((k + R2) R3 Vi)/(-k R2 + R2 R3 + R1 (R2 + R3)), 
  R3 -> Infinity]]

Out[3]=((k + R2) Vi)/(R1 + R2)

So, in your circuit we get:

$$\text{V}_\text{out}=\frac{\left(20\cdot10^3+10\cdot1000\right)\cdot75}{40\cdot1000+10\cdot1000}=45\space\text{V}\tag8$$

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