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Consider the DC circuit of Figure $ 4$. Inductance L satisfies $L = (R^2) C / 2$. Calculate:

a) The differential equation for the charge $ Q (t) $ contained in the capacitor;

b) The solution of diferential equation of the preceding item;

c) the power dissipated in the resistance and the energy stored in the capacitor for sufficiently long times so that only the stationary solution is present.

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closed as off-topic by Claude Leibovici, Watson, gebruiker, John B, Daniel W. Farlow Jun 10 '16 at 18:58

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  • 2
    $\begingroup$ Notice that this is not an RLC circuit because L and C are in parallel while they should be in series for it to be an RLC circuit... $\endgroup$ – valerio Jun 10 '16 at 9:20
  • $\begingroup$ you could use the complex impedances of each element, which combine in circuits like resistances... $\endgroup$ – danimal Jun 10 '16 at 9:24
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I am going to give you a sketch on how you can find the solution.

As I stated in the comments, notice that this is not an RLC circuit, because C and L are in parallel whereas they are in series in an RLC circuit.

Since C and L are in parallel, we have

$$v_C(t) = v_L(t)$$

and

$$i(t) = i_R(t) = i_L(t) + i_C(t)$$

From Kirchhoff's voltage law:

$$v_R(t)+v_L(t) = v_R(t)+v_C(t) = -\epsilon$$

Moreover,

$$v_L(t) = L \frac{di_L(t)}{dt}$$

and

$$ v_C(t) = \frac {Q(t)}{C} = \frac 1 C \int_{-\infty}^{t} i_C(\tau) d \tau$$

This is all you need to solve the problem.

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  • $\begingroup$ This is not a serial RLC circuit, but it is obviously RLC. $\endgroup$ – Yves Daoust Jun 10 '16 at 9:54
  • $\begingroup$ @YvesDaoust You are technically right, but usually RLC means the series configuration. Anyway, the solution that the OP quotes is the solution for the RLC series. $\endgroup$ – valerio Jun 10 '16 at 10:07
  • $\begingroup$ Yep, and by the way this is a "pathological circuit" as there is no resistance in the LC loop, which can sustain permanent oscillations. $\endgroup$ – Yves Daoust Jun 10 '16 at 10:16
  • $\begingroup$ -Rdi/dt + (1/c)*i=0? I have done Vr+Vl=Vr+Vc=-e --> iR+Ldi/dt + -(1/c)integral- Ldi/dt (cause Vc=Vl). Differentiating them, we have: -Rdi/dt+(1/c)*1=0? $\endgroup$ – user157308 Jun 10 '16 at 13:47
  • $\begingroup$ I assume there is some switch which is closed at $t = 0$, or perhaps $t = -\infty$? After a very long time, shouldn't the p.d. across the capacitor be zero? After a sufficiently long time, the current is constant so the p.d. across the inductor is zero, and so the p.d. across the capacitor is zero. $\endgroup$ – jim Jun 10 '16 at 18:42

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