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I'm working on finding the 2nd order differential equation for a RLC circuit. I'm stuck on finding the KCL equation for node 1. I understand how the current through the R2 resistor and the current through the capacitor is found, but I don't quite understand how the current flowing into the node is found. It looks like we are treating the resistor as if it was connected in series to the voltage source at the positive terminal. enter image description here

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  • \$\begingroup\$ Where is the inductor? \$\endgroup\$
    – Andy aka
    Dec 2, 2022 at 9:07
  • \$\begingroup\$ Sorry I meant a RCC circuit \$\endgroup\$
    – YS KIM
    Dec 2, 2022 at 9:43

1 Answer 1

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Well, it's not hard to see that:

$$\text{I}_\text{source}\left(t\right)=\text{I}_{\text{C}_1}\left(t\right)+\text{I}_{\text{R}_2}\left(t\right)\tag1$$

Where:

  • $$\text{I}_\text{source}\left(t\right)=\text{I}_{\text{R}_3}\left(t\right)\tag2$$
  • $$\text{I}_{\text{R}_2}\left(t\right)=\text{I}_{\text{C}_2}\left(t\right)\tag3$$

Using the current voltage relations in a capacitor and resistor we can write:

$$\frac{\text{V}_3\left(t\right)-0}{\text{R}_3}=\left(\text{V}_\text{s}'\left(t\right)-\text{V}_3' \left(t\right)\right)\cdot\text{C}_1+\frac{\text{V}_\text{s}\left(t\right)-\text{V}_3\left(t\right)}{\text{R}_2}\tag4$$

Now, you can solve for \$\text{V}_3\left(t\right)\$.

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