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I'm currently studying for second order circuits, and I'm stuck understanding this example in my book pictured at the bottom.

  1. I got the same KCL equation, but I don't understand why the KVL equation on the right-hand mesh isn't $$v = \frac{1}{C}\int i(t)dt +6i+{L}\frac{di}{dt}$$ The book just has: $$v = 6i+{L}\frac{di}{dt}$$ and I don't understand why you don't account for the voltage across the capacitor. Even if it doesn't initially have voltage across it, it would still be changing, right? Because I think that we're concerned about the rate of change, not just the initial value or the infinity value.
  2. I am also a little confused why you cannot just take the derivative of the KCL equation like this: $$\frac{v}{4}+i_{L}+i_{C}= \frac{v_{s}}{4}$$ $$\frac{d}{dt}(\frac{v}{4}+\frac{1}{L}\int v(t)dt +C\frac{dv}{dt}=\frac{v_{s}}{4})$$

$$\frac{d^{2}v}{dt^{2}}+\frac{dv}{dt}+4 = \frac{dv_{s}}{dt}$$ which leads to the characteristic equation for the problem $$\lambda ^{2}+\lambda +4 =0$$

Thank you in advance for any help! enter image description here

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  • \$\begingroup\$ Voltage across an inductor is \$L\frac{di}{dt}\$. The capacitor does have an initial voltage at t=0. \$\endgroup\$
    – Chu
    Apr 19, 2020 at 16:54
  • \$\begingroup\$ oops thank you fixed it \$\endgroup\$ Apr 19, 2020 at 16:58

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I don't understand why you don't account for the voltage across the capacitor.

\$v\$ is the voltage across the capacitor. They've accounted for it by putting it in the left hand side of the equation. They just haven't written it in terms of the mesh current and constitutive relation of the capacitor yet.

I am also a little confused why you cannot just take the derivative of the KCL equation like this ...

Your version has

$$i_L = \frac{1}{L}\int v(t)dt$$

This isn't correct because it ignores the effect of the 6 ohm resistor.

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  • \$\begingroup\$ Thank you so much for your speedy response! I think that's starting to make sense, I was kind of double dipping. Is the reason I even have to have the KVL equation in additional to the KCL equation instead of just differentiating the KCL equation because the inductor and resistor are in series in parallel with the capacitor? \$\endgroup\$ Apr 19, 2020 at 17:03
  • \$\begingroup\$ @EmilyJohnson, I'm not sure exactly what you mean. In (9.8-5) you have two unknowns, so you need another equation to eliminate one of them. You could use the node equation like the book did, or you could use the mesh equation for the other mesh. You will get the same result either way. \$\endgroup\$
    – The Photon
    Apr 19, 2020 at 17:08
  • \$\begingroup\$ Oh that finally makes sense after an hour of staring at it. Thank you so much! \$\endgroup\$ Apr 19, 2020 at 17:13

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