Trying to get v(t) for discharging RLC battery

passive-networks

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.

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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!

Best Answer

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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