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An RLC circuit consists of a voltage source in series with a resistor, a capacitor, and an inductor. An inductor is a coil of wire. When the current passing through the coil changes, a magnetic field is generated that induces a voltage across the inductor. With current $i(t)$, the voltage drop across the inductor is given by $L\frac{di}{dt}$ where $L$ is the inductance of the inductor, measured in henries.

(a) Use Kirchhoff's voltage law to determine the 2nd order differential equation that describes the charge q on the capacitor of the circuit.

(b) Suppose that $R = 50$ ohms, $L = 0.1$ henries, and $C = 5 × 10^{-4}$ farad. At time, $t = 0$, when the charge $q$ and the current $i = \frac{dq}{dt}$ are zero, and a voltage source of $110 V$ is connected. Describe the charge on the capacitor as a function of time.

I never learnt anything about Kirchhoff's law, but could anyone give me any tips on where to start?

Thanks!

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  • $\begingroup$ Could anyone tell me if im correct? I ended up with: $$0.1I''(t)+50I'=V'(t)$$ $\endgroup$ – user184692 Oct 26 '14 at 2:13
  • $\begingroup$ What do those mean? im not sure $\endgroup$ – user184692 Oct 26 '14 at 17:53
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Are you an engineering student? If so, be sure to check out this video:https://www.youtube.com/watch?v=WBfAEeEzDlg Applying what is being said in the video we end up with the general solution for the current in the series circuit: $$\frac{V'}{L}\left ( t \right )=i\frac{1}{LC}+i^{'}\frac{R}{L}+i^{''}$$ Important note: The coefficients on the right side of the above equation remain the same no matter what variable you're solving for. However the input signal on the left hand side, in general, will vary. Go try yourself and set up an ode for the voltage drop on the resistor.

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