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I'm having trouble with the following problem. I want to find \$V(t)\$ (part (d)) and have been working for quite a long time on it.

enter image description here

My first question is, if I do the following, are all the voltages in my differential equations the same voltage? It seems like they aren't, therefore this method wouldn't work.

What I tried doing was using Kirchoff's current rule so I had $$i_C + i_R + i_L = 0$$ $$\frac{d^2V}{dt^2} + \frac{1}{RC} \frac{dV}{dt} + \frac{1}{LC}V = 0$$

Second off, if this does all solve for the same voltage, then it is problematic because I'm getting a complex value for $$\omega_1 = \sqrt{\frac{1}{LC} - \left(\frac{1}{2RC}\right)^2}$$ in the following equation of voltage that we Ansatz $$v(t) = Ae^{-\gamma t} \cos (\omega_1 t - \phi)$$

So I feel like something is going wrong with what I'm doing. Please help me out in setting this equation up, thanks!

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  • \$\begingroup\$ For (b), we can reason like this. At time t=0, the inductor wants to keep pushing the same current. The voltage on C is 50, and that voltage appears across the resistor too which spans the same two nodes. So we know from Ohm's what current is instantaneously going through the resistor. Knowing the L current and R current, we can get the C current. From the existing voltage on C, current through C, and capacitance of C, we can figure out how fast it is charging at that moment (dv/dt at just after t=0). \$\endgroup\$
    – Kaz
    Commented Nov 21, 2013 at 7:03
  • \$\begingroup\$ Thanks, but I didn't necessarily have a question on part (b), see below the picture of the problem for the actual question I have. It refers mostly to part (d) \$\endgroup\$
    – spaderdabomb
    Commented Nov 21, 2013 at 7:09
  • \$\begingroup\$ Shouldn't the first equation be -iC + iR + iL = 0? As the capacitor discharges the current relations looks more like iC = iL + iR. \$\endgroup\$
    – jsrmalvarez
    Commented Nov 21, 2013 at 12:11
  • \$\begingroup\$ @jsrmalvarez, at the top node, the KCL equation is: \$i_R + i_L + i_C = 0\$. Recall that, according to the passive sign convention, current enters the positive labelled terminal of a circuit element. As these elements are in parallel, they have identical voltage across and thus, the current reference directions should be the same, e.g., from top to bottom. Now, if the capacitor is discharging, it is supplying power to the circuit, the current will exit the positive terminal and thus, \$i_C < 0\$. But this what we want since the power should be negative when the cap supplies power. \$\endgroup\$
    – Alfred Centauri
    Commented Nov 21, 2013 at 15:08
  • \$\begingroup\$ @jsrmalvares It could be that equation, but all you are doing is reversing the direction in which iC is measured. The circuital laws are expressed as terms which sum to zero. Of course, this means that if any of those terms are positive, some of them have to be negative to compensate. This balancing is played out in the values, not by putting minus signs into the symbolic formula. \$\endgroup\$
    – Kaz
    Commented Nov 21, 2013 at 20:37

1 Answer 1

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My first question is, if I do the following, are all the voltages in my differential equations the same voltage?

Yes, there is just one voltage in this circuit, the voltage across the parallel connected elements.

$$v_R = v_L = v_C = V $$

Second off, if this does all solve for the same voltage, then it is problematic because I'm getting a complex value

The solutions to this homogeneous 2nd order linear differential equation are of the form:

$$Ae^{s_1t} + Be^{s_2t}$$

where \$s_1\$ and \$s_2\$ are the roots of the quadratic equation:

$$s^2 + \dfrac{1}{RC}s + \dfrac{1}{LC} = 0 $$

Now, there are three possibilities:

  1. There are two distinct real roots - this is the overdamped case
  2. There are two identical real roots - this is the critically damped case
  3. There are two complex conjugate roots - this is the underdamped case

Can you take it from here?

Hint: recall Euler's formula: \$e^{st} =e^{\alpha t}(\cos\omega t + j\sin\omega t\$) where \$s = \alpha + j \omega\$

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