Electronic – Find Io(t) of RLC circuit (from Electric Circuits, Nilsson )

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My question is from the Electric Circuits, Nilsson 10th edition.

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My solution is

  1. For \$t < 0\$, we get \$i_o(0) = 2 \mathrm{A}\$ and \$v_o(0) = -400 \mathrm{V}\$.

  2. For \$t > 0\$, the circuit becomes a series of C, R, L. (suppose \$i_c\$ go up, \$i_L = i_o\$)
    $$
    i_c =C {dv_o \over dt} = i_o = i_L.
    $$

    So I get
    $$
    v_o+CR {dv_o \over dt} + LC {d^2v_o \over dt^2} = 0,
    $$

    So I get
    $$
    {d^2v_o \over dt^2} + {R \over L}{dv_o \over dt} + {1 \over LC}v_o = 0
    $$

    By solving 2nd ODE, we get

\$v_o = A_1 e^{-250t} + A_2 e^{-1000t}\$, and then by initial condition, we can get answer.

Is the above true?

The solution gives the same ODE for \$i_o\$ not for \$v_o\$. Not sure which one is true.

Best Answer

The given RLC circuit is overdamped as: $$ \frac{R}{2L} > \frac{1}{\sqrt{LC}}$$ Hence the general solution is the sum of two decaying exponentials: $$i(t)= K_1e^{xt} + K_2e^{yt}$$ where x,y are roots of the characteristic equation: $$s^2+\frac{R}{L}s+\frac{1}{LC}=0$$ ie., $$s^2+1250s+250000 = 0$$ $$\implies (s+1000)(s+250)=0$$ Hence, $$i(t)=K_1e^{-250t}+K_2e^{-1000t}$$

Now consider the capacitor,$$v_o(t) =\int{\frac{i(t)}{C}}$$ Hence \$v_o(t)\$ should be in the form: $$v_o(t)=\frac{K_1}{-250C}e^{-250t}+\frac{K_2}{-250C}e^{-1000t}+K_3$$

Analysing the initial conditions will help you solve the constants ...

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